Find the indicated term of the given geometric sequence. a1 = 14, r = –2, n = 11

Answer:

[tex]351 \pi[/tex] (about 1102.69902)

Step-by-step explanation:

The volume of a cone is represented by the formula [tex]V=\pi r^2 \frac{h}{3}[/tex], where [tex]V[/tex] is the volume, [tex]r[/tex] is the radius, and [tex]h[/tex] is the height.

This is as simple as the solution can get without estimating, but we can estimate with a calculator.  [tex]351 \pi[/tex] is approximately equal to 1102.69902.

Answer is provided in the image attached.

Answer:

B. The system may have infinitely many solutions

D. The system may have no solution

Step-by-step explanation:

we know that

If a system of linear equations contains two equations with the same slope

then

we may have two cases

case 1) The two equations are identical, in this case we are going to have infinite solutions

case 2) The two equations have the same slope but different y-intercept, (parallel lines) in that case the system has no solution.

Answer:

Its 30%

Step-by-step explanation:

I used guess and check.

(25%)

3100*.25 = 775

(30%)

3100*.3 = 930

Answer:

30%.

Step-by-step explanation:

It is the fraction 930/3100 multiplied by 100.

Percentage that is rent = (930 * 100 ) / 3100

=  30%.

Answer:

A. g(x) = 6|x| +10

Step-by-step explanation:

The parent function is given as:

f(x) = 3|x| + 5

Applying transformation:

function f is vertically stretched by a factor of 2 to give function g.

To stretch a function vertically we multiply the function by the factor:

2*f(x) = 2[3|x| + 5]

g(x) = 2*3|x| + 2*5

g(x) = 5|x| + 10

Answer: Option A.

Step-by-step explanation:

There are some transformations for a function f(x).

One of the transformations is:

If [tex]kf(x)[/tex] and [tex]k>1[/tex], then the function is stretched vertically by a factor of "k".

Therefore, if the function provided [tex]f(x) = 3|x| + 5[/tex] is vertically stretched by a factor or 2, then the transformation is the following:

[tex]2f(x)=g(x)=2(3|x| + 5)[/tex]

Applying Disitributive property to simplify, we get that the function g(x) is:

[tex]g(x)=6|x| +10[/tex]

Answer:

False

Step-by-step explanation:

We first write the equation in the form ax² + bx + c=0 which gives us:

3x² - 6x + 9=0

Given the quadratic formula,

x= [-b ±√(b²- 4ac)]/2a ,the discriminant proves whether the equation has real roots or not.

The discriminant, which is the value under the root sign, may either be positive, negative or zero.

Positive discriminant- the equation has two real roots

Negative discriminant- the equation has no real roots

Zero discriminant - The equation has two repeated roots.

In the provided equation, b²-4ac results into:

(-6)²- (4×3×9)

=36-108

= -72

The result is negative therefore the equation has no real solutions.

Answer: FALSE

Step-by-step explanation:

Rewrite the given equation in the form [tex]ax^2+bx+c=0[/tex], then:

[tex]3x^2 = 6x - 9\\3x^2-6x +9=0[/tex]

Now, we need to calculate the Discriminant with this formula:

[tex]D=b^2-4ac[/tex]

We can identify that:

[tex]a=3\\b=-6\\c=9[/tex]

Then, we only need to substitute these values into the formula:

 [tex]D=(-6)^2-4(3)(9)[/tex]

 [tex]D=-72[/tex]

Since [tex]D<0[/tex] the equation has no real solutions.

Answer:

2. the process of breeding only organisms with desirable traits

Step-by-step explanation:

Answer:

the answer is the second one

Step-by-step explanation:

Answer:

[tex]\sigma=6.16[/tex]

Step-by-step explanation:

By definition, the variance V of a population is defined as:

[tex]V = \sigma^2[/tex]

Where [tex]\sigma[/tex] is the standard deviation

We know that [tex]V = 38[/tex], then we can solve the equation for the standard deviation [tex]\sigma[/tex]

[tex]38 = \sigma^2[/tex]

[tex]\sigma^2=38[/tex]

[tex]\sigma=\sqrt{38}[/tex]

