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labwork [276]
3 years ago
5

14) write the equation to the line parallel to y=3x-5 that goes through the point (-2,-7) using any form

Mathematics
1 answer:
xeze [42]3 years ago
3 0

Answer:

The equation of this line would be y = 3x - 1

Step-by-step explanation:

In order to find this equation we must first find the slope of the original line. The original slope (the coefficient of x) is 3, which means the new slope will also be 3 because parallel lines have the same slope. Now, we can use this slope along with the point in point-slope form to find the equation of the line.

y - y1 = m(x - x1)

y + 7 = 3(x + 2)

y + 7 = 3x + 6

y = 3x - 1


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In circle L with KNM= 53, find the angle measure of minor arc KM
Anestetic [448]

Answer:

106

Step-by-step explanation:

8 0
3 years ago
Find an equation of a line that goes through the points (1,6) and (4, -2). Write your answer in the form y=mx + y.
zmey [24]

Answer:

The desired equation is y = (-8/3)x + 26/3.

Step-by-step explanation:

Moving from (1,6) to (4, -2) involves an increase of 3 in x and a decrease of 8 in y.  Thus, the slope of the line thru these two points is m = rise / run = -8/3.

Using the slope-intercept form of the eq'n of a straight line  and inserting the data given (slope = m = -8/3, x = 4,  y = -2), we get:

y = mx + b => -2 = (-8/3)(4) + b, or  -2 = -32/3 + b

Multiply all terms by 3 to clear out the fraction:

-6 = -32 + 3b.

Then 26 = 3b, and b = 26/3.

The desired equation is y = (-8/3)x + 26/3.

8 0
2 years ago
A N S W E R Q U I C K P L E A S E
chubhunter [2.5K]

Answer:

1. A

2. D

3. D

Step-by-step explanation:

The standard form of a parabola is

y=\frac{1}{4p}(x-h)^2+k            ..... (1)

Where, (h,k) is vertex, (h,k+p) is focus and y=k-p is directrix.

1. The directrix of a parabola is y=−8 . The focus of the parabola is (−2,−6) .

k-p=-8                   ...(a)

(h,k+p)=(-2,-6)

k+p=-6            .... (b)

h=-2

On solving (a) and (b),  we get k=-7 and p=1.

Put h=-2, k=-7 and p=1 in equation (1).

y=\frac{1}{4(1)}(x-(-2))^2+(-7)

y=\frac{1}{4}(x+2)^2-7

Therefore option A is correct.

2 The directrix of a parabola is the line y=5 . The focus of the parabola is (2,1) .

k-p=5                   ...(c)

(h,k+p)=(2,1)

k+p=1            .... (d)

h=2

On solving (c) and (d),  we get k=3 and p=-2.

Put h=2, k=3 and p=-2 in equation (1).

y=\frac{1}{4(-2)}(x-(2))^2+(3)

y=-\frac{1}{8}(x-2)^2+3

Therefore option D is correct.

3. The focus of a parabola is (0,−2) . The directrix of the parabola is the line y=−3 .

k-p=-3                   ...(e)

(h,k+p)=(0,-2)

k+p=-2            .... (f)

h=0

On solving (e) and (f),  we get k=-2.5 and p=0.5.

Put h=0, k=-2.5 and p=0.5 in equation (1).

y=\frac{1}{4(0.5)}(x-(0))^2+(-2.5)

y=\frac{1}{2}(x)^2-2.5

y=\frac{1}{2}(x)^2-\frac{5}{2}

Therefore option D is correct.

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2 years ago
The vertex of the function y= 32x-4x^2
mafiozo [28]

I'm very sure the answer would be (4, 64)

7 0
2 years ago
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