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1 year ago

I need help with this math i am not very good at it

Mathematics

1 answer:

Gwar [14]1 year ago

6 0

Answer:

44 sq. in.

Step-by-step explanation:

Area (total) = (1/2)(3)(4) + 10(3) + (1/2)(8)(2)

= 6 + 30 + 8

= 44 sq. in.

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Complete the equation of the line through (3, -1) and (4, 7)

nikitadnepr [17]

Answer:

y = 8x -25

Step-by-step explanation:

The two-point form of the equation of a line can be used.

y = (y2 -y1)/(x2 -x1)(x -x1) +y1

y = (7-(-1))/(4-3)(x -3) +(-1)

y = 8(x -3) -1

y = 8x -25 . . . . slope-intercept form of the equation

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We don't know what you're starting with, so we don't know how to "complete the equation." Perhaps the above equation is sufficient for the task.

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3 years ago

To find the slope of the two points

S_A_V [24]

Answer:

slope is -7/5x-4

Step-by-step explanation:

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3 years ago

A 5 pound bag of apples costs 4.50 and an 8 pound bag of the same type of apples costs 7.52 greg found the unit price which is t

irga5000 [103]

Answer: 0.04 per pound

Step-by-step explanation:

A 5 pound bag of apples costs 4.50.

The unit price will be 4.50 divided by 5 which will be:

4.50 ÷ 5 = 0.9

The unit price is 0.9 per pound

A 8 pound bag of the same type of apples costs 7.52.

The unit price will be 7.52 divided by 8 which will give:

7.52 ÷ 8 = 0.94

The unit price is 0.94 per pound

The difference in the unit prices will be the subtraction of 0.90 from 0.94.

0.94 - 0.90 = 0.04

The difference in the unit prices is 0.04 per pound.

6 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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2 years ago

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