7. If 40% of a given number is 8, then what is 15% of the
given number?

Answer:

[tex]3\sqrt{2} \sqrt{x}[/tex]

Answer: 10.5

Step-by-step explanation:

Hello the answer is 10.5 you just have to multiply 3 1/2 times 3 and there's your answer

Have a great day hope this helps!:):):

Answer:

10 1/2

Step-by-step explanation:

convert 3 1/2 into an improper fraction

3 1/2 = 7/2

if the flour is tripled that means

3 1/2 x 3

= 7/2 x 3

= (7x3) / 2

= 21/2  (convert to mixed fraction, by long division)

= 10 1/2

Answer:

Step-by-step explanation:

3x - 6 + 2x + 108 + 88 = 360     Gather like terms

5x + 108 + 88 - 6 = 360

5x = 190                                      Divide by 5

x = 190/5

x = 38  

Answer:

A because it is increasing

Step-by-step explanation:

Answer:

I would say c

Step-by-step explanation:

you think if you go down a hill what happen is you speed up so you would need to be go down or heading in a negative direction.

Answer:

Please, find the image with the diagram of the lines corresponding to this question:

The answers are:

Explanation:

First question:

You must use the fact that the angles DCE and BCD are adjacent angles, because they share the vertex (C) and have a common side (CD).

Thus, as a first fact, the measure of the angle BCE is equal to the sum of the measures of the angles BCD and DCE.

On the other hand, BE is a straight line, thus the measure of angle BCE is 180º.

Hence, you can write:

Second question:

Solve for x:

Substitute m ∠DCE with 120º:

Hence, for m ∠DCE = 120º, x = 60º.

Answer:

63

Step-by-step explanation:

Answer:

65

Step-by-step explanation:

75% of 65 = 48

Answer:

[tex]\frac{1}{4}[/tex] of circumference of the circle

Step-by-step explanation:

We are given that

Arc NOP=Central angle=[tex]\theta=90^{\circ}[/tex]

Radius of circle=5 units

We have to find the statement which describes best the length of arc NOP.

Arc length=[tex]\frac{central\;angle}{360}\times 2\pi r[/tex]

Using the formula

Arc length=[tex]\frac{90}{360}\times circumference\;of\;circle[/tex]

Where Circumference of circle=[tex]2\pi r[/tex]

Arc length NOP=[tex]\frac{1}{4}[/tex]circumference of circle

Hence, the arc length NOP=[tex]\frac{1}{4}[/tex] of circumference of the circle

Answer:

1/4 the circumference of circle L

Step-by-step explanation:

Only if they are each the same distance away from zero, but on opposite sides of the number line.

Hope this help! :)

Answer:If you add a negative and a positive you do the the integers step by changing the add operation to a subtraction.

Step-by-step explanation:-4+3

integers property     -4 - 3 = 1

negative plus a positive is a positive.

Answer:

2

Step-by-step explanation:

Answer:

30

Step-by-step explanation:

It states that it rained twice as much in 5 days all you do is multiply the 2 times 3 which is 6 so then u multiply 6 times 5 which is 30.

The river rose a total of 93 inches over five days.

The question asks how much a river rises in five days given a certain pattern of increase. To find the solution, we use a geometric sequence because the river rises by a constant multiple each day. On the first day, the river rises three inches. Since on each subsequent day the river rises by twice the amount it did the previous day, the sequence of increases over five days is: 3 inches, 6 inches, 12 inches, 24 inches, and 48 inches.

The total rise of the river is the sum of these increases:

3 + 6 + 12 + 24 + 48 = 93 inches.

Therefore, the river rose a total of 93 inches over the span of five days.

Answer:

The bigger pot is a 9.1-quart pot

Step-by-step explanation:

Initial pot size = 6.5-quart

New pot is 40% bigger than the initial pot size

Bigger pot size = 6.5-quart + (0.4 × 6.5-quart) = 6.5-quart + 2.6-quart = 9.1-quart

Answer:

v = 2

Step-by-step explanation:

ALL OF THESE WILL BE SOLVE FOR X AND Y!!!

