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lara31 [8.8K]
3 years ago
9

What is the slope of a line that contains the points (-8,7) and (-4,-5)?

Mathematics
1 answer:
Pani-rosa [81]3 years ago
7 0

Answer:

-3

Step-by-step explanation:

To find the slope given two points

m = (y2-y1)/(x2-x1)

   = (-5-7)/(-4- -8)

  = (-12)/(-4+8)

  =-12/4

  = -3

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Ocean waves move in parallel lines toward the shore. The figure shows the path that a windsurfer takes
Minchanka [31]

Answer:

nopw

Step-by-step explanation:

4 0
2 years ago
​(6x2−2x+16​)+​(4x2+7x−55​)
AysviL [449]
(6 • 2 - 2x + 16) + (4 • 2 + 7x - 55)
(-2x + 16 + 6 • 2) + (7x - 55 + 4 • 2)
(-2x + 16 + 12) + (7x - 55 + 8)
-2x + 28 + 7x - 47
-2x + 7x + 28 - 47
5x + (-19)
5x - 19
So the answer is 5x - 19.

8 0
3 years ago
I need help with 8x+3-10=-2 (x-2)+3
astraxan [27]

0 ≠ 4. The equation is wrong, and has no true solution.

8x + 3 - 10x = -2(x - 2) + 3

<em><u>Distributive property.</u></em>

8x + 3 - 10x = -2x + 4 + 3

<em><u>Combine like terms.</u></em>

-2x + 3 = -2x + 7

<em><u>Cancel like terms.</u></em>

<em><u>Subtract 3 from both sides.</u></em>

0  ≠ 4

5 0
3 years ago
4 days I have 140 copies of a book, 9 days I only have 50 left. Write an equation where c(t)=my+b where c is number of copies st
alexandr402 [8]

Answer:

c(t)=-18t+212.

Step-by-step explanation:

We have been given that 4 days I have 140 copies of a book, 9 days I only have 50 left. We are asked to write an equation c(t)=my+b, where c is number of copies still on hand and t days being available.

We have two points on line (4,140) and (9,50). Now, we will use these points to find slope as:

m=\frac{140-50}{4-9}

m=\frac{90}{-5}

m=-18

Now, we will use point-slope form of equation y-y_1=m(x-x_1) and substitute m=-18 and coordinates of point (9,50) as:

y-50=-18(x-9)

y-50=-18x+162

y-50+50=-18x+162+50

y=-18x+212

Therefore, our required equation would be c(t)=-18t+212.

5 0
3 years ago
Prove the trigonometric identity
Annette [7]

Answer:

Proved See below

Step-by-step explanation:

Man this one is a world of its own :D Just a quick question are you a fellow Add Math student in O levels i remember this question from back in the day :D Anyhow Lets get started

For this question we need to know the following identities:

1+tan^{2}x=sec^2x\\\\1+cot^2x=cosec^2x\\\\sin^2x+cos^2x=1

Lets solve the bottom most part first:

1-\frac{1}{1-sec^2x} \\\\

Take LCM

1-\frac{1}{1-sec^2x} \\\\\frac{1-sec^2x-1}{1-sec^2x} \\\\\frac{-sec^2x}{1-sec^2x} \\\\\frac{-(1+tan^2x)}{-tan^2x}

now break the LCM

\frac{-1}{-tan^2x}+\frac{-tan^2x}{-tan^2x}\\\\\frac{1}{tan^2x}+1\\\\cot^2x+1

because 1/tan = cot x

and furthermore,

cot^2x+1\\cosec^2x

now we solve the above part and replace the bottom most part that we solved with cosec^2x

\frac{1}{1-\frac{1}{cosec^2x} } \\\\\frac{1}{1-sin^2x} \\\\\frac{1}{cos^2x}\\\\sec^2x

Hence proved! :D

4 0
3 years ago
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