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Dominik [7]
3 years ago
14

What is the equation of the line in point-slope form?

Mathematics
2 answers:
Novosadov [1.4K]3 years ago
7 0

Answer:

y + 4 =1/2 (x + 4)

Step-by-step explanation:

Your welcome uwu

Alisiya [41]3 years ago
6 0

Answer: y=-2x+2/4

Step-by-step explanation:

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Which is a right triangle?<br> 10, 13,8<br> 7, 9, 16<br> 4. 8. 12<br> 36, 27, 45
geniusboy [140]

Answer:

get a piece of graphing paper and count out the squares and then when you make the triangles you should have your answer.

Step-by-step explanation:

all i know is that it's not the third one hope this helps!

8 0
3 years ago
Read 2 more answers
WILL GIVE BRAINLIEST need help asap
Aleonysh [2.5K]

The location of the y value of R' after using the translation rule is -10

<h3>What will be the location of the y value of R' after using the translation rule? </h3>

The translation rule is given as:

(x + 4, y - 7)

The pre-image of R is located at (-17, -3)

Rewrite as

R = (-17, -3)

When the translation rule is applied, we have:

R' = (-17 + 4, -3 - 7)

Evaluate

R' = (-13, -10)

Remove the x coordinate

R'y = -10

Hence, the location of the y value of R' after using the translation rule is -10

Read more about translation at:

brainly.com/question/26238840

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2 years ago
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2 years ago
I need help with a question that has to do with Ratio and Proportion. I want to know how to do it, not just only the answer.
Tema [17]

Answer:

y is = 15

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7 0
3 years ago
Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt
polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt

=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k

The first integral is trivial since (\tan t)'=\sec^2t.

The second can be done by substituting u=t^2-1:

u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C

The third can be found by integrating by parts:

u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t

\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8

\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C

8 0
3 years ago
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