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3 years ago

11

I don't know how to do these 2

1 answer:

m_a_m_a [10]3 years ago

6 0

52. (x - 2) cause the graph to shift to the right 2 units
So key points from y = -3x - 1 will have 2 added to the x-coordinate
(0, -1) will turn into (2, -1) and (1, -4) will turn into (3, -4)

The correct answer is J.
You could also substitute (x - 2) into the equation y =-3(x - 2) -1 and simplify
y = -3x + 6 - 1
y = -3x + 5
plot that b (5) on the y-axis and drop down 3, move right 1 and plot another point

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denis-greek [22]

Answer:

Part 1) The radius of the circle is r=17 units

Part 2) The points (-15,14) and (-15,-16) lies on this circle

Step-by-step explanation:

step 1

Find the radius of the circle

we know that

The distance between the center of the circle and any point on the circle is equal to the radius of the circle

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

(-7, -1) and (8, 7)

substitute

$r=\sqrt{(7+1)^{2}+(8+7)^{2}}$

$r=\sqrt{(8)^{2}+(15)^{2}}$

$r=\sqrt{289}$

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step 2

Find out the y-coordinate of point (-15,y)

The equation of the circle in standard form is equal to

$(x-h)^2+(y-k)^2=r^2$

where

(h,k) is the center

r is the radius

substitute the values

$(x+7)^2+(y+1)^2=17^2$

$(x+7)^2+(y+1)^2=289$

Substitute the value of x=-15 in the equation

$(-15+7)^2+(y+1)^2=289$

$64+(y+1)^2=289$

$(y+1)^2=289-64$

$(y+1)^2=225$

square root both sides

$(y+1)=(+/-)15$

$y=-1(+/-)15$

$y=-1(+)15=14$

$y=-1(-)15=-16$

therefore

we have two solutions

point (-15,14) and point (-15,-16)

see the attached figure to better understand the problem

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