Please help will mark brainliest (answer must be right)

True or false

The points on a line can be paired one to one with real numbers ?

Answers

Answer 1

Answer: TRUE

Step-by-step explanation: This is true, because there is an infinite amount of real numbers in both, and they are both countably infinite (so these infinities are equal). Hope this helps!


Related Questions

A coach is dividing a soccer team of 28 players into groups. If each group has the same number of players, what is the greatest number of groups there can be if each group has no more than 10 players?

Answers

Answer:

The only possible groups that could be made if each group have the same number of people is 2 groups of 14 or 4 groups of 7. Since each group cannot have more than 10 people, the only group left is 4 groups of 7.

Marti is filling a 10– inch diameter ball with sand to make a medicine ball that can be used for exercising. To determine if the medicine ball will be too heavy after it is completely full of sand, she did some research and found that there is approximately 100 pounds of sand per cubic foot. How heavy will the medicine ball be after it is filled with sand, rounded to the nearest pound? A.30 pounds B.58 pounds C.24 pounds D.83 pounds

Answers

Answer:

Option A. [tex]30\ pounds[/tex]

Step-by-step explanation:

step 1

Find the volume of the sphere ( medicine ball)

The volume is equal to

[tex]V=\frac{4}{3}\pi r^{3}[/tex]

we have

[tex]r=10/2=5\ in[/tex] ----> the radius is half the diameter

Convert inches to feet

Remember that

1 ft=12 in

[tex]r=5\ in=5/12\ ft[/tex]

assume

[tex]\pi=3.14[/tex]

substitute

[tex]V=\frac{4}{3}(3.14)(5/12)^{3}[/tex]

[tex]V=0.3029\ ft^{3}[/tex]

step 2

Find the weight of the ball

Multiply the volume in cubic foot by 100

[tex]0.3029*100=30.29\ pounds[/tex]

Round to the nearest pound

[tex]30.29=30\ pounds[/tex]

The medicine ball would be A. 30 pounds

Volume of the medicine ball

Since the medicine ball is a sphere, its volume is

V = πd³/6 where d = diameter of medicine ball = 10 in = 10 in × 1 ft/12 in = 0.8333 ft

So, substituting the value of the variable into the equation, we have

V = πd³/6

V = π(0.8333 ft)³/6

V = π(0.5787 ft³)/6

V = 1.818 ft³/6

V = 0.303 ft³

Mass of medicine ball

The mass of the medicine ball = mass of sand per cubic foot × volume of medicine ball

where

mass of sand per cubic foot = 100 lb/ft³ and volume of medicine ball = 0.303 ft³

So,

mass of the medicine ball = 100 lb/ft³ × 0.303 ft³

= 30.3 lb

≅ 30 pounds

So, the medicine ball would be A. 30 pounds

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a box contains four $1 and six $5 bills. if three bills are selected at random without replacement, find the probability that all three are $5 bills.
A.27/125 B.1/4 C.1/6 D3/5

Answers

Answer:

C.1/6

Step-by-step explanation:

Initially the box has four $1 and six $5 bills. The probability of selecting a $5 bill in the first trial would be given as;

(number of $5 bills) / (total number of bills)

= (6)/(4+6) = 3/5

If in the first attempt we actually pick a $5 bill, the number of $5 bills will reduce by one to 5. Now, the probability of picking a $5 bill in the second attempt will be given as;

(new number of $5 bills) / (new total number of bills)

= (5)/(4+5) = 5/9

The new number of $5 bills will now be; 6 - 2 = 4 since we have already picked 2 without replacing them.

Now, the probability of picking a $5 bill in the third attempt will be given as;

(new number of $5 bills) / (new total number of bills)

= (4)/(4+4) = 1/2

Since the three attempts are independent, the probability of picking  all three $5 bills is;

3/5 * 5/9 * 1/2 = 1/6

Final answer:

The probability of drawing three $5 bills from a box containing four $1 bills and six $5 bills, when the bills are drawn without replacement, is 1/6.

Explanation:

The question is asking for the probability of drawing three $5 bills from a box containing four $1 bills and six $5 bills, given that the bills are drawn without replacement. This is a problem of combinatorial probability. We first find the total ways of selecting three bills from the box, then find the ways of selecting three $5 bills.

