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Is the relationship shown by the data linear? If so, model the data with an equation.

Answer:

113.1 cm²

Step-by-step explanation:

V = [tex]\frac{1}{3}[/tex] [tex]\pi[/tex] r² h

h = 12 cm , r = 3 cm , [tex]\pi[/tex] = [tex]\frac{22}{7}[/tex]

V ≈ 113.1 cm²

Answer:

see explanation

Step-by-step explanation:

Given

f(x) = [tex]3^{x}[/tex] , then

f(x + 1) = [tex]3^{x+1}[/tex] = [tex]3^{x}[/tex] × 3

Then

f(x) + f(x + 1)

= [tex]3^{x}[/tex] + 3. [tex]3^{x}[/tex] ← factor out [tex]3^{x}[/tex] from each term

= [tex]3^{x}[/tex] (1 + 3)

= 4. [tex]3^{x}[/tex]

= 4f(x)

Answer: 122 square feet

Explanation:

The larger rectangle has area of 14*15 = 210 square feet.

The smaller rectangle has a horizontal dimension of 15-4 = 11 ft and its vertical dimension is 14-6 = 8 ft. Therefore, the area is 11*8 = 88 square feet.

The non-shaded region has area that is the difference of the two previous areas calculated earlier. So that would be 210-88 = 122 square feet

Answer:

5 minutes

Step-by-step explanation:

5 * $0.99 = $4.95

$\sf\underline\bold{Question:1-}$

A ball is thrown upwards and it goes to the height of 100meter and comes down. What is the net displacement?

$\sf\underline\bold{Solution}$

$\sf{According \: to\:the\: question,}$

Displacement for the above situation is 0. As we know, that displacement is the shortest path from the initial to the final point. Here, the initial and the final points are the same, and henceforth, it takes no time to travel. So the displacement is 0.

$\sf\underline\bold{Question:2-}$

An athlete completes one round of a circular track of diameter 200m in 30 seconds. Find the distance covered by the athlete at the end of 30 second.

$\sf\underline\bold{Solution:}$

$\sf\bold{Given\:parameters:}$

$\sf\small{☆The\:diameter\:of\:the\:circular\:track:200m}$

$\sf{Radius=}$ $\sf\dfrac{200}{2}$ → $\sf\underline{Radius = 100m}$

☆Time taken by an athlete to complete one round : 30 seconds.

$\space$

$\sf\bold{To\:find:}$

❍Distance travelled by an athlete in 30 seconds.

$\space$

❍ AND,Distance travelled by the athlete will be equal to the cumference of the circle.

$\space$

$\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\sf{So,}$

$\mapsto$ $\sf{Circumference\:of\:the\:circle: 2 πr}$

$\space$

$\mapsto$ $\sf{Circumference=2\times}$ $\sf\dfrac{22}{7}$ $\sf{\times 100}$

$\space$

$\mapsto$ Circumference of the circle : $\sf\dfrac{4400}{7}$

$\space$

[tex]\sf\underline\bold{∴Circumference = 628.57m}[/tex]

$\space$

||Therefore,The distance travelled in 30 seconds, by the athlete is 628.57m.||

0 = 1st answer

628.57 m = Question 2 answer.

Answer:

Step-by-step explanation:

a) 9.5 feet above the ground

b)  max height at  of 34.5 feet  5 feet horizontal distance

c) 10.87 feet away

Answer:

[tex] \frac{21}{4} + \frac{35}{8} \\ = \frac{77}{8} \\ = 9 \frac{5}{8} [/tex]

[tex]A = \begin{bmatrix}1&2\\1&1\end{bmatrix} \implies A^{-1} = \dfrac1{\det(A)}\begin{bmatrix}1&-1\\-2&1\end{bmatrix} = \begin{bmatrix}-1&1\\2&-1\end{bmatrix}[/tex]

where det(A) = 1×1 - 2×1 = -1.

[tex]B = \begin{bmatrix}0&-1\\1&2\end{bmatrix} \implies B^{-1} = \dfrac1{\det(B)}\begin{bmatrix}2&1\\-1&0\end{bmatrix} = \begin{bmatrix}2&1\\-1&0\end{bmatrix}[/tex]

where det(B) = 0×2 - (-1)×1 = 1. Then

[tex]B^{-1}A^{-1} = \begin{bmatrix}2&1\\-1&0\end{bmatrix} \begin{bmatrix}-1&1\\2&-1\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}[/tex]

On the other side, we have

[tex]AB = \begin{bmatrix}1&2\\1&1\end{bmatrix} \begin{bmatrix}0&-1\\1&2\end{bmatrix} = \begin{bmatrix}2&3\\1&1\end{bmatrix}[/tex]

and det(AB) = det(A) det(B) = (-1)×1 = -1. So

[tex](AB)^{-1} = \dfrac1{\det(AB)}\begin{bmatrix}1&-3\\-1&2\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}[/tex]

and both matrices are clearly the same.

