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3 years ago

The perimeter of a rectangle is 80 feet.Find the dimensions if the length is 5 feet longer than the width

Mathematics

1 answer:

Airida [17]3 years ago

7 0

I believe it would be 25+25+15+15 = 80ft

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What is 2,621 divided by 37 with remainder

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Answer: 209

Step-by-step explanation:

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3 years ago

The expression πr² is used to determine the area of a circle, where r is the length of the radius. What does the expression π(5b

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8 0

3 years ago

Please help me with this

Drupady [299]

The answer is d 9,891

3 0

3 years ago

Please can someone help me?

Debora [2.8K]

If the perimeters are equal, you can form an equation to find x:
x + x + 12 + x + x + 12 = x + 11 + x + 11 + x + 16
4x + 24 = 3x + 38
- 3x
x + 24 = 38
- 24
x = 14
Now you can substitute the x value into the perimeter:
(4 × 14) + 24
56 + 24 = 80 cm

(3 × 14) + 38
42 + 38 = 80 cm

So the perimeter would be 80 cm.
I hope this helps!

6 0

3 years ago

Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is

$\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6$

4 0

3 years ago