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My name is Ann [436]
3 years ago
5

write an equation in slope-intercept form for the line that satisfies the following condition. passes through (12,-4), perpendic

ular to the graph of y = 7/12x + 23
Mathematics
1 answer:
Oksana_A [137]3 years ago
7 0
Since we are given the equation of the line in which our line of interest is perpendicular with, we can obtain the slope of our line which is basically the negative reciprocal of the slope of the given line. 

m1 = 7/12

therefore

m2 = -12/7

the form of the equation is

(y - y1) = m2 (x - x1)

substituting values 

(y + 4) = (-12/7) (x -12)

y = (-12/7)x + 116/7
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Find a so that the point (-1, 2) is on the graph of f(x)=ax^2+5.
Gnesinka [82]
We are asked to determine the value of a such that the function f(x) = ax^2 + 5 is fit for the point given (-1,2). In this case, we substitute 2 to y and -1 to x. The result is then 2 = a*(-1)^2 + 5 ; 2 = a + 5; a is then equal to -3
4 0
3 years ago
HELP PLEASE..........
serg [7]

Answer:

19.47

Step-by-step explanation:

Use the law of sines.

sin(90)/6 = sin(x)/2

Multiply both sides by 2

2 * sin(90)/6 = sin(x)

Use a calculator

0.33333 = sin(x)

sin^-1 0.3333 = 19.47

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2 years ago
What is the slope of the line
mart [117]
Slope = 4/1 = 4

hope it helps
3 0
3 years ago
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HELP WILL GIVE BRAINLIEST TO THE FIRST ANSWER!!!!<br> is 0=0 no solution or all real numbers???
kenny6666 [7]

Answer:

All real numbers

Step-by-step explanation:

Since 0 does equal 0, and value would be able to work. So any x-value will work. Infinite values for <em>x</em>, resulting in all real numbers.

7 0
3 years ago
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What is the distance between the points (-6,5) and (8,-2) on a coordinate plane?
Naddik [55]

Distance Formula: D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2

Apply the points: D=\sqrt{(8-(-6))^2+((-2)-5)^2

Solve: D=\sqrt{(8-(-6))^2+((-2)-5)^2}\\D=\sqrt{14^2+(-7)^2}\\D=\sqrt{169+49}\\D=\sqrt{218}

Since the square root of 218 cannot be simplified, the answer is

D=\sqrt{218}

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✉️ If any further questions, inbox me! ✉️

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3 years ago
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