More generally, we have by definition of inverse,

[tex](AB)(AB)^{-1} = I[/tex]

where [tex]I[/tex] is the identity matrix. Multiply on the left by A ⁻¹ to get

[tex]A^{-1}(AB)(AB)^{-1} = A^{-1}I = A^{-1}[/tex]

Multiplication of matrices is associative, so we can regroup terms as

[tex](A^{-1}A)B(AB)^{-1} = A^{-1} \\\\ B(AB)^{-1} = A^{-1}[/tex]

Now multiply again on the left by B ⁻¹ and do the same thing:

[tex]B^{-1}\left(B(AB)^{-1}\right) = (B^{-1}B)(AB)^{-1} = B^{-1}A^{-1} \\\\ (AB)^{-1} = B^{-1}A^{-1}[/tex]

The question is incomplete, the complete question is;

A factory produces 1,250,000 toys each year. The number of toys is expected to increase by about 150% per year. Which model can be used to find the number of toys being produced, n (in millions), in t years?

A. n= 2.5(1.5)/t, t cannot = 0

B. n= 1.5t^2 + 1.25

C. n= 1.5t + 1.25

D. n= 1.25(2.5^t)

Answer:

D. n= 1.25(2.5^t)

Step-by-step explanation:

This is an exponential growth problem. In mathematics, for an exponential growth problem;

P = Po(1 + r)^t

Where;

P= amount at time t

Po =initial amount

r= rate

t= time

In the context of the question we have;

n= 1.25(1 + 150/100)^t

n= 1.25(2.5)^t

Answer:

>

Step-by-step explanation:

Answer:

the Mode Is 3

Step-by-step explanation:

You Have To Put the numbers from ascending order to descending order..The Numbers that appears the most is the mode

Answer:

The number of pictures changes from 8 to 4.

Step-by-step explanation:

I assume the first question was the number of pictures that could be placed when the pictures are 6 inches wide.

Let's first answer the first question.

Wall length: 9.75 ft

Window: 5¾ ft = 5.75 ft

Available wall space for pictures:

9.75 ft - 5.75 ft = 4 ft

Now we convert 4 ft to inches.

1 ft = 12 inches

4 ft = 4 * 12 inches = 48 inches

There is 48 inches of wall space to place pictures that measure 6 inches each.

48 inches / 6 inches = 8

Originally, there is room for 8 pictures.

Question 4.

Each picture is 10 inches wide.

There is still 48 inches of wall space for the pictures.

48 inches / 10 inches = 4.8

There is room for 4.8 pictures. Since Mr. Hamilton will place only whole picture, he can only place 4 pictures now.

Answer: The number of pictures changes from 8 to 4.

Answer:

D) 9:00 am

Step-by-step explanation:

Because 9:00 is the midpoint of 7 and 12

Answer:

4

Step-by-step explanation:

A=ch

then,

2ft×2ft

4ft

Answer:

C

Step-by-step explanation:

First, let's say 4³ is a and 5⁻² is b. We know that (a/b)ⁿ = aⁿ/bⁿ for any n, so

(a/b)⁵ = a⁵/b⁵

=  (4³)⁵ /(5⁻²)⁵

Next, one power rule states that (4³)⁵ = 4 ⁽³ˣ⁵⁾ = 4¹⁵ and (5⁻²)⁵ = 5 ⁽⁻²ₓ⁵⁾=5⁻¹⁰, so

(4³)⁵ /(5⁻²)⁵ = 4¹⁵ / 5⁻¹⁰

Next, anything to a negative power (e.g. x⁻ⁿ) is equal to 1 over the absolute value of the power, so x⁻ⁿ = 1/xⁿ. Applying that here, we can say that

5⁻¹⁰ = 1/5¹⁰

4¹⁵ / 5⁻¹⁰ = 4¹⁵ / (1/5¹⁰) = (4¹⁵/1) / (1/5¹⁰) = 4¹⁵ * 5¹⁰

Answer:

97/4752

Step-by-step explanation:

So, this problem sounds hard but it is actually simple.

First, add

65/12 + 8/3 = 97/12

Next, subtract

65/12 - 8/3 = 33/12

Then, divide

97/12 ÷ 33/12 = 97/4752

Hope this helps! :)

Answer:

c)3pi/4

Step-by-step explanation:

Answer:

I assume that we want to complete the statement:

"The maximum profit of $__ dollars is reached when __ items are produced"

We know that the profit equation is defined between x = 0 and x = 100, which are the two roots of the equation (so the profit is equal to zero for x = 0 and for x = 100).

Then we can assume that the profit will be positive in this range.

Thus, the quadratic equation should have a negative leading coefficient, which would mean that the arms of the graph go downwards.

