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2 years ago

13

A triangular piece of rubber is stretched equally from all sides, without distorting its shape, such that each side of the enlar

ged triangle is twice the length of the original side.
The area of the triangle to times the original area.

2 answers:

3241004551 [841]2 years ago

6 0

<span>The area increases to 2 times the original value. </span>

butalik [34]2 years ago

4 0

Answer:

The area of the new enlarged triangle is 4 times the original triangle.

Step-by-step explanation:

Lets take this triangle to be an equilateral triangle.

Area is given by : $\frac{\sqrt{3} }{4} \times a^{2}$

And let us assume the side to be 7 units.

Area when side is 7 units:

So, area = $\frac{\sqrt{3} }{4} \times(7)^{2}$

= 21.22 square units.

When each side is enlarged to twice. That gives side is 14 units.

So, area = $\frac{\sqrt{3} }{4} \times(14)^{2}$

= 84.87 square units

So, the enlarged area is : $\frac{84.87}{21.22}$ = 3.999≈ 4 times the original area.

Answer: The area of the new enlarged triangle is 4 times the original triangle.

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Drag the cards to create a two-digit number that matches these rules. Note: Not all cards will be used 6 is one of the digits 20

Greeley [361]

Answer:

As i understand this:

We have cards with the values:

0, 1, 2 and 6.

We want to create a two digit number such that:

Not all the cards are used.

6 is one digit of our number.

When we round to the nearest ten, we get 20.

Now, first remember how rounding works.

If we have a number like ab, where a and b are single digits (a is in the tens place and b is in the units place), then:

if b is 5 or more, we round up

if b is less than 5, we round down.

Now in this case we want to have a digit equal 6, if we use this in the units place, then when rounding, we will round up.

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1 year ago

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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elena-14-01-66 [18.8K]

Answer:

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(PT)^2 + (CT)^2 = (PC)^2
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(4)^2 + (4)^2 = (PC)^2
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