Problem 1 Solve For x: x=−6y−7

Show work: Add -6y to both sides.

x+6y+−6y=−7+−6y

x=−6y−7

Problem 1 Solve For y: y = -1/6x + -7/6

Show work: Add -x to both sides.

x+6y+−x=−7+−x

6y=−x−7

Step 2: Divide both sides by 6.

6y/6 = -x -7/6

Problem 2 Solve For x: x=−6y−7

Show work:  Add -12y to both sides.

2x+12y+−12y=−14+−12y

2x=−12y−14

Step 2: Divide both sides by 2.

2x/2 = -12y - 14/2

Problem 2 Sovle For y: y=-1/6x + -7/6

Show work: Add -2x to both sides.

2x+12y+−2x=−14+−2x

12y=−2x−14

Step 2: Divide both sides by 12.

12y/12 = -2x-14/12

Answer: The length is 62 inches and the width is 13 inches

Step-by-step explanation: The perimeter of the rectangular portrait has been given as 150 inches. We also know that the perimeter of a rectangle is given as

Perimeter= 2(L + W)

However we don't have the measurements for the length and width. What we do have are descriptions of both. The length is given as W, while the length is ten more than four times the width. That is, the length equals

10 + 4W

Therefore we have the length and the width as

L = 4W + 10 and

W = W

If the perimeter is 150, and

Perimeter = 2(L + W) then,

150 = 2(4W + 10 + W)

150 = 2(5W + 10)

150 = 10W + 20

Subtract 20 from both sides of the equation

130 = 10W

Divide both sides of the equation by 10

13 = W

With that in mind we can now calculate the length as

L = 4W + 10

Substitute for the value of W

L = 4(13) + 10

L = 52 + 10

L = 62

Therefore, the length is 62 inches and the width is 13 inches

Answer:

Length = 62 inches, Width = 13 inches

Step-by-step explanation:

Let L represent the length of the frame, W, the width and P, the perimeter

Perimeter of a rectangle, P = 2 (L + W) .....eq 1

Also, P = 150 inches

and

L = 10 + 4W .....eq 2

Slotting in the respective values of P and L in eq 1

150 = 2 {(10 + 4W) + W}

Expanding the bracket

150 + 2 (10 + 5W)

150 = 20 + 10W

Subtracting 20 from both sides of the equation

150 - 20 = 20 - 20 + 10W

130 = 10W

Dividing both sides by the coefficient of W which is 10

13 = W

Therefore, W = 13 inches

Slotting in the value of W in eq 2

L = 10 + 4 (13)

L = 10 + 52

L = 62 inches

Lets ensure that the values of L and W are correct

P = 2 (L + W)

150 = 2 (13 + 62)

150 = 2(75)

150 = 150

Hence, L = 62 inches, W = 13 inches

[tex]\boxed{sinA=\frac{a}{c}} \\ \\ \boxed{cosA=\frac{b}{c}}[/tex]

The trigonometric functions are very important in math and physics. Sound, light and electricity all involve oscillations and are modeled by sine and cosine functions. So, we can provide a trig ratio for the sine and cosine function by taking a right triangle as shown in the figure below. So the relationships are as follows:

[tex]sinA=\frac{Opposite \ side}{Hypotenuse} \\ \\ cosA=\frac{Adjacent \ side}{Hypotenuse} \\ \\ \\ For \ the \ figure: \\ \\ a:Opposite \ side \\ \\ b:Adjacent \ side \\ \\ c:Hypotenuse[/tex]

So we can write:

[tex]\boxed{sinA=\frac{a}{c}} \\ \\ \\ \boxed{cosA=\frac{b}{c}}[/tex]

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2x^2 + 5x + 3

What are the factors of the polynomial?