If this is the case, we know that the maximum will be at the vertex.

Here we know that the vertex is:

(50, 1000)

Where remember, x represents the number of items and y represents the profit.

So, given that the maximum is at the vertex, and we know that the vertex is (50, 1000) we can conclude that the maximum profit is $1000, and this happens when the number of produced items is 50.

Then the complete statement is:

"The maximum profit of $1000 dollars is reached when 50 items are produced"

Fencing required to go around the whole garden = 368 feet

To calculate the fencing needed, given measures in the question,

Since, area of a rectangular garden is given by the expression,

Area = Length × Width  

4864 = 152 × Width

Width = [tex]\frac{4864}{152}[/tex]

          = 32 feet

Since, length of the fence needed = Perimeter of the garden

And Perimeter of a rectangular garden is given by the expression,

Perimeter = 2(length + width)

By substituting the values of length and width of the garden in the expression,

Perimeter = 2(152 + 32)

                 = 368 feet

       Therefore, 368 feet of the fence will be required to go around the whole garden.

Learn more,

https://brainly.com/question/22073676

Answer:

I HOPE IT WILL HELP

EXPLANATION IS IN THE PHOTO .

Answer:

The polynomial is 2x^3 - 8x^2 - 24x

And we can factor out a 2x from each of the three terms:

2x(x^2 - 4x - 12)

Lastly, factor the remaining quadratic:

2x(x+(-2))(x+6)

And we have our answer:

a=2

b=-2

c=6

Let me know if this helps!

Answer:

a =2, b =2, and c = -6

Step-by-step explanation:

We factor the polynomial and then see which value corresponds to what.

2x^3-8x^2-24x

As we see it, all terms are factorable by 2x. So if we take out 2x from every term, we get

2x(x^2 - 4x - 12)

Now we factor the quadratic, which we can do mentally to get

2x(x+2)(x-6)

ax(x+b)(x+c)

Comparing that to ax(x+b)(x+c), we can tell that a =2, b =2, and c = -6.

Answer:

c = 12

Step-by-step explanation:

i'm Attaching the work

Answer:

[tex]x = 0[/tex] or [tex]x = 1[/tex].

Step-by-step explanation:

Start by adding [tex]1[/tex] to both sides of this equation:

[tex](x - 1) + 1 = (\sqrt{x} - 1) + 1[/tex].

[tex]x = \sqrt{x}[/tex].

If two numbers are equal, their square should also be equal. Therefore, since[tex]x = \sqrt{x}[/tex], it must be true that [tex]x^{2} = (\sqrt{x})^{2}[/tex]. That is: [tex]x^{2} = x[/tex].

Notice that since [tex]x[/tex] is under a square root, the result must ensure that [tex]x \ge 0[/tex].

Subtract [tex]x[/tex] from both sides of the equation:

[tex]x^{2} - x = x - x[/tex].

[tex]x^{2} - x = 0[/tex].

Factor [tex]x[/tex] out:

[tex]x\, (x - 1) = 0[/tex].

Hence, by the Factor Theorem, [tex]x = 0[/tex] and [tex]x = 1[/tex] would satisfy this rearranged equation. Because of the square root in the original equation, these two value must be non-negative ([tex]x \ge 0[/tex]) to qualify as actual roots of that equation.

In this example, both [tex]x = 0[/tex] and [tex]x = 1[/tex] qualify as roots of that equation.

x-1 = \sqrt{x} -1

#BrainliestBunch

Answer:

y = (-3/4)x + 2

Step-by-step explanation:

Find the slope of this line.  Note how the first x-coordinate (-8) becomes 4, a jump of 12, and how the first y-coordinate (8) becomes -1, a decrease of 9.  Then the slope is

m = (change in y) / (change in x) = -9/12 = m = -3/4

Find the y-intercept from this data using the slope-intercept form:

y = mx + b becomes  8 = (-3/4)(-8) + b when x = -8, y = 8 and m = -3/4.

Solving this equation for b, we get:

                                     8 = 6 + b, so that b must be 2.

The desired equation is y = (-3/4)x + 2.

If polygons are similar ratio of sides will be same

[tex]\\ \sf\longmapsto \frac{6}{14} = \frac{3}{x} \\ \\ \sf\longmapsto 6x = 14 \times 3 \\ \\ \sf\longmapsto 6x = 42 \\ \\ \sf\longmapsto x = \frac{42}{6} \\ \\ \sf\longmapsto x = 7[/tex]

Answer:

wea did tha r come from??

Step-by-step explanation:

it is supposed to be d

Answer:

1. Given

2. Vertical angle theorem

3. AB≅DB, EB≅CB, m<ABC≅m<DBE

4. △ABC ≅ △DBE

Hope this helps :D

Answer:

what?

Step-by-step explanation:

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