The total ways of selecting three bills is given by combination formula C(n, r) = n! / (r!(n-r)!), where n is the total number of bills and r is the number selected. In this case, n=10 (4 $1 bills and 6 $5 bills) and r=3. So, C(10,3) = 120.

The ways of selecting three $5 bills is C(6,3) = 20,

So the probability of drawing three $5 bills is 20/120 = 1/6.

Therefore, the correct answer is C. 1/6.

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Find (ƒ + g)(x) where ƒ(x) = 5x2 + 4, g(x) = 6x2 – x.

(ƒ + g)(x) = 11x2 – x + 4

(ƒ + g)(x) = 11x2 – 4x

Answers

To find (ƒ + g)(x) for the given functions ƒ(x) = 5x² + 4 and g(x) = 6x² - x, we add the like terms to get

(ƒ + g)(x) = 11x² - x + 4.

To find (ƒ + g)(x) where ƒ(x) = 5x²  + 4, and g(x) = 6x²  – x, we simply add the two functions together. We combine like terms to calculate the sum.

By adding the corresponding terms:

The x²  terms: 5x² + 6x² = 11x² The x terms: Since there is no x term in ƒ(x), we only have the x term from g(x), which is -x.The constant terms: 4 from ƒ(x) and there is no constant term in g(x).

Combining these, we get

(ƒ + g)(x) = f(x) + g(x)

(f + g)(x) = 11x² - x + 4.

Which hyperbola has one focus in common with the hyperbola x^2/16 - y^2/9 = 1

Answers

Answer:

The same focus is (-5 , 0) ⇒ Answer D

Step-by-step explanation:

* Lets study the equation of the hyperbola

# The standard form of the equation of a hyperbola with  

  center (0 , 0) and transverse axis parallel to the x-axis is

  x²/a² - y²/b² = 1

- The coordinates of the foci are (± c , 0),  where c² = a² + b²

# The standard form of the equation of a hyperbola with  

  center (h , k) and transverse axis parallel to the x-axis is

  (x - h)²/a² - (y - k)²/b² = 1

- the coordinates of the foci are (h ± c , k), where c² = a² + b²

# The standard form of the equation of a hyperbola with  

  center (h , k) and transverse axis parallel to the y-axis is

  (y - k)²/a² - (x - h)²/b² = 1

- the coordinates of the foci are (h , k ± c), where c² = a² + b²

* Now lets solve the problem

∵ x²/16 - y²/9 = 1

∴ a² = 16 and b² = 9

∵ c² = a² + b²

∴ c² = 16 + 9 = 25 ⇒ take √ to find the values of c

∴ c = ±√25 = ± 5

∴ The foci are (5 , 0) , (-5 , 0)

# Answer A:

∵ (y - 5)/16 - (x - 13)/9 = 1

∵  (y - k)²/a² - (x - h)²/b² = 1

∴ The foci are (h , k + c) , (h , k - c)

∴ h = 13 and k = 5

∵ a² = 16 and b² = 9

∵ c² = a² + b²

∴ c² = 16 + 9 = 25 ⇒ take √ to find the values of c

∴ c = ±√25 = ± 5

∴ The foci are (13 , 5+5) , (13 , 5-5)

∴ The foci are (13 , 10) , (13 , 0) ⇒ not the same

# Answer B:

∵ (x - 13)²/25 - (y - 5)²/144

∵ (x - h)²/a² - (y - k)²/b² = 1

∵ The foci are (h ± c , k)

∴ h = 13 and k = 5

∵ a² = 25 and b² = 144

∵ c² = a² + b²

∴ c² = 125 + 144 = 169 ⇒ take √ to find the values of c

∴ c = ±√169 = ± 13

∴ The foci are (13 + 13 , 5) , (13 - 13 , 5)

∴ The foci are (26 , 5) , (0 , 5) ⇒ not the same

# Answer C:

∵ (y - 5)/25 - (x - 13)/144 = 1

∵  (y - k)²/a² - (x - h)²/b² = 1

∴ The foci are (h , k + c) , (h , k - c)

∴ h = 13 and k = 5

∵ a² = 25 and b² = 144

∵ c² = a² + b²

∴ c² = 25 + 144 = 169 ⇒ take √ to find the values of c

∴ c = ±√169 = ± 13

∴ The foci are (13 , 5+13) , (13 , 5-13)

∴ The foci are (13 , 18) , (13 , -8) ⇒ not the same

# Answer D:

∵ (y + 13)/144 - (x + 5)/25 = 1

∵  (y - k)²/a² - (x - h)²/b² = 1

∴ The foci are (h , k + c) , (h , k - c)

∴ h = -5 and k = -13

∵ a² = 144 and b² = 25

∵ c² = a² + b²

∴ c² = 144 + 25 = 169 ⇒ take √ to find the values of c

∴ c = ±√169 = ± 13

∴ The foci are (-5 , -13+13) , (-5 , -13-13)

∴ The foci are (-5 , 0) , (-5 , -26) ⇒ one of them the same

* The same focus is (-5 , 0)

Does anyone know what this n does in the equation?

