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gladu [14]
2 years ago
15

Can you help with this? Length x width x height

Mathematics
2 answers:
lyudmila [28]2 years ago
6 0

Answer:

So first you want to multiply the first two fractions (on the first shape) 5/8 x 6/8= 15/32 and then you multiply that by your second number which

is 5/8. 5/8 x 15/32 is  75/256in^3

(And on the second shape) 4/5 x 4/5 is 16/25. 16/25 x 4/5 which is 64/125in^3

Gemiola [76]2 years ago
5 0

Answer:

75/256 in³    <    64/125 in³

Step-by-step explanation:

First, find the volumes of the two 3-D figures:

For Figure 1:

Using the volume formula:

V = l × w × h

   = 6/8 × 5/8 × 5/8

   = (6/8 × 5/8) × 5/8

   = 30/64 × 5/8

V = 150/512

Simplify:

V = 75/256

So Figure 1's volume is 75/256 in³.

For Figure 2:

V = l × w × h

   = 4/5 × 4/5 × 4/5

   = (4/5 × 4/5) × 4/5

   = 16/25 × 4/5

   = 64/125

So Figure 2's volume is 64/125 in³.

Now, comparing the two volumes:

You can cross multiply the fractions to figure out which fraction is greater:

Figure 1                       Figure 2

<u> 75 </u>                =                <u> 64 </u>

256                                  125

75 (125)          =             256 (64)

9375               =             16384

So Figure 2 has the greater volume.

Your final result should be:

75/256 in³    <    64/125 in³

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The vertices of ∆ABC are A(-2, 2), B(6, 2), and C(0, 8). The perimeter of ∆ABC is units is? What is the area?
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we have

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see the attached figure to better understand the problem

we know that

The perimeter of the triangle is equal to

P=AB+BC+AC

and

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in this problem

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we know that

The distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

Step 1

<u>Find the distance AB</u>

A(-2, 2),B(6, 2)

Substitute the values in the formula

d=\sqrt{(2-2)^{2}+(6+2)^{2}}

d=\sqrt{(0)^{2}+(8)^{2}}

dAB=8\ units

Step 2

<u>Find the distance BC</u>

B(6, 2),C(0, 8)

Substitute the values in the formula

d=\sqrt{(8-2)^{2}+(0-6)^{2}}

d=\sqrt{(6)^{2}+(-6)^{2}}

dBC=6\sqrt{2}\ units

Step 3

<u>Find the distance AC</u>

A(-2, 2),C(0, 8)

Substitute the values in the formula

d=\sqrt{(8-2)^{2}+(0+2)^{2}}

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dAC=2\sqrt{10}\ units

Step 4

<u>Find the distance DC</u>

D(0, 2),C(0, 8)

Substitute the values in the formula

d=\sqrt{(8-2)^{2}+(0-0)^{2}}

d=\sqrt{(6)^{2}+(0)^{2}}

dDC=6\ units

Step 5

<u>Find the perimeter of the triangle</u>

P=AB+BC+AC

substitute the values

P=8\ units+6\sqrt{2}\ units+2\sqrt{10}\ units

P=22.81\ units

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The perimeter of the triangle is equal to 22.81\ units

Step 6

<u>Find the area of the triangle</u>

A=\frac{1}{2}*base *heigth

in this problem

base=AB=8\ units\\heigth=DC=6\ units

substitute the values

A=\frac{1}{2}*8*6

A=24\ units^{2}

therefore

the area of the triangle is 24\ units^{2}

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