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3 years ago

What is the y-intercept when a line has a slope of 3 and passes through the point (2, 7)?

Mathematics

1 answer:

dybincka [34]3 years ago

4 0

Answer:

The y-intercept is (0,1)

Step-by-step explanation:

So, first you need to find the equation! Point slope form is (y-y1)=m(x-x1).

Plug the numbers in and you get (y-7)=3(x-2). Now, the y intercept is where x=0, so plug in zero for x.

y-7=3(0-2)

y-7=3(-2)

y-7= -6.

-6+7= 1, so y=1

The ordered pair is then (0,1). Hope this was useful :)

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If x is the midpoint of pq, write the relation between px and pq as an inequality

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3 years ago

Circle P is described by the equation (x−1)2+(y+6)2=9 and circle Q is described by the equation (x+4)2+(y+14)2=4. Select from th

xeze [42]

Answer:

(a) Circle Q is 9.4 units to the center of circle P

(b) Circle Q has a smaller radius

Step-by-step explanation:

Given

$P:(x - 1)^2 + (y + 6)^2 = 9$

$Q:(x + 4)^2 + (y + 14)^2 = 4$

Solving (a): The distance between both

The equation of a circle is:

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

Where

$Center: (h,k)$

$Radius:r$

P and Q can be rewritten as:

$P:(x - 1)^2 + (y + 6)^2 = 3^2$

$Q:(x + 4)^2 + (y + 14)^2 = 2^2$

So, for P:

$Center = (1,-6)$

$r = 3$

For Q:

$Center = (-4,-14)$

$r = 2$

The distance between them is:

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$

Where:

$Center = (1,-6)$ --- $(x_1,y_1)$

$Center = (-4,-14)$ --- $(x_2,y_2)$

So:

$d = \sqrt{(1 - -4)^2 + (-6 - -14)^2$

$d = \sqrt{(5)^2 + (8)^2$

$d = \sqrt{25 + 64$

$d = \sqrt{89$

$d = 9.4$

Solving (b): The radius;

In (a), we have:

$r = 3$ --- circle P

$r = 2$ --- circle Q

By comparison

$2 < 3$

<em>Hence, circle Q has a smaller radius</em>

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