(x + 5 < 4) ∩ (x - 3 > -6).

Answers

Answer:

The symbol ∩ signifies the intersection of the left operand and the right operand. Here, it means "and", as it often does.

Step-by-step explanation:

The solution to the left inequality is x < -1.

The solution to the right inequality is x > -3.

The intersection symbol (∩) means you are interested in the interval where these solutions overlap—the intersection of the solutions: -3 < x < -1; (-3, -1) in interval notation.

Triangle ABC is similar to triangle DEF. The length of AC is 10cm. The length of BC is 16 cm. The length of DF is 8cm. What is the length of EF?

Answers

Answer: The length of EF=15 cm.

Step-by-step explanation:

The woodlands middle school poll results show that about 79.3% of people who prefer pizza are students and about 81% of people who prefer burgers are students.​
a : there is not enough evidence to support a relationship between lunch preference and role at school
b : there is evidence to support a relationship between lunch preference at school

Answers

Answer: A: there is not enough evidence to support a relationship between lunch preference and role at school

The relationship between poll result and the student preference is: option A, not enough evidence

Why is there no evidence?

This due to the fact that the statistics in the question is incomplete. W do not have sufficient data that would be used for hypothesis testing.

Due to the fact above, we conclude that there is insufficient evidence to get the relationship between variables.

In conclusion, there is not enough evidence to support a relationship between lunch preference and role at school

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Eight less than seven times a number is the same as for l four more than three times a number. Find the number.

Answers

Question :- Eight less then seven times a number is the same as for l four more than three times a number. Find the number.

Answer :-

Let the number be x

Therefore,

[tex]=> 8-7x = 4+3x[/tex]

[tex]=> 8-4 = 3x+7x[/tex]

[tex]=> 4 = 10[/tex]

[tex]=> X = 5/4[/tex]

Hope it helps!

Choose all of the statements that correctly describe the transformation rule. Reflection over x-axis: (x, y) ? (?x, y) Reflection over y-axis: (x, y) ? (x, ?y) Rotation of 90° counter-clockwise about origin: (x, y) ? (?y, x) Rotation of 180° counter-clockwise about origin: (x, y) ? (?x, ?y) Rotation of 270° counter-clockwise about origin: (x, y) ? (y, ?x)

Answers

Answer:

Transformations are important subjects in geometry. In this exercise, these are the correct transformation rules:

1. Reflection over x-axis:

Consider the point [tex](x,y)[/tex], if you reflect this point across the x-axis you should multiply the y-coordinate by -1, so you get:

[tex]\boxed{(x,y)\rightarrow(x,-y)}[/tex]

2. Reflection over y-axis:

Consider the point [tex](x,y)[/tex], if you reflect this point across the y-axis you should multiply the x-coordinate by -1, so you get:

[tex]\boxed{(x,y)\rightarrow(-x,y)}[/tex]

3. Rotation of 90° counter-clockwise about origin:

Consider the point [tex](x,y)[/tex]. To rotate this point by 90° around the origin in counterclockwise direction, you can always swap the x- and y-coordinates and then multiply the new x-coordinate by -1. In a mathematical language this is as follows:

[tex]\boxed{(x,y)\rightarrow(-y,x)}[/tex]

4. Rotation of 180° counter-clockwise about origin:

Consider the point [tex](x,y)[/tex]. To rotate this point by 180° around the origin, you can flip the sign of both the x- and y-coordinates. In a mathematical language this is as follows:

[tex]\boxed{(x,y)\rightarrow(-x,-y)}[/tex]

5. Rotation of 270° counter-clockwise about origin:

Rotate a point 270° counter-clockwise about origin is the same as rotating the point 90° in clock-wise direction. So the rule is:

[tex]\boxed{(x,y)\rightarrow(y,-x)}[/tex]

Answer:Transformations are important subjects in geometry. In this exercise, these are the correct transformation rules:

Step-by-step explanation:

Write the limit as a definite integral on the interval [a, b], where ci is any point in the ith subinterval. limit interval lim ||δ|| → 0 n 5 ci4 i = 1 δxi [5, 9]

Answers

Seems like it would be

[tex]\displaystyle\int_5^9x^4\,\mathrm dx[/tex]

This system of equations has an infinite number of solutions. Define the solutions algebraically, and allow z to represent all real numbers.

