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3 years ago

HELP Write an equation of a line passing through (-8, 6) and parallel to the line y=-4x - 1

Mathematics

1 answer:

MArishka [77]3 years ago

4 0

The line is parallel so it has the same slope as the other line, -4.

Substitute the x and y variables with the given points and find the y-intercept.

(-8, 6)

y = -4x + b

6 = -4(8) + b

6 = -32 + b

6 + 32 = b

38 = b

So the equation is:

y = -4 + 38

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Solve this thing pls <img src="https://tex.z-dn.net/?f=3%5Cleft%28q-7%5Cright%29%3D27" id="TexFormula1" title="3\left(q-7\ri

evablogger [386]

Answer:

Solution of equation ( q ) = 16

Step-by-step explanation:

In this question we have given an equation that is 3 ( q - 7 ) = 27 and we have asked to solve this equation that means to find the value of  q .

Solution : -

$\quad \: \longmapsto \: 3(q - 7) = 27$

Step 1 : Solving parenthesis :

$\quad \: \longmapsto \:3q - 21 = 27$

Step 2 : Adding 21 on both sides :

$\quad \: \longmapsto \:3q - \cancel{ 21} + \cancel{21} = 27 + 21$

On further calculations we get :

$\quad \: \longmapsto \:3q = 48$

Step 3 : Dividing by 3 from both sides :

$\quad \: \longmapsto \: \frac{ \cancel{3}q}{ \cancel{3}} = \cancel {\frac{48}{3} }$

On further calculations we get :

$\quad \: \longmapsto \: \pink{\boxed{\frak{q = 16}}}$

Therefore, solution of equation ( q ) is 16 .

Verifying : -

Now we are very our answer by substituting value of q in the given equation . So ,

3 ( q - 7 ) = 27

3 ( 16 - 7 ) = 27

3 ( 9 ) = 27

27 = 27

L.H.S = R.H.S

Hence, Verified.

Therefore, our solution is correct .

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Slope of (27,2),(38,-15)

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I'm in college, I honestly don't know what it is but I hope you find it!

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The 3rd term of (a-b)4 is

fgiga [73]

The third term of the expansion is 6a^2b^2

<h3>How to determine the third term of the expansion?</h3>

The binomial term is given as

(a - b)^4

The r-th term of the expansion is calculated using

r-th term = C(n, r - 1) * x^(n - r + 1) * y^(r - 1)

So, we have

3rd term = C(4, 3 - 1) * (a)^(4 - 3 + 1) * (-b)^(3-1)

Evaluate the sum and the difference

3rd term = C(4, 2) * (a)^2 * (-b)^2

Evaluate the exponents

3rd term = C(4, 2) * a^2b^2

Evaluate the combination expression

3rd term = 6 * a^2b^2

Evaluate the product

3rd term = 6a^2b^2

Hence, the third term of the expansion is 6a^2b^2

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Step-by-step explanation:

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