[tex]\sigma=6.16[/tex]

Finally  the standard deviation is: [tex]\sigma=6.16[/tex]

Answer:

2x^2 +2x-2

Step-by-step explanation:

f(x)=-x^2+6x-1

g(x)=3x^2-4x-1

(f+g)(x)= -x^2+6x-1 +3x^2-4x-1

          =  2x^2 +2x-2

Answer:

x1=-11 x2=-1 y=11

Step-by-step explanation:

you can see the explanation in the pics

Answer:

[tex]\large\boxed{x^2+11x+24=0\Rightarrow(x+5.5)^2=6.25}[/tex]

Step-by-step explanation:

[tex](a+b)^2=a^2+2ab+b^2\qquad(*)\\\\x^2+11x+24=0\qquad\text{subtract 24 from both sides}\\\\x^2+(2)(x)(5.5)=-24\qquad\text{add}\ 5.5^2\ \text{to both sides}\\\\\underbrace{x^2+(2)(x)(5.5)+5.5^2}_{(*)}=-24+5.5^2\\\\(x+5.5)^2=-24+30.25\\\\(x+5.5)^2=6.25\Rightarrow x+5.5=\pm\sqrt{6.25}\\\\x+5.5=-2.5\ \vee\ x+5.5=2.5\qquad\text{subtract 5.5 from both sides}\\\\x=-8\ \vee\ x=-3[/tex]

To rewrite x^2 + 11x + 24 = 0 by completing the square, we first organize terms, then add the square of half the coefficient of x to both sides to create a perfect square. Taking the square root of both sides then provides the solution for x, resulting in x = -5.5 ± √6.25.

To rewrite the equation x^2 + 11x + 24 = 0 by completing the square, we first need to make the quadratic and linear terms to create a square.

1. Rewrite the equation as: x^2 + 11x + __ = -24 + __

2. Take half of the coefficient of x, (11/2) and square it. (11/2)^2 = 30.25

3. Add this value on both sides of the equation:

x^2 + 11x + 30.25 = -24 + 30.25

4. Now, the left side of the equation is a perfect square and it can be written as:

(x + 5.5)^2 = 6.25

5. Finally, to solve for x, take the square root of both sides to get:

x + 5.5 = ± √6.25

x = -5.5 ± √6.25

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[tex]\bf \textit{surface area of a sphere}\\\\ SA=4\pi r^2~~ \begin{cases} r=radius\\ \cline{1-1} SA=400\pi \end{cases}\implies 400\pi =4\pi r^2 \implies \cfrac{\stackrel{100}{~~\begin{matrix} 400\pi \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~} }{~~\begin{matrix} 4\pi \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}=r^2 \\\\\\ 100=r^2\implies \sqrt{100}=r\implies 10=r \\\\[-0.35em] ~\dotfill[/tex]

[tex]\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}\qquad \implies V=\cfrac{4\pi (10)^3}{3}\implies V=\cfrac{4000\pi }{3}\implies V\approx 4188.79[/tex]

Answer:

D=15 so N=10

Step-by-step explanation:

N+D=25 where the coins make up the number of nickels, N and the number of dimes, N

Now each nickel is worth .05   (not .5)

Each time is worth .10  or .1

So the two equations are .05N+.1D=2  and N+D=25

I'm going to multiply 100 on both sides so I can clear the decimals from first equation giving me 5N+10D=200.

So I'm going to multiply the second by 5 giving my 5N+5D=125

Line up equations and you should see we can solve this system by elmination by subtracting the equations

5N+10D=200

5N+5D=125

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

       5D=75

         D=75/5=15

So N+D=25 and D=15 so N=10.

Step-by-step explanation:

The product of a and b is equal to a · b.

Let w - width and l - length, then the product of the width and the lenght is

w · h = wh

The product of width(w) and height(h) is equal to w.h.