3x − 4y + 4z = 7
x − y − 2z = 2
2x − 3y + 6z = 5
x =

y =

z = all real numbers

Answers

Answer:

x = 1+12z, y = -1+10z, and z = z

Step-by-step explanation:

Step 1: Convert the system into the augmented matrix form:

• 3 -4 4 | 7  

• 1 -1 -2 | 2

• 2 -3 6 | 5

Step 2: Multiply row 2 with -2 and add it into row 3:

• 3 -4 4 | 7  

• 1 -1 -2 | 2

• 0 -1 10 | 1

Step 3: Multiply row 2 with -3 and add it into row 1:

• 0 -1 10 | 1  

• 1 -1 -2 | 2

• 0 -1 10 | 1

Step 4: Replace row 1 with row 2 and multiply the updated row 2 with -1 and add it into row 3:

• 1 -1 -2 | 2

• 0 -1 10 | 1  

• 0 0 0 | 0

Step 5: Multiply row 2 with -1 and add it in row 1:

0 1 -10 -1

• 1 0 -12 | 1

• 0 -1  10 | 1  

• 0 0  0 | 0

Step 6: It can be seen that there are infinite solutions of this system since the last row is all zeroes. It can be seen that when this updated augmented matrix is converted into a system, it comes out to be:

• x - 12z = 1

• -y + 10z = 1

Step 7: Make x and y the subject of their respective equations:

• x = 1 + 12z

• y = -1 + 10z

So final answer is x = 1+12z, y = -1+10z, and z = z!!!

help ASAP please and thank you

Answers

A. addition:

[tex]3x^{2}  + 2x - 6\\2x^{2}  - 2x + 10\\------\\5x^{2} + 4[/tex]

B. subtraction:

[tex]3x^{2} + 2x - 6\\2x^{2} - 2x + 10\\-------\\x^{2} + 4x - 16[/tex]

Answer:

1. 5x²+4         2.-x²-4x+16

Step-by-step explanation:

The question is on operations in quadratic equations

1. Addition

P=3x²+2x-6

Q= 2x²-2x+10

P+Q= 3x²+2x-6 +2x²-2x+10

Collect like terms

3x²+2x²+2x-2x-6+10

5x²+4

2.Subtraction

Q-P

(2x²-2x+10) -(3x²+2x-6)

open brackets

2x²-2x+10-3x²-2x+6

collect like terms

2x²-3x²-2x-2x+10+6

-x²-4x+16

-x²+4x+16

What are the coordinates of side HI? (–8, 1) and (–4, –3) (–4, 7) and (–2, 5) (1, –8) and (–3, –4) (7, –4) and (5, –2)

Answers

Answer:

Step-by-step explanation:

the answer is b defenitly

Find the value of the matrices
A) 17
B) 12
C) 15
D) 13

Answers

Answer:

option C is correct

Step-by-step explanation:

We need to find the determinant.

= -1(7*9 - 3*9)-( -3)(4*9 - 3*3) -2(4*9 -7*3)

= -1(63-27) +3 (36 - 9) -2( 36- 21)

= -1(36)+3(27)-2(15)

= -36+81-30

= 15

Option C is correct

Ten students are asked to visit a college admissions counselor. The counselor can meet with one student at a time. In how many ways can four time slots be assigned?


5040


24


210


151,200

Answers

Answer:

5040

Step-by-step explanation:

at least thats what it is on GP

Rewrite the parametric equations by eliminating the parameter:

x= 4t+1 and y=t-3

a. y=x-13/4
b. 5x-2
c. y=3x+4
d. y= x-4/4

Answers

I believe the answer is c. I hope that’s right! Good luck!

The parametric equation is : [tex]\frac{x-13}{4}[/tex]

The correct option is (a).

What is  Equation?

In algebra, an equation can be defined as a mathematical statement consisting of an equal symbol between two algebraic expressions that have the same value.

