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3 years ago

9

Which of the following is the correct point-slope equation for the line that passes through the point (-4,-2) and is parallel to

the line given by y=5x+44?

2 answers:

SVEN [57.7K]3 years ago

8 0

Answer:

D

Step-by-step explanation:

drek231 [11]3 years ago

8 0

Answer:

Step-by-step explanation:

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If the length of the hypotenuse of a right triangle is 39 millimeters and the length of one leg is 15 millimeters what is the le

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Answer:

36 millimeters

Step-by-step explanation:

From Pythagoras theorem, the square of the hypotenuse is equal to the sum of the square of the two other legs

In mathematical terms;

a^2 = b^2 + c^2

Let a represent the hypotenuse = 39 mm and the length of one leg, say c, is 15 mm

Slotting in the values of a and b

39^2 = b^2 + 15^2

1521 = b^2 + 225

collect like terms

1521 - 225 = b^2

1296 = b^2

Take the square root of both sides

36 = b

Therefore b = 36 mm

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2 years ago

Find the equation of the circle with a diameter whose end points are (-1,-2) and (-3,2)

Sergeeva-Olga [200]

Answer:

The equation of the circle with a diameter whose end points are (-1,-2) and (-3,2) is

$x^{2} +y^{2}+4x-1=0$

Step-by-step explanation:

<u>Explanation:</u>-

<u>Step 1:</u>-

The equation of the circle having center and radius is

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

here center is (h,k) and radius is r

Given diameter whose end points are (-1,-2) and (-3,2)

The diameter of the circle is passing through the center of the circle

so center of the circle = midpoint of two end points

$(\frac{-1 +(-3) }{2} ,\frac{-2+2 }{2} )$

$(-2,0)$

therefore center (h,k) = (-2,0)

<u>Step 2:-</u>

we have to find the radius of the circle

the radius of the circle = the distance from center to the one end point

i.e., C P = r

Given one end point is P(-3,2) and center C(-2,0)

The distance formula of two points are

$\sqrt{(x_{2}-x_{1} ) ^{2}+ (y_{2}-y_{1} ) ^{2}}$

$r=\sqrt{{(-3)-(-1) ) ^{2}+ (2-(-2)) ^{2}}$

$r=\sqrt{5}$

<u>Step 3</u>:-

center (h,k) = (-2,0) and

radius $r=\sqrt{5}$

The standard form of circle equation

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

$(x-(-2))^2+(y-0)^2=\sqrt{5} ^2$

on simplification is

$x^{2} +y^{2}+4 x-1=0$

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