(2x+3)(x+1)

(2x-3)(x-1)

(3x+2)(x+1)

(3x-2)(x-1)

Option A

The factors are:

[tex]2x^2+5x+3 = (2x+3)(x+1)[/tex]

Given that, the quadratic equation is:

[tex]2x^2 + 5x + 3[/tex]

We have to find the factors of polynomial

[tex]2x^2+5x+3[/tex]

Split 5x as 2x and 3x

[tex]2x^2+5x+3 = 2x^2 +2x + 3x + 3[/tex]

[tex]\mathrm{Break\:the\:expression\:into\:groups}[/tex]

[tex]2x^2+5x+3=\left(2x^2+2x\right)+\left(3x+3\right)[/tex]

[tex]\mathrm{Factor\:out\:}2x\mathrm{\:from\:}2x^2+2x\mathrm{:\quad }2x\left(x+1\right)[/tex]

Thus we get,

[tex]2x^2+5x+3 = 2x(x+1) + (3x+3)[/tex]

[tex]\mathrm{Factor\:out\:}3\mathrm{\:from\:}3x+3\mathrm{:\quad }3\left(x+1\right)[/tex]

Thus we get,

[tex]2x^2+5x+3 = 2x(x+1) + 3(x+1)[/tex]

[tex]\mathrm{Factor\:out\:common\:term\:}x+1[/tex]

Thus we get,

[tex]2x^2+5x+3 = (2x+3)(x+1)[/tex]

Thus the factors are found for given polynomial

Final answer:

The correct factors of the polynomial +5x+3 are (2x+3)(x+1), as they multiply to give the original polynomial. None of the other provided options yield the correct polynomial upon multiplication.

Explanation:

The question asks to identify the factors of the polynomial +5x+3. Factors are expressions that, when evaluated, produce the value of the polynomial. Let's examine the provided options to find which pair of binomials gives us the correct polynomial upon multiplication:

(2x+3)(x+1) = 2x² + 2x + 3x + 3 = 2x² + 5x + 3, which matches the original polynomial.

(2x-3)(x-1) = 2x² - 2x - 3x + 3 = 2x² - 5x + 3, which does not match the original polynomial.

(3x+2)(x+1) = 3x² + 3x + 2x + 2 = 3x² + 5x + 2, which does not match the original polynomial.

(3x-2)(x-1) = 3x² - 3x - 2x + 2 = 3x² - 5x + 2, which does not match the original polynomial.

Therefore, the correct factors of the polynomial +5x+3 are (2x+3)(x+1).

By substituting x = -2 in each expression, we find that x-2 equals -4, x-1 equals -3, x₀ equals 1, x₁ equals -2, and x₂ equals 4.

To solve this, we need to substitute x = -2 into each expression. Here are the results:

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Answer:

About

step 1 highlight numbers

step 2 do ctrl+f

step 3 type in 9

ENJOY

1051541451431641621571721571511441234567881234567812345678123678326470547 2999999259923478990124999995689902993413269916749953349999914649932724997 2994567809912568990139956799809929936781467998299634699818991169966144990 2999994569970124995699801323459999012615302799995324993243699019923412993 2994567801993569980299356780239999456725634569974326992644399243992369936 2994567801992689901239967899029939945745315319931253399436998011992349950 2998012345299999388352999991039953991232012479934673289999982640499999415l

Step-by-step explanation:

Answer:

[tex]\lim _{x\to \infty \:}\left(\frac{\sqrt{9x^2+x+1}-\sqrt{4x^2+2x+1}}{x+1}\right)=1[/tex]

Step-by-step explanation:

Considering the expression

[tex]\lim _{x\to \infty \:}\frac{\sqrt{9x^2+x+1}-\sqrt{4x^2+2x+1}}{x+1}[/tex]

Steps to solve

[tex]\lim _{x\to \infty \:}\frac{\sqrt{9x^2+x+1}-\sqrt{4x^2+2x+1}}{x+1}[/tex]

[tex]\mathrm{Divide\:by\:highest\:denominator\:power:}\:\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}[/tex]

[tex]\lim _{x\to \infty \:}\left(\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}\right)[/tex]