Given parametric equation:

x= 4t+1 and y=t-3

From, y=t-3

         t= y+3

put t in x= 4t+1

x= 4(y+3) +1

x= 4y +12+1

x= 4y  + 13

x-13=4y

y= [tex]\frac{x-13}{4}[/tex]

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You are working on an air conditioning system. A roll of cylindrical copper tubing has an outside diameter of 7/8 inch and an inside diameter of 3/4 inch. How much refrigerant can 12 feet of the tubing hold?

Answers

Answer: A

Step-by-step explanation:

The correct answer is A. 0.037 cubic feet.

Convert the lengths to inches:

12 feet is equal to 12 * 12 inches = 144 inches.

Calculate the area of the annulus (the space between the inner and outer circles):

First, convert the diameters to radii:

Outer radius = 7/8 inches * 0.5 = 7/16 inches

Inner radius = 3/4 inches * 0.5 = 3/8 inches

Then, calculate the area of the annulus:

Area = π * (outer radius^2 - inner radius^2)

Area ≈ π * ((7/16)^2 - (3/8)^2) ≈ 0.0875 square inches

Calculate the volume of the refrigerant:

Multiply the area by the length of the tubing:

Volume = Area * Length

Volume ≈ 0.0875 square inches * 144 inches ≈ 12.48 cubic inches

Convert the volume to cubic feet:

Remember that 1 inch^3 = 1/12^3 cubic feet.

Therefore, the volume in cubic feet is:

Volume (ft^3) = Volume (in^3) / (12^3)

Volume (ft^3) ≈ 12.48 cubic inches / (12 * 12 * 12) ≈ 0.037 cubic feet

The closest answer choice to 0.037 cubic feet is A. 0.037 cubic feet.

Complete Question:

You are working on an air conditioning system. A roll of cylindrical copper tubing has an outside diameter of 7/8 inch and an inside diameter of 3/4 inch. How much refrigerant can 12 feet of the tubing hold?

A. 0.037 cubic feet

B. 0.065 cubic feet

C. 0.147 cubic feet

D. 5.30 cubic feet

A rectangle has a length of 6X +3 units and a width of eight units write a simplified expression for the area in square are you friends of this rectangle

Answers

Answer:

A = 48x + 24 (square units)

Step-by-step explanation:

L = 6x + 3

W = 8

A = L * W

A = 8(6x + 3)

A = 48x + 24

My answers keep getting erased, even the ones that were actually perfectly fine and explained. The person who asked the question even said that I was right.

I'm just annoyed with Brainly at this point.

I need to add a question so.......

Find the answer to 749x832=?

Answers

Answer:

623168

Step-by-step explanation:

749x832=623168

Circle M is circumscribed about right triangle ABC with legs 6 meters and 8 meters.
What is the exact circumference of ⊙M

Answers

ABC is a right triangle, so AC has length given by

[tex]AC^2=(6\,\mathrm m)^2+(8\,\mathrm m)^2\implies AC=\sqrt{100\,\mathrm m^2}=10\,\mathrm m[/tex]

Then the circumference of circle M is [tex]10\pi\,\mathrm m[/tex].

huh.. can someone please help me, i honestly really need this rn.. :(

Answers

Answer:

If

p(x) is a polynomial, the solutions to the equation

p(x) = 0 are called the zeros of the

polynomial. Sometimes the zeros of a polynomial can be determined by factoring or by using the

Quadratic Formula, but frequently the zeros must be approximated. The real zeros of a polynomial

p(x) are the x-intercepts of the graph of

y = p(x).

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

The Factor Theorem: If

(x − k) is a factor of a polynomial, then

x = k is a zero of the polynomial.

Conversely, if

x = k is a zero of a polynomial, then

(x − k) is a factor of the polynomial.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Example 1: Find the zeros and x-intercepts of the graph of

p(x) =x

4−5x

2 + 4.

x

4−5x

2 + 4 = 0

(x

2 − 4)(x

2 −1) = 0

(x + 2)(x − 2)(x +1)(x −1) = 0

x + 2 = 0 or x − 2 = 0 or x +1= 0 or x −1= 0

x = −2 or x = 2 or x = −1 or x =1

So the zeros are –2, 2, –1, and 1 and the x-intercepts are (–2,0), (2,0), (–1,0), and (1,0).