Let ;

w - width

l - length

∴ the product of the width and the length is w · h = wh

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Answer:

[tex]a=\frac{2S -2v_ot-2s_o}{t^2}[/tex]

Step-by-step explanation:

We have the equation of the position of the object

[tex]S = \frac{1}{2}at ^2 + v_ot+s_o[/tex]

We need to solve the equation for the variable a

[tex]S = \frac{1}{2}at ^2 + v_ot+s_o[/tex]

Subtract [tex]s_0[/tex] and [tex]v_0t[/tex] on both sides of the equality

[tex]S -v_ot-s_o = \frac{1}{2}at ^2 + v_ot+s_o - v_ot- s_o[/tex]

[tex]S -v_ot-s_o = \frac{1}{2}at ^2[/tex]

multiply by 2 on both sides of equality

[tex]2S -2v_ot-2s_o = 2*\frac{1}{2}at ^2[/tex]

[tex]2S -2v_ot-2s_o =at ^2[/tex]

Divide between [tex]t ^ 2[/tex] on both sides of the equation

[tex]\frac{2S -2v_ot-2s_o}{t^2} =a\frac{t^2}{t^2}[/tex]

Finally

[tex]a=\frac{2S -2v_ot-2s_o}{t^2}[/tex]

E(-3, -4), F(1, -3), G(3, -6), and H(1, -6); a reflection across the x-axis

E(-3, -1), F(1, -2), G(3, 1), and H(1, 1); a translation 5 units down

E(3, 4), F(-1, 3), G(-3, 6), and H(-1, 6); a reflection across the y-axis

E(4, 4), F(8, 3), G(10, 6), and H(8, 6);a translation 7 units right

Transformation is the movement of a point from its initial location to a new location. Types of transformation are rotation, translation, reflection and dilation.

Quadrilateral ABCD has vertices A(-3, 4), B(1, 3), C(3, 6), and D(1, 6). Hence:

E(-3, -4), F(1, -3), G(3, -6), and H(1, -6); a reflection across the x-axis

E(-3, -1), F(1, -2), G(3, 1), and H(1, 1); a translation 5 units down

E(3, 4), F(-1, 3), G(-3, 6), and H(-1, 6); a reflection across the y-axis

E(4, 4), F(8, 3), G(10, 6), and H(8, 6);a translation 7 units right

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Answer:

A it's is located halfway between the parabolas focus and directrix

Step-by-step explanation:

hope this helps

Answer:

Option A is correct that is it's is located halfway between the parabolas focus and directrix.

Step-by-step explanation:

We are given a parabola.

To find: Best statement which describes the location of the vertex on the parabola.

Standard Equation of Parabola which open on the Right hand direction.

( y - k )² = 4a( x - h )

Here, ( h , k ) is the vertex of the parabola and ( h + a , k ) is the focus.

Distance of the focus from vertex = a

Equation of the Directrix of the Parabola is , y =  k - a

Distance of the Directrix from vertex = a

Therefore, Option A is correct that is it's is located halfway between the parabolas focus and directrix.

Answer:

Part 2) Option b. 5.4

Part 3) Option c. 8

Part 4) Option a. 15

Part 5) Option d. 8.9

Step-by-step explanation:

Part 2) Find the value of x to the nearest tenth

we know that

x is the radius of the circle

Applying the Pythagoras Theorem

[tex]x^{2}=3.6^{2}+(8/2)^{2}[/tex]

[tex]x^{2}=28.96[/tex]

[tex]x=5.4\ units[/tex]

Part 3) Find the value of x

In this problem

x=8

Verify

step 1

Find the radius of the circle

Let

r -----> the radius of the circle

Applying the Pythagoras Theorem

[tex]r^{2}=8^{2}+(15/2)^{2}[/tex]

[tex]r^{2}=120.25[/tex]

[tex]r=\sqrt{120.25}[/tex]

step 2

Find the value of x

Applying the Pythagoras Theorem

[tex]r^{2}=x^{2}+(15/2)^{2}[/tex]

substitute

[tex]120.25=x^{2}+56.25[/tex]

[tex]x^{2}=120.25-56.25[/tex]

[tex]x^{2}=64[/tex]

[tex]x=8\ units[/tex]

Part 4) Find the value of x

In this problem

x=OP=15

Verify

step 1

Find the radius of the circle

Let

r -----> the radius of the circle

In the right triangle FPO

Applying the Pythagoras Theorem

[tex]r^{2}=15^{2}+(40/2)^{2}[/tex]

