Y = mx + b
slope(m) = -3
(2,7)...x = 7 and y = 2
now we sub and fund b, the y int
2 = -3(7) + b
2 = -21 + b
2 + 21 = b
23 = b <== ur y int
Answer & Step-by-step explanation:
i = 2
b = 150
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The equations:
i= b-148
or
b= i+148
or
b-i = 148
I am not really sure which equation you want, but I hope this helps!!!
I will use the letter x instead of theta.
Then the problem is, given sec(x) + tan(x) = P, show that
sin(x) = [P^2 - 1] / [P^2 + 1]
I am going to take a non regular path.
First, develop a little the left side of the first equation:
sec(x) + tan(x) = 1 / cos(x) + sin(x) / cos(x) = [1 + sin(x)] / cos(x)
and that is equal to P.
Second, develop the rigth side of the second equation:
[p^2 - 1] / [p^2 + 1] =
= [ { [1 + sin(x)] / cos(x) }^2 - 1] / [ { [1 + sin(x)] / cos(x)}^2 +1 ] =
= { [1 + sin(x)]^2 - [cos(x)]^2 } / { [1 + sin(x)]^2 + [cos(x)]^2 } =
= {1 + 2sin(x) + [sin(x)^2] - [cos(x)^2] } / {1 + 2sin(x) + [sin(x)^2] + [cos(x)^2] }
= {2sin(x) + [sin(x)]^2 + [sin(x)]^2 } / { 1 + 2 sin(x) + 1} =
= {2sin(x) + 2 [sin(x)]^2 } / {2 + 2sin(x)} = {2sin(x) ( 1 + sin(x)} / {2(1+sin(x)} =
= sin(x)
Then, working with the first equation, we have proved that [p^2 - 1] / [p^2 + 1] = sin(x), the second equation.
Answer: D. (-2, -1)
Step-by-step explanation:
Here we do two reflections to the point (-1, 2).
First, we do a reflection over the line x = y. Remember that a reflection over a line keeps constant the distance between our point and the given line, so we have that for a pint (x, y), the reflection over the line y = x is:
Ry=x (x, y) = (y, x)
so for our point, we have:
Ry=x (-1, 2) = (2, -1)
Now we do a reflection over the y-axis, again, a reflection over a line keeps constant the distance between our point and the given line, so if we have a point (x,y) and we do a reflection over the y-axis, our new point will be:
Ry-axis (x,y) = (-x, y)
Then in our case:
Ry-axis (2, -1) = (-2, -1)
The correct option is D.
<h3>
Answer: 3x - 4y = 40</h3>
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Work Shown:
y = (3/4)x - 10
4y = 3x - 40 .... multiply everything by 4 to clear out the fraction
4y+40 = 3x
40 = 3x-4y
3x-4y = 40
This is in the form Ax+By = C
A = 3, B = -4, C = 40
This is not the only solution possible since we can multiply both sides by some number to scale the equation.