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

The number of times a factor occurs in a polynomial is called the multiplicity of the factor. The

corresponding zero is said to have the same multiplicity. For example, if the factor

(x − 3) occurs to

the fifth power in a polynomial, then

(x − 3) is said to be a factor of multiplicity 5 and the

corresponding zero, x=3, is said to have multiplicity 5. A factor or zero with multiplicity two is

sometimes said to be a double factor or a double zero. Similarly, a factor or zero with multiplicity

three is sometimes said to be a triple factor or a triple zero.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Example 2: Determine the equation, in factored form, of a polynomial

p(x) that has 5 as double

zero, –2 as a zero with multiplicity 1, and 0 as a zero with multiplicity 4.

p(x) = (x − 5)

2(x + 2)x

4

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Example 3: Give the zeros and their multiplicities for

p(x) = −12x

4 + 36x3 − 21x

2.

−12x

4 + 36x3 − 21x

2 = 0

−3x

2(4x

2 −12x + 7) = 0

−3x

2 = 0 or 4x

2 −12x + 7 = 0

x

2 = 0 or x = −(−12)± (−12)

2−4(4)(7)

2(4)

x = 0 or x = 12± 144−112

8 = 12± 32

8 = 12±4 2

8 = 12

8 ± 4 2

8 = 3

2 ± 2

2

So 0 is a zero with multiplicity 2,

x = 3

2 − 2

2 is a zero with multiplicity 1, and

x = 3

2 + 2

2 is a zero

with multiplicity 1.

(Thomason - Fall 2008)

Because the graph of a polynomial is connected, if the polynomial is positive at one value of x and

negative at another value of x, then there must be a zero of the polynomial between those two values

of x.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Example 4: Show that

p(x) = 2x3 − 5x

2 + 4 x − 7 must have a zero between

x =1 and

x = 2.

p(1) = 2(1)

3 − 5(1)

2 + 4(1) − 7 = 2(1) − 5(1) + 4 − 7 = 2 − 5 + 4 − 7 = −6

and

p(2) = 2(2)3 − 5(2)

2 + 4(2) − 7 = 2(8) − 5(2) + 8 − 7 =16 −10 + 8 − 7 = 7.

Because

p(1) is negative and

p(2) is positive and because the graph of

p(x) is connected,

p(x)

must equal 0 for a value of x between 1 and 2.

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If a factor of a polynomial occurs to an odd power, then the graph of the polynomial actually goes

across the x-axis at the corresponding x-intercept. An x-intercept of this type is sometimes called an

odd x-intercept. If a factor of a polynomial occurs to an even power, then the graph of the

polynomial "bounces" against the x-axis at the corresponding x-intercept, but not does not go across

the x-axis there. An x-intercept of this type is sometimes called an even x-intercept.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Example 5: Use a graphing calculator or a computer program to graph

y = 0.01x

2(x + 2)3(x − 2)(x − 4)

4 .

x

y

–2 2 4

5

Because the factors

(x + 2) and

(x − 2) appear to odd

powers, the graph crosses the x-axis at

x = −2

and

x = 2.

Because the factors x and

(x − 4) appear to even

powers, the graph bounces against the x-axis at

x = 0

and

x = 4.

Note that if the factors of the polynomial were

multipled out, the leading term would be

0.01x10.

This accounts for the fact that both tails of the graph

go up; in other words, as

x → −∞,

y

Step-by-step explanation:

Find the fifth roots of 243(cos 240° + i sin 240°).

Answers

Answer:

See below.

Step-by-step explanation:

Fifth root of 243 = 3,

Suppose r( cos Ф + i sinФ) is the fifth root of 243(cos 240 + i sin 240),

then r^5( cos  Ф  + i sin  Ф )^5 = 243(cos 240 + i sin 240).

Equating equal parts and using de Moivre's theorem:

r^5 =243  and  cos  5Ф  + i sin  5Ф = cos 240 + i sin 240

r = 3 and  5Ф = 240 +360p so Ф =  48 + 72p

So Ф = 48, 120, 192, 264, 336  for   48 ≤ Ф < 360

So there are 5 distinct solutions given by:

3(cos 48 + i sin 48),

3(cos 120 + i sin 120),

3(cos 192 + i sin 192),

3(cos 264 + i sin 264),

3(cos 336 + i sin 336).. (Answer).