[tex]\lim _{x\to a}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\lim _{x\to a}f\left(x\right)}{\lim _{x\to a}g\left(x\right)},\:\quad \lim _{x\to a}g\left(x\right)\ne 0[/tex]

[tex]\mathrm{With\:the\:exception\:of\:indeterminate\:form}[/tex]

[tex]\frac{\lim _{x\to \infty \:}\left(\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}\right)}{\lim _{x\to \infty \:}\left(1+\frac{1}{x}\right)}.....[1][/tex]

As

[tex]\lim _{x\to \infty \:}\left(\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}\right)=1[/tex]

Solving

[tex]\lim _{x\to \infty \:}\left(\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}\right)....[A][/tex]

[tex]\lim _{x\to a}\left[f\left(x\right)\pm g\left(x\right)\right]=\lim _{x\to a}f\left(x\right)\pm \lim _{x\to a}g\left(x\right)[/tex]

[tex]\mathrm{With\:the\:exception\:of\:indeterminate\:form}[/tex]

[tex]\lim _{x\to \infty \:}\left(\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}\right)-\lim _{x\to \infty \:}\left(\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}\right)[/tex]

Also

[tex]\lim _{x\to \infty \:}\left(\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}\right)=3[/tex]

Solving

[tex]\lim _{x\to \infty \:}\left(\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}\right)......[B][/tex]

[tex]\lim _{x\to a}\left[f\left(x\right)\right]^b=\left[\lim _{x\to a}f\left(x\right)\right]^b[/tex]

[tex]\mathrm{With\:the\:exception\:of\:indeterminate\:form}[/tex]

[tex]\sqrt{\lim _{x\to \infty \:}\left(9+\lim _{x\to \infty \:}\left(\frac{1}{x}+\lim _{x\to \infty \:}\left(\frac{1}{x^2}\right)\right)\right)}[/tex]

[tex]\lim _{x\to \infty \:}\left(9\right)=9[/tex]

[tex]\lim _{x\to \infty \:}\left(\frac{1}{x}\right)=0[/tex]

[tex]\lim _{x\to \infty \:}\left(\frac{1}{x^2}\right)=0[/tex]

So, Equation [B] becomes

⇒ [tex]\sqrt{9+0+0}[/tex]

⇒ [tex]3[/tex]

Similarly, we can find

[tex]\lim _{x\to \infty \:}\left(\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}\right)=2[/tex]

So, Equation [A] becomes

⇒ [tex]3-2[/tex]

⇒ 1

Also

[tex]\lim _{x\to \infty \:}\left(1+\frac{1}{x}\right)=1[/tex]

Thus, equation becomes

[tex]\frac{\lim _{x\to \infty \:}\left(\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}\right)}{\lim _{x\to \infty \:}\left(1+\frac{1}{x}\right)}=\frac{1}{1}=1[/tex]

Therefore,

[tex]\lim _{x\to \infty \:}\left(\frac{\sqrt{9x^2+x+1}-\sqrt{4x^2+2x+1}}{x+1}\right)=1[/tex]

Keywords: limit

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13d + 25 = 8d

13d - 8d = -25

5d = -25

d = -5

so the answer is (-5)

Answer:

Step-by-step explanation:

13d + 25 = 8d

13d = 8d -25

13d - 8d = -25

5d = -25

d = -5

Answer:

There's no picture. Is it D.√3 / 2

Step-by-step explanation:

Answer:

Susan had $41.

Step-by-step explanation:

The amount of money they have is enough for $99 spending. The answer can be found by testing two numbers and seeing whether the amounts need to be increased or decreased to make the distance between them 17, with a sum of 99.

40 and 57: distance of 17, but not enough for $99. $41 and $58 equal 99, with a distance of 17 between them. Jenny, with $17 more, has 58. Susan has 41.

