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2 years ago

The area of a rectangle with width x and length 5x is 5x2. What does the coefficient 5 mean in terms of the problem?

Mathematics

1 answer:

ludmilkaskok [199]2 years ago

8 0

Answer:

The length is 5 times the width

Step-by-step explanation:

Let

W ----> the width of the rectangle

L ----> the length of the rectangle

we know that

The area of rectangle is equal to

$A=LW$

In this problem we have

$L=5x$ ----> equation A

$W=x$ ----> equation B

substitute equation B in equation A

$L=5W$

therefore

The length is 5 times the width

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What is the solution to the equation below?

MakcuM [25]

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2 years ago

Read 2 more answers

!SUBSTITUTION METHOD!<br><br>6y - 5z = -8 <br>3z = -6 <br>4x - 3y - 2z= 21

hichkok12 [17]

Answer:

Step-by-step explanation:

6y - 5z = -8 .......... Equation 1

3z = -6 ................... Equation 2

4x - 3y - 2z= 21...... Equation 3

<u>First solve for z in Equation 2</u>

That's

3z = - 6

Divide both sides by 3

Next substitute the value of z into Equation 1 in order to find y

We have

6y - 5(-2) = - 8

6y + 10 = - 8

6y = - 8 - 10

6y = - 18

Divide both sides by 6

Finally substitute the values of y and z into Equation 3 to find the value of x

That's

4x - 3(-3) - 2(-2) = 21

4x + 9 + 4 = 21

4x + 13 = 21

4x = 21 - 13

4x = 8

Divide both sides by 4

So the solutions are

Hope this helps you

3 0

3 years ago

Rewrite the following integral in spherical coordinates.

lora16 [44]

In cylindrical coordinates, we have $r^2=x^2+y^2$ , so that

$z = \pm \sqrt{2-r^2} = \pm \sqrt{2-x^2-y^2}$

correspond to the upper and lower halves of a sphere with radius $\sqrt2$ . In spherical coordinates, this sphere is $\rho=\sqrt2$ .

$1 \le r \le \sqrt2$ means our region is between two cylinders with radius 1 and $\sqrt2$ . In spherical coordinates, the inner cylinder has equation

$x^2+y^2 = 1 \implies \rho^2\cos^2(\theta) \sin^2(\phi) + \rho^2\sin^2(\theta) \sin^2(\phi) = \rho^2 \sin^2(\phi) = 1 \\\\ \implies \rho^2 = \csc^2(\phi) \\\\ \implies \rho = \csc(\phi)$

This cylinder meets the sphere when

$x^2 + y^2 + z^2 = 1 + z^2 = 2 \implies z^2 = 1 \\\\ \implies \rho^2 \cos^2(\phi) = 1 \\\\ \implies \rho^2 = \sec^2(\phi) \\\\ \implies \rho = \sec(\phi)$

which occurs at

$\csc(\phi) = \sec(\phi) \implies \tan(\phi) = 1 \implies \phi = \dfrac\pi4+n\pi$

where $n\in\Bbb Z$ . Then $\frac\pi4\le\phi\le\frac{3\pi}4$ .

The volume element transforms to

$dx\,dy\,dz = r\,dr\,d\theta\,dz = \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi$

Putting everything together, we have

$\displaystyle \int_0^{2\pi} \int_1^{\sqrt2} \int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta = \boxed{\int_0^{2\pi} \int_{\pi/4}^{3\pi/4} \int_{\csc(\phi)}^{\sqrt2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta} = \frac{4\pi}3$

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Answer:

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3 years ago