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3 years ago

Are the graphs of −5y=2x+3 and y=25x+4 parallel, perpendicular, or neither?

1 answer:

6 0

Parallel graphs have the same slope. perpendicular has opposite reciprocal slopes. These equations have slopes of -2/5 and 25 so they are neither.

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Will mark brainliest !!!

Wittaler [7]

Answer:

just mark points it is very easy

5 0

3 years ago

Hey what's the answer ? <img src="https://tex.z-dn.net/?f=%28x%20-%20y%29" id="TexFormula1" title="(x - y)" alt="(x - y)"

Varvara68 [4.7K]

Answer:

 -1

Step-by-step explanation:

here's your solution

=> The value of x = 1

=> The value of y = 2

=> we need to find ( x - y)

=> so, put the value of above variables in it

=> hence, (1 - 2) = - 1

=> - 1 is correct answer

hope it helps

3 0

3 years ago

Need help on this calculus

N76 [4]

$\displaystyle\int\sin^3t\cos^3t\,\mathrm dt$

One thing you could do is to expand either a factor of $\sin^2t$ or $\cos^2t$ , then expand the integrand. I'll do the first.

You have

$\sin^2t=1-\cos^2t$

which means the integral is equivalent to

$\displaystyle\int\sin t(1-\cos^2t)\cos^3t\,\mathrm dt$

Substitute $u=\cos t$ , so that $\mathrm du=-\sin t\,\mathrm dt$ . This makes it so that the integral above can be rewritten in terms of $u$ as

$\displaystyle-\int(1-u^2)u^3\,\mathrm du=\int(u^5-u^3)\,\mathrm du$

Now just use the power rule:

$\displaystyle\int(u^5-u^3)\,\mathrm du=\dfrac16u^6-\dfrac14u^4+C$

Back-substitute to get the antiderivative back in terms of $t$ :

$\dfrac16\cos^6t-\dfrac14\cos^4t+C$

4 0

3 years ago

Read 2 more answers

HELP WILL MARK BRAINIEST!!!

maria [59]

Answer:

23 feet

Step-by-step explanation:

The absolute value of -14 is fourteen.

The absolute value of 23 is twenty-three

23>14

4 0

2 years ago

Find the slope of each line.

REY [17]

The slope is 4/3.
I hope this helps :)

4 0

3 years ago