[tex]r^{2}=625[/tex]

[tex]r=25[/tex]

step 2

Find the value of x

In the right triangle RQO

Applying the Pythagoras Theorem

[tex]25^{2}=x^{2}+(40/2)^{2}[/tex]

[tex]625=x^{2}+400[/tex]      

[tex]x^{2}=625-400[/tex]

[tex]x^{2}=225[/tex]

[tex]x=15\ units[/tex]

Part 5) Find the value of x

Applying the Pythagoras Theorem

[tex]6^{2}=4^{2}+(x/2)^{2}[/tex]

[tex]36=16+(x/2)^{2}[/tex]

[tex](x/2)^{2}=36-16[/tex]

[tex](x/2)^{2}=20[/tex]

[tex](x/2)=4.47[/tex]

[tex]x=8.9[/tex]

ANSWER

shifts up 7 units.

EXPLANATION

The given function is

[tex]f(x) = {x}^{2} [/tex]

This is the base of the quadratic function without any transformation.

It is also refer to as the parent function.

The transformed function has equation:

[tex]f(x) = {x}^{2} + 7[/tex]

This transformation is of the form

[tex]y = f(x) + k[/tex]

This transformation shifts the graph of the base function up by k units.

Since k=7, the base function is shifted up by 7 units.

The pattern is plus 7

6 + 7 = 13

13 + 7 = 20

20 + 7 = 27

This means to find the next term you must add 7 to 27

27 + 7 = 34

To find the term after 34 add seven to that as well

34 + 7 = 41

so...

6, 13, 20, 27, 34, 41

Hope this helped!

~Just a girl in love with Shawn Mendes

Answer:

[tex]\large\boxed{V_A=1512\ m^3}[/tex]

Step-by-step explanation:

[tex]\text{If a prism A is similar to a prism B with a scale k, then:}\\\\\text{1.\ The ratio of the lengths of the corresponding edges is equal to the scale k}\\\\\dfrac{a}{b}=k\\\\\text{2. The ratio of the surface area of the prisms is equal}\\\text{to the square of the scale k}\\\\\dfrac{S.A._A}{S.A._B}=k^2\\\\\text{3. The ratio of the prism volume is equal to the cube of the scale k}\\\\\dfrac{V_A}{V_B}=k^3[/tex]

[tex]\text{We have}\\\\k=6:5=\dfrac{6}{5}\\\\V_B=875\ m^3\\\\V_A=x\\\\\text{Substitute to 3.}\\\\\dfrac{x}{875}=\left(\dfrac{6}{5}\right)^3\\\\\dfrac{x}{875}=\dfrac{216}{125}\qquad\text{cross multiply}\\\\125x=(875)(216)\qquad\text{divide both sides by 125}\\\\x=\dfrac{(875)(216)}{125}\\\\x=\dfrac{(7)(216)}{1}\\\\x=1512\ m^3[/tex]

Prism A is similar to Prism B with a scale factor of 6:5. If the volume of Prism B is 875 m2. The volume of prism B is 1512 meter square.

Suppose the initial measurement of a figure was x units.

And let the figure is scaled and the new measurement is of y units.

Since the scaling is done by multiplication of some constant, that constant is called the scale factor.

Let that constant be 's'.

Then we have:

[tex]s \times x = y\\s = \dfrac{y}{x}[/tex]

Thus, the scale factor is the ratio of the new measurement to the old measurement.

Prism A is similar to Prism B with a scale factor of 6:5.

If the volume of Prism B is 875 m2, find the volume of Prism A.

scale factor = 6/5

The ratio of the surface area of the prism A to the prism B

A1 / A2 = k^2

The ratio of the prism is equal to the cube of the scale k.

V1 / V2 = k^3

Let x be the volume of Prism A.

x / 875 = (6/5)^2

x / 875 = 216 / 125

x = 875 * 216 / 125

x = 1512

Therefore, the volume of prism B is 1512 meter square.

