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stealth61 [152]

4 years ago

6

The length of a rectangle is four times its width. If the perimeter is at most 100 centimeters, what is the greatest possible va

lue for the width?

1 answer:

PIT_PIT [208]4 years ago

8 0

P=2(L+W)
100≥P

L is 4 times W
L=4W
sub 4W for L

100≥2(4W+W)
100≥2(5W)
100≥10W
divide both sides by 10
10≥W

the greatest possible width is 10cm

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2 years ago

Solve for y. -2y=3x+10

patriot [66]

Answer:

y= -(3/2)x -5

Step-by-step explanation:

Divide both sides by negative 2

-2y=3x+10

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y= -(3/2)x -5

4 0

3 years ago

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Sindrei [870]

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4 years ago

Danielle construct a scale model of a building with a rectangular base her model is 2 inches in length and 1 inch in width the s

Kitty [74]

Given:

a.) The scale of the model is 1 in equals 47 ft.

b.) Danielle constructs a scale model of a building with a rectangular base her model is 2 inches in length and 1 inch in width.

To be able to determine the actual area of the base, let's first identify the actual dimensions of the base.

We get,

$\text{Length: 2 (inches) x }\frac{47\text{ ft.}}{1\text{ (inch)}}\text{ = 2 x }\frac{47\text{ ft.}}{1}\text{ = 94 ft.}$ $\text{Width: 1 (inch) x }\frac{47\text{ ft.}}{1\text{ (inch)}}\text{ = 1 x }\frac{47\text{ ft.}}{1}\text{ = 47 ft.}$

Let's now solve for the area of the base, since it is a rectangle, we will be using the formula below:

$\text{ Area = L x W ; where L = Length and W = Width}$

We get,

$\text{ Area = L x W}$ $\text{ = 94 ft. x 47 ft.}$ $\text{ Area = 4,418 ft.}^2$

Therefore, the actual area of the base is 4,418 ft.².

7 0

1 year ago

Examplelt: A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the wall

harina [27]

Answer :

4167 bricks.

Explanation :

Since the wall with all its bricks makes up the space occupied by it, we need to find the volume of the wall, which is nothing but a cuboid.

Here,

${\qquad \dashrightarrow{ \sf{Length=10 \: m=1000 \: cm}}}$

$\qquad \dashrightarrow{ \sf{Thickness=24 \: cm}}$

$\qquad \dashrightarrow{ \sf{Height=4 m=400 \: cm}}$

Therefore,

${\qquad \dashrightarrow{ \bf{Volume \: of \: the \: wall = length \times breadth \times height}}}$

${\qquad \dashrightarrow{ \sf{Volume \: of \: the \: wall = 1000 \times 24 \times 400 \: {cm}^{3} }}}$

Now, each brick is a cuboid with Length = 24 cm, Breadth = 12 cm and height = 8 cm.

So,

${\qquad \dashrightarrow{ \bf{Volume \: of \: each \: brick = length \times breadth \times height}}}$

${\qquad \dashrightarrow{ \sf{Volume \: of \: each \: brick = 24 \times 12 \times 8 \: {cm}^{3} }}}$

So,

${\qquad \dashrightarrow{ \bf{Volume \: of \: bricks \: required = \dfrac{volume \: of \: the \: wall}{volume \: of \: each \: brick} }}}$

${\qquad \dashrightarrow{ \sf{Volume \: of \: bricks \: required = \dfrac{1000 \times 24 \times 400}{24 \times 13 \times 8} }}}$

${\qquad \dashrightarrow{ \sf{Volume \: of \: bricks \: required = \bf \: 4166.6} }}$

Therefore,

<u>The wall requires 4167 bricks. </u>

6 0

2 years ago