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ss7ja [257]
3 years ago
14

If HI is not parallel to JK, what is m

Mathematics
1 answer:
dsp733 years ago
5 0

Answer:

no coco concho kicks sub chick lux 8jzjxjfxktx

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Read 2 more answers
Find the equation of a circle with a center at (–7, –1) where a point on the circle is (–4, 3).
gogolik [260]

Answer:

(x + 7)^2 + (y + 1)^2 = 25.

Step-by-step explanation:

The center of a circle is easy to set up. According to the formula below, the formula for the circle will be (x - a)^2 + (y - b)^2 = r^2.

In this case, a = -7 and b = -1, so we have...

(x - (-7))^2 + (y - (-1))^2 = r^2

(x + 7)^2 + (y + 1)^2 = r^2

To get the radius, we need to find the distance between the center and the point on the circle. The distance formula is d = sqrt((x2 - x1)^2 + (y2 - y1)^2).

In this case, x2 = -4, x1 = -7, y2 = 3, and y1 = -1.

sqrt((-4 - -7)^2 + (3 - -1)^2) = sqrt((-4 + 7)^2 + (3 + 1)^2) = sqrt((3)^2 + (4)^2) = sqrt(9 + 16) = sqrt(25) = plus or minus 5.

Since distance can only be positive, the distance is 5 units, meaning that the radius is 5 units.

5^2 = 25

So, your equation should be (x + 7)^2 + (y + 1)^2 = 25.

Hope this helps!

8 0
3 years ago
Please anyone answer me
ollegr [7]

Let's divide the shaded region into two areas:

area 1: x = 0 ---> x = 2

ares 2: x = 2 ---> x = 4

In area 1, we need to find the area under g(x) = x and in area 2, we need to find the area between g(x) = x and f(x) = (x - 2)^2. Now let's set up the integrals needed to find the areas.

Area 1:

A\frac{}{1}  = ∫g(x)dx =  ∫xdx =  \frac{1}{2}  {x}^{2}  | \frac{2}{0}  = 2

Area 2:

A\frac{}{2}  = ∫(g(x) - f(x))dx

= ∫(x -  {(x - 2)}^{2} )dx

=  ∫( - {x}^{2}  + 5x - 4)dx

= ( - \frac{1}{3}{x}^{3} +   \frac{5}{2} {x}^{2}  - 4x)   | \frac{4}{2}

= 2.67 - ( - 0.67) = 3.34

Therefore, the area of the shaded portion of the graph is

A = A1 + A2 = 5.34

3 0
3 years ago
What is 14/26 simplified?
Reika [66]
14/26 simplified is 7/13
4 0
3 years ago
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