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trasher [3.6K]
3 years ago
9

what is an equation of the line that is perpendicular to y = - 3/4 x + 6 and passes through point (3, 9)?​

Mathematics
1 answer:
nika2105 [10]3 years ago
4 0

Answer:

y=-3/4x +11.25

Step-by-step explanation:

You keep the -3/4x the same because it is also your slope so that the graphs never touch. But to get it to cross the point (3,9) you need to move the graph up. Also known as shifting the graph. You would change the y-intercept and follow the constant slope to check whether the point is on the graph. You do it a couple times with different y-intercepts to find the correct graph. :3

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PLEASE ANSWER QUICKLY
Rudiy27

Answer:

3

Step-by-step explanation:

f(x) = 3x² - 2x + 1

f(x) = ax² + bx + c

a = 3

7 0
2 years ago
Is 1/4 bigger than 0
Helga [31]
Yes, it is bigger than 0
3 0
3 years ago
What is the equation of a line with a slope of -2 that passes through the point (6, 8)?
iren [92.7K]

Answer:

2x + y = 20

Step-by-step explanation:

gradient = -2

x = 6, y = 8

y - 8 = -2(x - 6)

y - 8 = -2x + 12

2x + y = 12 + 8

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6 0
3 years ago
Read 2 more answers
Please someone help me on this <33
dangina [55]

Answer:

B

Step-by-step explanation:

The initial value for a function is the value of y when x=0. For function A, we can see from the table that when x = 0, we have y = 4. For function B, we plug in 0 in our equation and solve: y = 3(0) + 5, or y = 5. So, because function B has a larger value of y when x = 0, function B has the greater initial value because the initial value for Function A is 4 and the initial value for function B is 5.

7 0
3 years ago
Given that sin theta = 1/4, 0→theta→π/2, what<br>. what is the exact value of cos 8?​
natali 33 [55]
<h3>Answer: Choice B \frac{\sqrt{15}}{4}</h3>

===========================================================

Work Shown:

Angle theta is between 0 and pi/2, so this angle is in quadrant Q1.

Square both sides of the given equation

\sin \theta = \frac{1}{4}\\\\\sin^2 \theta = \left(\frac{1}{4}\right)^2\\\\\sin^2 \theta = \frac{1}{16}

Then use the pythagorean trig identity to get

\sin^2 \theta + \cos^2 \theta = 1\\\\\cos^2 \theta = 1-\sin^2 \theta\\\\\cos \theta = \sqrt{1-\sin^2 \theta} \ \ \ \text{cosine is positive in Q1}\\\\\cos \theta = \sqrt{1-\frac{1}{16}}\\\\\cos \theta = \sqrt{\frac{16}{16}-\frac{1}{16}}\\\\\cos \theta = \sqrt{\frac{16-1}{16}}\\\\\cos \theta = \sqrt{\frac{15}{16}}\\\\\cos \theta = \frac{\sqrt{15}}{\sqrt{16}}\\\\\cos \theta = \frac{\sqrt{15}}{4}\\\\

3 0
3 years ago
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