At a historical landmark, candles are used to simulate an authentic atmosphere. A volunteer is currently putting new candles in the candle holders. On the east side, he replaced candles in 10 small candle holders and 4 large candle holders, using a total of 52 candles. On the west side, he replaced the candles in 2 small candle holders and 4 large candle holders, for a total of 36 candles. How many candles does each candle holder hold?

Answers

Final answer:

To solve this problem, we can set up a system of equations to represent the given information. Solving this system of equations, we find that each small candle holder holds 2 candles and each large candle holder holds 8 candles.

Explanation:

To solve this problem, let's represent the number of candles each small candle holder holds as x and the number of candles each large candle holder holds as y. We can set up a system of equations to represent the given information:

For the east side: 10x + 4y = 52For the west side: 2x + 4y = 36

We can solve this system of equations by eliminating one variable. Subtracting the equations, we get 8x = 16. Dividing both sides by 8, we find that x = 2. Substituting this value back into one of the equations, we can solve for y. Using the first equation, we have 10(2) + 4y = 52, which simplifies to 20 + 4y = 52. Subtracting 20 from both sides, we obtain 4y = 32. Dividing by 4, we find that y = 8. Therefore, each small candle holder holds 2 candles and each large candle holder holds 8 candles.

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Final answer:

To find out how many candles each candle holder holds, set up a system of equations using the given information and solve for the variables. Each small candle holder holds 2 candles and each large candle holder holds 8 candles.

Explanation:

To find out how many candles each candle holder holds, we need to set up a system of equations using the given information. Let's say the number of candles a small candle holder can hold is 's' and the number of candles a large candle holder can hold is 'l'.

From the east side, we have the equation 10s + 4l = 52. From the west side, we have the equation 2s + 4l = 36. Solving these equations, we can find the values of 's' and 'l', which will give us the number of candles each candle holder can hold.

We can eliminate 'l' by multiplying the first equation by 2 and the second equation by 5. This gives us 20s + 8l = 104 and 10s + 20l = 180. By subtracting the second equation from the first, we get 10l = 76. Dividing both sides by 10, we find that l = 7.6. Since we cannot have a fraction of a candle, we can approximate l to the nearest whole number, which is 8.

Substituting the value of l back into one of the original equations, we can find the value of s. Using the first equation, we have 10s + 4(8) = 52. Simplifying this equation gives us 10s + 32 = 52. Subtracting 32 from both sides, we find that 10s = 20. Dividing both sides by 10, we find that s = 2. Therefore, each small candle holder holds 2 candles and each large candle holder holds 8 candles.

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At an elementary school carnival, students can draw a rubber duck out of a tub of water. Each duck has a number written on the bottom of it which correlates with a prize. There are a total of 8 ducks in the tub. Two ducks have a 5 on them, four ducks have a 6 on them, and two duck has a 7 on it. What is the expected value of a duck?


a. 8.25

b. 6.85

c. 25

d. 4


Please help!!!!

Answers

Answer: 4

Step-by-step explanation: I Am Almost Sure The Answer Is 4. I Did The Math Out And That Is What I Got.

cylinder a has a radius of 1 m and a height of 4 m cylinder B has a radius of 1 m and a height of 8 m what is the ratio of the volume of cylinder a to the volume of cylinder B

a. 1:2
b. 2:1
c. 1:1
d. 1:4

Answers

Answer: option a

Step-by-step explanation:

The volume of a cylinder can be calculated with this formula:

[tex]V=\pi r^2h[/tex]

Where the radius is "r" and the height is "h"

Calculate the volume of the Cylinder A:

[tex]V_A=\pi (1m)^2(4m)\\\\V_A=4\pi\ m^3[/tex]

Calculate the volume of the Cylinder B:

[tex]V_B=\pi (1m)^2(8m)\\\\V_B=8\pi\ m^3[/tex]

Now, the ratio of the volume of the Cylinder A to the volume of the Cylinder B can be calculated with:

[tex]ratio=\frac{V_A}{V_B}[/tex]

Substituting values, you get:

[tex]ratio=\frac{4\pi\ m^3}{8\pi\ m^3}[/tex]

[tex]ratio=\frac{1}{2}[/tex] or 1:2

Solve: log2(x-4) = 4

Answers

Answer:

D

Step-by-step explanation:

log₂(x-4) = 4

Undo the log by raising 2 to both sides:

2^(log₂(x-4)) = 2^4

x - 4 = 2^4

x - 4 = 16

x = 20

Answer D.

log2(x-4) =4. The answer is D) 20

Help me with B please!