MI ≅ MR proved by using ASA postulate of congruence

Step-by-step explanation:

Let us revise the cases of congruence

∵ M is the mid-point of TE

∴ MT = ME

In Δs TMI and EMR

∵ ∠T ≅ ∠E ⇒ given

∵ ∠TMI ≅ EMR ⇒ vertical opposite angles

∵ MT = ME ⇒ proved

∴ Δ TMI ≅ ΔEMR by ASA postulate of congruence

- From congruence, corresponding sides are equal

∴ MI ≅ MR

MI ≅ MR proved by using ASA postulate of congruence

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The question is about a geometry proof related to congruent angles and midpoints, but additional information is needed to provide a specific solution.

The question appears to be a geometry proof involving congruent angles and midpoints. The goal is to prove that two line segments MI and MR are congruent given ∠E ≅ ∠T, and M is the midpoint of TE.

To prove this, one would need to provide more information about the points I and R, such as their positions relative to point E, T, and M.

Without this additional information, it is difficult to give a specific proof. However, typical proof strategies might involve showing that triangles EMI and TMR are congruent by Side-Angle-Side (SAS), Angle-Side-Angle (ASA), or Angle-Angle-Side (AAS) postulates, depending on the position of points I and R.

Answer:

Plane B is closer.

The distance for Plane A is 7.9 km.

The distance for Plane B is 7.4 km.

Step-by-step explanation:

To find the distance to the tower, altitude must be converted to kilometers. Each 1000 ft is 0.3048 km, so the heights of the planes are ...

-- Plane A: (20 thousand ft)*(0.3048 km/thousand ft) = 6.096 km

-- Plane B: (8 thousand ft)*(0.3048 km/thousand ft) = 2.4384 km

The Pythagorean theorem can be used to find the distance from each plane to the tower:

-- distance = √((ground distance)² + (height)²)

-- Plane A distance = √(5² +6.096²) ≈ √62.16 ≈ 7.9 . . . km

-- Plane B distance = √(7² +2.4384²) ≈ √54.95 ≈ 7.4 . . . km

The distance for Plane B is shorter, so Plane B is closer to the tower.

Answer:

1.26 liters.

Step-by-step explanation:

Given that, a hummingbird feeder holds [tex]\frac{2}{3}[/tex] of a cup of hummingbird solution.

Again, a bottle of solution will fill the feeder 8 times.

Therefore, a bottle of solution will measure [tex](\frac{2}{3}\times 8) = \frac{16}{3} = 5.33[/tex] cups of hummingbird solution.

Now, one cup is equivalent to 0.236588 liters.

So, 5.33 cups is equal to (0.236588 × 5.33) = 1.26 liters. (Answer)

Answer:

approximately 2,800

Step-by-step explanation:

you would round 691 to 700 then multiply by 4

Answer:

2,800

Step-by-step explanation:

In saying "largest number," I believe the question means that you should round 691 up to 700 (nearest hundreds). Multiple 700 by 4, and you will get 2,800! Hope this helps! :)

Answer:

8:30 PM

Step-by-step explanation:

Let's say that x is minutes since 6:20 PM, and y is position in miles.

Since Jonathan drives at a constant rate, the position vs time graph is linear.  Two points on the line are (0, 75) and (20, 85).  The slope of the line is:

m = Δy / Δx

m = (85 − 75) / (20 − 0)

m = ½

So the equation of the line is:

y = ½ x + 75

We want to find x when y is 140.

140 = ½ x + 75

65 = ½ x

x = 130

So 130 minutes after 6:20 PM is when Jonathan's position will be 140 miles.  130 minutes is 2 hours and 10 minutes.  So the time will be 8:30 PM.

By calculating the speed at which Jonathan was driving and applying that rate to the total distance, it can be concluded that Jonathan would have driven a total of 140 miles by about 8:30 PM.

Jonathan drove 10 miles from 6:20 PM to 6:40 PM, giving a 20-minute duration. We can say the rate of speed he was driving at is 10 miles/20 minutes. We can simplify this rate down to 1 mile/2 minutes to make it easier to work with.
To calculate when he would've driven a total of 140 miles we follow these steps:

Subtract the 75 miles he had already driven from the total of 140 miles we want to find out about, which gives us 65 miles.  

Therefore, Jonathan would've driven a total of 140 miles by about 8:30 PM.

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