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3 years ago

Don’t know where to start

Mathematics

2 answers:

Sonbull [250]3 years ago

8 0

Answer:

y = -3x; the const. of prop. is -3.

Step-by-step explanation:

Recall: a proportional relationship satisfies y = kx, where k is the constant of proportionality.

In this particular example we see that (-3)(-2) = 6, (-3)(1) = -3, and (-3)(2) = -6.

This confirms adherence to the equation of a proportional relationship. The constant of proportionality here is -3.

The function governing the table is y = -3x.

NemiM [27]3 years ago

7 0

This might help but I say in my mind what letter came first in the alphabet and that’s the letter I go with first

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Solve for x.<br><br> 2x+3y=20<br> -2x+y=4

Mama L [17]

Answer:

x = 1

Step-by-step explanation:

-2x + y = 4

y = 2x + 4

2x + 3y = 20

2x + 3(2x + 4) = 20

2x + 6x + 12 = 20

8x + 12 = 20

8x = 8

x = 1

3 0

2 years ago

Find the volume of the solid generated when R (shaded region) is revolved about the given line. x=6−3sec y, x=6, y= π 3, and

Dmitrij [34]

Answer:

$V=9\pi\sqrt{3}$

Step-by-step explanation:

In order to solve this problem we must start by graphing the given function and finding the differential area we will use to set our integral up. (See attached picture).

The formula we will use for this problem is the following:

$V=\int\limits^b_a {\pi r^{2}} \, dy$

where:

$r=6-(6-3 sec(y))$

$r=3 sec(y)$

a=0

$b=\frac{\pi}{3}$

so the volume becomes:

$V=\int\limits^\frac{\pi}{3}_0 {\pi (3 sec(y))^{2}} \, dy$

This can be simplified to:

$V=\int\limits^\frac{\pi}{3}_0 {9\pi sec^{2}(y)} \, dy$

and the integral can be rewritten like this:

$V=9\pi\int\limits^\frac{\pi}{3}_0 {sec^{2}(y)} \, dy$

which is a standard integral so we solve it to:

$V=9\pi[tan y]\limits^\frac{\pi}{3}_0$

so we get:

$V=9\pi[tan \frac{\pi}{3} - tan 0]$

which yields:

$V=9\pi\sqrt{3}$ ]

6 0

3 years ago

Help me pleasee<br> Questions r in picture

jok3333 [9.3K]

<h2>Answer:</h2>

$\large \bf\implies\frac{12}{61}$

<h2>Step-by-step explanation:</h2>

<h2>Given :</h2>

$\tt \frac{8}{101} , \frac{9}{91} , \frac{10}{81} , \frac{11}{71} ...$

nth term

<h2>Solution :</h2>

We have to add 1 in numerator and -10 in denominator because

$\tt \frac{8}{101} , \frac{9}{91} , \frac{10}{81} , \frac{11}{71} ...[Given]$

$\frac{8 \: + \: 1}{101 \: - \: 10} = \frac{9}{91} \\\\ \frac{9 + 1}{91 - 10} = \frac{10}{81} \\ \\ \frac{10 + 1}{81 - 10} = \frac{11}{71} \\ \\ \frac{11 + 1}{71 - 10} = \frac{12}{61} ...$

The difference is 1 in numerator so we add 1 and the difference is -10 in denominator so we subtract -10.

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Leto [7]

Answer:

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Step-by-step explanation:

A dilation has center (0,0). Find the image of the point L (-5,0) for the scale factor 4.

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Answer:

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Step-by-step explanation:

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