Answers

Answer:

Step-by-step explanation:

Even though the teams appear from the mean to be similar in their winnings, they are not.  This is mostly because the Tigers have a greater range (difference between the highest and lowest values) than do the Foxes.  The Tigers' range is 16 - 1 = 15 while the Foxes' range is 6 - 3 = 3.

In other words, using the mean to determine how closely matched these teams are is worthless.

Use the drawing tool(s) to form the correct answer on the provided graph.

Graph the following system of equations in the coordinate plane. Use the Mark Feature tool to indicate the solution to the system on the graph.

Answers

Answer:

The solution of the system of equations is (-3 , 5)

Step-by-step explanation:

* Lets describe the drawing of each line

- The form of the equation of any line is y = mx + c, where m

  is the slope of the line and c is the y-intercept (the point of

  intersection between the line and the y-axis is (0 , c))

* The line y = -x + 2 represented by the red line

- The line intersect the y-axis at point (0 ,2)

- The line intersect the x-axis at point (2 , 0)

- The slope of the line is -1, so the angle between the positive

  part of x-axis and the line is obtuse

* The line x - 3y = -18 represented by blue line

- Put the line in the form y = mx + c

- The line is x - 3y = -18⇒ add 18 and 3y to both sides

- The line is 3y = x + 18 ⇒ ÷ 3 both sides

- The line is y = 1/3 x + 6

- The line intersect the y-axis at point (0 ,6)

- The line intersect the x-axis at point (-18 , 0)

- The slope of the line is 1/3, so the angle between the positive

  part of x-axis and the line is acute

* Look to the attached graph

- The point of intersection between the two line is the solution

 of the system of equation

- From the graph the point of intersection is (-3 , 5)

* The solution of the system of equations is (-3 , 5)

Answer:

one of the lines will pass (0,2) and (2,0) and intersect at (5,3) and pass through (0,5)

Step-by-step explanation:

edit: it actually passes through (0,6) mark as brainliest if right

Jerry pours 86 milliliters of water into 8 tiny beakers he measures an equal amount of water into the first 7 beakers he pours the remaining water into the eight beaker it measure 16 milliliters how many milliliters of water are in each of the first 7 beakers. Using the RDW process show how you got the answer.

Answers

Answer:

There are 7 milliliters of water in each of the first 7 beakers

Step-by-step explanation:

* At first lets read the problem

- There are 86 milliliters of water to pour into tiny beakers

- Jerry has 8 tiny beakers

- He pours same amount in seven of them

- He pours 16 milliliters in the 8th one

- we want to know how many milliliters in each of the 7 beakers

* Look to the attached drawing

- There are 8 shapes represent the tiny beakers

- Seven of them have same amount x

- The last one has amount 16 milliliters

- All of them have 86 milliliters

* Now lets write the steps to answer the problem

∵ The amount of water is 86 milliliters

∵ The 8th tiny beaker has 16 milliliters

∴ The amount of water in the 7 beakers = 86 - 16 = 70 milliliters

∵ Each one of the 7 beakers has x milliliters of water

∴ 7 × (x) = 70 ⇒ divide each side by 7

∴ x = 7 milliliters

* There are 7 milliliters of water in each of the first 7 beakers

Each of the first 7 beakers contains 10 milliliters of water.

1. First, we identify the total amount of water Jerry poured, which is 86 milliliters.

 2. We know that the eighth beaker contains 16 milliliters of water.

3. To find out how much water was poured into the first 7 beakers, we subtract the amount in the eighth beaker from the total amount:

[tex]\[ \text{Water in first 7 beakers} = \text{Total water} - \text{Water in 8th beaker} \] \[ \text{Water in first 7 beakers} = 86 \text{ ml} - 16 \text{ ml} \] \[ \text{Water in first 7 beakers} = 70 \text{ ml} \][/tex]

4. Now, we divide the total water in the first 7 beakers by 7 to find out how much water is in each beaker:

[tex]\[ \text{Water per beaker} = \frac{\text{Water in first 7 beakers}}{\text{Number of beakers}} \] \[ \text{Water per beaker} = \frac{70 \text{ ml}}{7} \] \[ \text{Water per beaker} = 10 \text{ ml} \][/tex]

Therefore, each of the first 7 beakers contains 10 milliliters of water.

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