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2 years ago

Which graph represents a line with a slope of -2/3 and a y-intercept equal to that of the line y = 2/3x – 2?

Mathematics

1 answer:

Aleks04 [339]2 years ago

8 0

Answer:

graph 1

Step-by-step explanation:

Graph using table of values

x y

2 -2/3

1 -4/3

0 -2

they are all on that graph

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Help me with this. i am not smart ..

marta [7]

I’m not that smart... but let me use first as example.. 240 divided by 4

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3 years ago

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Hey y’all I don’t understand, can you help me pls ASAP and show steps if u could

Shkiper50 [21]

Answer:

(x, y) = (-2, -6)

Step-by-step explanation:

Learn more at brainly.com/question/15168004

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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

A. 34cm^3<br> b. 1560cm^3<br> c. 104cm^3<br> d 676cm^3

Vanyuwa [196]

Answer:

676 cubic cm choice d

Step-by-step explanation:

Base area times height gets volume

Base area = (1/2)*13 * 8 = 52 sq. cm (area of the right triangle)

height = 13 cm here

Volume = 13 cm * 52 sq. cm = 676 cubic cm

4 0

3 years ago

Find the term indecent of x in the expansion of (x^2-1/x)^6

Mars2501 [29]

By the binomial theorem,

$\displaystyle \left(x^2-\frac1x\right)^6 = \sum_{k=0}^6 \binom 6k (x^2)^{6-k} \left(-\frac1x\right)^k = \sum_{k=0}^6 \binom 6k (-1)^k x^{12-3k}$

I assume you meant to say "independent", not "indecent", meaning we're looking for the constant term in the expansion. This happens for k such that

12 - 3k = 0 ===> 3k = 12 ===> k = 4

which corresponds to the constant coefficient

$\dbinom 64 (-1)^4 = \dfrac{6!}{4!(6-4)!} = \boxed{15}$

3 0

2 years ago