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Virty [35]
3 years ago
12

Find the area of the shape shown below

Mathematics
1 answer:
ValentinkaMS [17]3 years ago
5 0

Answer:

I don't have a calculator with me, and I'm lazy to take it, but here is how it's done

Step-by-step explanation:

First let's just take it as the shape is a perfect rectangle without any folds. Therefore just take

9 x (3.5 + 2) = 63

Now just count the area of those two folded triangles. Then just take 63 and minus of the area of those two triangles, that's your anwer

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Answer:

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Step-by-step explanation:

8 0
2 years ago
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One side of a rectangle is 3 inches shorter than the other side, and the perimeter is 54 inches. What are the dimensions of the
vitfil [10]

Answer:

2x + 2(x - 3)

Step-by-step explanation:

4 0
3 years ago
A pharmacist wants to mix a 30% saline solution with a 10% saline solution to get 200 mL of a 12% saline solution. How much of e
Bess [88]

Answer:

30% constituents=20 mL

10% constituents=180 mL

Step-by-step explanation:

x= 30% volume

y=10% volume

For our first equation, we know the total volume is 200 mL and is the sum

x+y=200

y=200-x (1)

For our second equation, we do a mass balance for 200 mL of final solution.

12% w/v = 0.12 g/mL

This means that in 1 mL of solution, we have 0.12 g of NaCl.

For any solution, concentration multiplied by volume will give the mass of NaCl:

Mass in x mL= C*V (g/mL) (mL)

So in 200 mL, we have

0.12*200 (g/mL) (mL)

=24g of NaCl

Cx*Vx + Cy*Vy=24

0.3x+0.1y=24 (2)

Substitute y=200-x into (2)

0.3x+0.1(200-x)=24

0.3x+20-0.1x=24

0.2x=24-20

0.2x=4

Divide both sides by 0.2

0.2x/0.2=4/0.2

x=20

Substitute x=20 into (1)

y=200-x

y=200-20

y=180

30% constituents=20 mL

10% constituents=180 mL

6 0
3 years ago
If the volume of the solid below is 490 cubic inches, what is the height of the solid?
aleksklad [387]

Answer:The percent change in volume between cylinders A and B is 50%.

Step-by-step explanation:

7 0
3 years ago
Find the derivative of f(t)=2√t - 2/√t
Semmy [17]
The answer is t=1.

Hope this helps.
8 0
3 years ago
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