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1 year ago

5

Box 1: Dimensions: x by 3x by x3 Area of base = x(3x) = 3x2 Box 2: Dimensions: x by 4x – 1 by x3 Area of base = x(4x – 1) = 4x2

– x Complete the statements about the number of terms in the polynomial representing the volume of each box. Box 1's volume will be a Box 2's volume will be a . Explain your reasoning.

1 answer:

e-lub [12.9K]1 year ago

8 0

Answer:

box 1 is a monomial

box 2 is a binomial

Step-by-step explanation:

reasoning is a monomial times a monomial remains a monomial and a binomial times a monomial becomes a binomial when it cant be farther simplified. (if you want to add the answers for box 1 is 3x^5 and the answer for box 2 is 4x^5-x^4)

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Answer:

the answer is D

Step-by-step explanation:

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2 years ago

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A conical pile of road salt has a diameter of 112 feet and a slant height of 65 feet. After a storm, the linear dimensions of th

QveST [7]

we know that

the volume of a cone is equal to

$V= \frac{1}{3} \pi r^{2}h$

in this problem

the radius is equal to

$r= \frac{112}{2}= 56ft$

1) <u>Find the height of the cone before the storm</u>

Applying the Pythagorean Theorem find the height

$h^{2} = l^{2}-r^{2}$

$l=65 ft$

$h^{2} = 65^{2}-56^{2}$

$h^{2} = 1,089$

$h=33 ft$

2) <u>Find the volume before the storm</u>

$V= \frac{1}{3}*\pi* 56^{2}*33$

$V=34,496\pi\ ft^{3}$

3) <u>Find the volume after the storm</u>

After a storm, the linear dimensions of the pile are 1/3 of the original dimensions

so

r=(56/3) ft

h=(33/3)=11 ft

$V= \frac{1}{3}*\pi* (56/3)^{2}*11$

$V= 1,277.63\pi\ ft^{3}$

<u>4) Find how this change affect the volume of the pile</u>

Divide the volume after the storm by the volume before the storm

$\frac{1,277.63 \pi }{34,496 \pi } = \frac{1}{27}$

therefore

<u>the answer part a) is</u>

The volume of the pile after the storm is $\frac{1}{27}$ times the original volume

<u>Part b)</u> Estimate the number of lane miles that were covered with salt

5) <u>Find the amount of salt that was used during the storm</u>

$=34,496 \pi - 1,277.63 \pi \\= 33.218.37 \pi \\= 104,358.59\ ft^{3}$

6) <u>Find the pounds of road salt used</u>

$104,358.59*80=8,348,687.2\ pounds$

7) <u>Find the number of lane miles that were covered with salt</u>

$8,348,687.2/350=23,853.39 \ lane\ miles$

therefore

<u>the answer part b) is</u>

the number of lane miles that were covered with salt is $23,853.39 \ lane\ miles$

<u>Part c) </u>How many lane miles can be covered with the remaining salt? Round your answer to the nearest lane mile

the remaining salt is equal to $1,277.63\pi\ ft^{3}$

$1,277.63\pi\ ft^{3}=4,013.79\ ft^{3}$

8) <u>Find the pounds of road salt </u>

$4,013.79*80=321,103.20\ pounds$

9) <u>Find the number of lane miles </u>

$321,103.20/350=917.44 \ lane\ miles$

therefore

<u>the answer part c) is</u>

the number of lane miles is $917 \ lane\ miles$

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2 years ago

In June 2012, Twitter was reporting 400 million tweets per day. Each tweet can consist of up to 140 characters (letter, numbers,

gulaghasi [49]

Each tweet would cost between 1 and 140 pennies you can write that as an inequality [1 ≤ x ≤ 140] Since you can't tweet 0 characters or above 140.

if you want to change it to pounds divide by 100.

A) 1 ≤ x ≤ 140

B) Each character costs a penny, you can't tweet less than 1 character or above 140.

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1 year ago

The coordinates of the midpoint of line GH are M(−132,−6) and the coordinates of one endpoint are G(−4, 1). The coordinates of t

Darina [25.2K]

Answer:

Since, the coordinates of the midpoint of line GH are M(\frac{-13}{2}, -6)(

2

−13

,−6) .

The coordinates of endpoint G are (-4,1)

We have to determine the coordinates of endpoint H.

The midpoint of the line segment joining the points (x_1, y_1)(x

1

,y

1

) and (x_2, y_2)(x

2

,y

2

) is given by the formula (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})(

2

x

1

+x

2

,

2

y

1

+y

2

) .

Here, The endpoint G is (-4,1) So, x_1 = -4 , y_1=1x

1

=−4,y

1

=1

Let the endpoint H be (x_2,y_2)(x

2

,y

2

)

The midpoint coordinate M is (\frac{-13}{2}, -6)(

2

−13

,−6) .

So, \frac{-13}{2} = \frac{-4+x_2}{2}

2

−13

=

2

−4+x

2

{-13} = {-4+x_2}−13=−4+x

2

{-13}+4 = {x_2}−13+4=x

2

{x_2}=9x

2

=9

Now, -6 = \frac{1+y_2}{2}−6=

2

1+y

2

-12 = {1+y_2}−12=1+y

2

y_2= -13y

2

=−13

So, the other endpoint H is (-9,-13).

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Effectus [21]

The answer i got is B

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