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saw5 [17]
3 years ago
9

If a rectangle has an area of 36 square units, what could the dimensions be?

Mathematics
1 answer:
Oxana [17]3 years ago
6 0

The area of a rectangle is length times width. Since a square's length is equal to its width, a square's area is equal to the length of one side times itself. So, reversed, the square root of the square's area gives the length of one side. In this case, the square root of 36 square centimeters is 6 centimeters. Again, since all four sides of a square are the same, its perimeter is equal to the length of one side times four. For a square with one side equal to 6 centimeters, the perimeter equals 24 centimeters.

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koban [17]

Answer: It would be 4b+44

Step-by-step explanation:

This is the distributive property! so a(b+c)=ab+ac

In this case–4b+44!

6 0
2 years ago
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Please help Ill give Brainly !!!!
Gwar [14]

Answer:

The first answer is 87 and the second question answer is 229

Step-by-step explanation:

How i got the first question is by subtracting 122 and 35 which got me 87.

How i got the second question is by 18 - 3 = 15 and 244 - 15 equals 229. hope this helps :)

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2 years ago
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How do you find the inverse of a exponet
faust18 [17]
"Inverse of an exponent?"  Unsure of what you mean.  Do you want to find the inverse of an exponential function?

Suppose  y = 2e^x.  Find the inverse function:

1.  Interchange x and y:  x = 2e^y
2.  Solve this for y:  ln x = ln 2 + y, so y = ln x - ln 2, or   y = ln (x/2)  (answer)
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3 years ago
Sam applies a force at 30° to the horizontal to move a football tackling dummy 10 m horizontally.
Basile [38]

Answer:

F=\frac{32\sqrt{3}}{3}\text{\:or\:} \approx 18.48\:\text{N}

Step-by-step explanation:

We can find the horizontal component of force he applied by using the formula for work:

W=F\Delta x

Solving for force, we have:

160=F\cdot 10,\\F=16\:\text{N}

However he applied the original force at an angle of 30^{\circ} to the horizontal. This force of 16 newtons is only the horizontal component of that force. To find the magnitude of the original force, we can use basic trigonometry:

\cos 30^{\circ}=\frac{16}{x},\\x=\frac{16}{\cos 30^{\circ}},\\\\x=\frac{16}{\frac{\sqrt{3}}{2}},\\\\x=16\cdot\frac{2}{\sqrt{3}}=\frac{32}{\sqrt{3}}=\boxed{\frac{32\sqrt{3}}{3}}

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3 years ago
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