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Answer:

20

Step-by-step explanation:

use the Pythagorum  therum (a^2+b^2=c^2

so you get 12^2+16^2=C^2

so that Equtes to 144+256=C^2

400=c^2

so square 400 and you get 20

For this case we have that by definition, the Pythagorean theorem states that:

[tex]c = \sqrt {a ^ 2 + b ^ 2}[/tex]

Where:

c: It is the hypotenuse of the triangle

a, b: Are the legs

According to the figure, the hypotenuse is represented by BC, then:

[tex]BC = \sqrt {12 ^ 2 + 16 ^ 2}\\BC = \sqrt {144 + 256}\\BC = \sqrt {400}\\BC = 20[/tex]

Thus, the hypotenuse of the triangle is 20

ANswer:

Option D

For this case we have the following equation:

[tex]2 (x + 2) = x-4[/tex]

Applying distributive property to the terms within the parentheses on the left side of the equation we have:

[tex]2x + 4 = x-4[/tex]

Subtracting "x" on both sides of the equation we have:

[tex]2x-x + 4 = -4\\x + 4 = -4[/tex]

Subtracting 4 on both sides of the equation we have:

[tex]x = -4-4\\x = -8[/tex]

Answer:

[tex]x = -8[/tex]

Answer:

x = -8

Step-by-step explanation:

We are given that Manuela solved following equation and we are to find its solution:

[tex] 2 ( x + 2 ) = x - 4 [/tex]

Expanding the left side of the equation by multiplying the terms inside the bracket by 2:

[tex]2x+4=x-4[/tex]

Arranging the equation in a way such that like terms are on each side (variables on the left and constants on the right):

[tex]2x-x=-4-4[/tex]

x = -8

Answer:

47.91 units²

Step-by-step explanation:

This can be solved using Heron's triangle (see attached)

in this case, your lengths are

a = 3+9=12

b=3+5=8

c=5+9=14

Hence,.

S = (1/2) x (a + b + c) = (1/2) (12+8+14) = 17

(s - a) = 17 -12 = 5

(s - b) = 9

(s - c) = 3

Area = √ [  s (s-a) (s-b) (s-c) ]

= √ [  17 x 5 x 9 x 3 ] = √2295 = 47.9061 = 47.91 units² (rounded to nearest hundreth)

Answer: The value of x = 1

Step-by-step explanation:

Given : Parallelogram ABCD, diagonals AC and BD intersect at point E.

such that

[tex]AE = 11x -3 \text{ and} CE = 12- 4x.[/tex]

We know that the diagonal of a parallelogram bisects each other.

Therefore , we have the following equation :-

[tex]11x -3= 12- 4x\\\\\Righatrrow11x+4x=12+3\\\\\rightarrow\ 15x=15\\\\\Rightarrow\x=\dfrac{15}{15}=1[/tex]

Hence, the value of x = 1

a = interest rate for first bond.

b = interest rate for second bond.

we know the rates add up to 3%, so a + b = 3.

we also know that investing the same amount hmm say $X gives us the amounts of 360 and 480 respectively.

let's recall that to get a percentage of something we simply [tex]\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}[/tex]

so then, "a percent" of X is just (a/100)X = 360.

and "b percent" of X is just (b/100)X = 480.

[tex]\bf a+b=3\qquad \implies \qquad \boxed{b}=3-a~\hfill \begin{cases} \left( \frac{a}{100} \right)X=360\\\\ \left( \frac{b}{100} \right)X=480 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \left( \cfrac{a}{100} \right)X=360\implies X=\cfrac{360}{~~\frac{a}{100}~~}\implies X=\cfrac{36000}{a} \\\\\\ \left( \cfrac{b}{100} \right)X=480\implies X=\cfrac{480}{~~\frac{b}{100}~~}\implies X=\cfrac{48000}{b} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\bf X=X\qquad thus\qquad \implies \cfrac{36000}{a}=\cfrac{48000}{b}\implies \cfrac{36000}{a}=\cfrac{48000}{\boxed{3-a}} \\\\\\ (3-a)36000=48000a\implies \cfrac{3-a}{a}=\cfrac{48000}{36000}\implies \cfrac{3-a}{a}=\cfrac{4}{3} \\\\\\ 9-3a=4a\implies 9=7a\implies \cfrac{9}{7}=a\implies 1\frac{2}{7}=a\implies \stackrel{\mathbb{LOWER~RATE}}{\blacktriangleright 1.29\approx a \blacktriangleleft}[/tex]

[tex]\bf \stackrel{\textit{since we know that}}{b=3-a}\implies b=3-\cfrac{9}{7}\implies b=\cfrac{12}{7}\implies b=1\frac{5}{7}\implies \blacktriangleright b \approx 1.71 \blacktriangleleft[/tex]

Answer:

Correct answer is "A"

Step-by-step explanation:

It is a tranformation about 90° in anti-clock wise direction

In geometry, transformations are used to move a point or points from one position to another. The transformation of [tex](x,y) \to (-y,x)[/tex] is a 90 degrees rotation about the origin.

Given that:

[tex]A(-1,1) \to A'(-1,-1)[/tex]

[tex]B(1,1) \to B'(-1,1)[/tex]

[tex]C(1,4) \to C'(-4,1)[/tex]

The transformation rule is:

[tex](x,y) \to (-y,x)[/tex]

When a point is rotated through [tex](x,y) \to (-y,x)[/tex]

Such point has undergone a 90 degrees counterclockwise rotation.

Hence, option (a) is correct.

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well, lateral area means only the area of the sides, namely just the four triangular faces.

[tex]\bf \textit{area of an equilateral triangle}\\\\ A=\cfrac{s^2\sqrt{3}}{4}~~ \begin{cases} s=sides\\ \cline{1-1} s=8 \end{cases}\implies A=\cfrac{8^2\sqrt{3}}{4}\implies A=16\sqrt{3} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of the 4 triangles}}{4(16\sqrt{3})}\implies 64\sqrt{3}[/tex]

Answer:

The product is -55x³ - 24x² + 22x - 3

Step-by-step explanation:

* Lets revise how to find the product of trinomial and binomial

- If (ax² ± bx ± c) and (dx ± e) are two binomials, where a , b , c , d , e

  are constant, their product is:

# Multiply (ax²) by (dx) ⇒ 1st term in the trinomial and 1st term in the

  binomial

# Multiply (ax²) by (e) ⇒ 1st term in the trinomial and 2nd term in

  the binomial

# Multiply (bx) by (dx) ⇒ 2nd term the trinomial and 1st term in

  the binomial

# Multiply (bx) by (e) ⇒ 2nd term in the trinomial and 2nd term in

  the binomial

# Multiply (c) by (dx) ⇒ 3rd term in the trinomial and 1st term in

  the binomial

# Multiply (c) by (e) ⇒ 3rd term the trinomial and 2nd term in

  the binomial

# (ax² ± bx ± c)(dx ± e) = adx³ ± aex² ± bdx² ± bex ± cdx ± ce

- Add the terms aex² and bdx² because they are like terms

- Add the terms bex and cdx because they are like terms

* Now lets solve the problem

- There are a trinomial and a binomials (11x² + 7x - 3) and (-5x + 1)

- We can find their product by the way above

∵ (11x²)(-5x) = -55x³ ⇒ 1st term in the trinomial and 1st term in the binomial

∵ (11x²)(1) = 11x² ⇒ 1st term in the trinomial and 2nd term in the binomial

∵ (7x)(-5x) = -35x² ⇒ 2nd term the trinomial and 1st term in the binomial

∵ (7x)(1) = 7x ⇒ 2nd term in the trinomial and 2nd term in the binomial

∵ (-3)(-5x) = 15x ⇒ 3rd term in the trinomial and 1st term in the binomial

∵ (-3)(1) = -3 ⇒ 3rd term the trinomial and 2nd term in the binomial

∴ (11x² + 7x - 3)(-5x + 1) = -55x³ + 11x² - 35x² + 7x + 15x - 3

- Add the like terms ⇒ 11x² - 35x² = -24x²

- Add the like terms ⇒ 7x + 15x = 22x

∴ The product is -55x³ - 24x² + 22x - 3