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3 years ago

write each of the following in the form of a x + B Y + C is equal to zero and find the values of a b and c

Mathematics

1 answer:

Alex_Xolod [135]3 years ago

4 0

Step-by-step explanation:

(i) 2x = 5

compare with ax+by+c = 0

→ 2x + 0.y - 5 = 0

» a = 2 ,b = 0 ,c = -5

(ii) y - 2 = 0

compare with ax + by + c = 0

→ 0.x + 1y -2 = 0

» a = 0 , b = 1 , c = -2

(iii) y/7 = 3

y = 3×7

y = 21

y - 21 = 0

compare with ax + by + c = 0

→ 0.x + 1y -21 = 0

» a = 0 , b = 1 , c = -21

(iv). x = -14/13

13x = -14

13x + 14 = 0

compare with ax + by + c = 0

→ 13x + 0.y + 14 = 0

» a = 13 , b = 0 , c = 14

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Explain the circumstances for which the interquartile range is the preferred measure of dispersion

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3 years ago

Type the correct answer in each box. A circle is centered at the point (5, -4) and passes through the point (-3, 2). The equatio

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Answer:

$(x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}$

Step-by-step explanation:

Given:

Center of circle is at (5, -4).

A point on the circle is $(x_1,y_1)=(-3, 2)$

Equation of a circle with center $(h,k)$ and radius 'r' is given as:

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

Here, $(h,k)=(5,-4)$

Radius of a circle is equal to the distance of point on the circle from the center of the circle and is given using the distance formula for square of the distance as:

$r^2=(h-x_1)^2+(k-y_1)^2$

Using distance formula for the points (5, -4) and (-3, 2), we get

$r^2=(5-(-3))^2+(-4-2)^2\\r^2=(5+3)^2+(-6)^2\\r^2=8^2+6^2\\r^2=64+36=100$

Therefore, the equation of the circle is:

$(x-5)^2+(y-(-4))^2=100\\(x-5)^2+(y+4)^2=100$

Now, rewriting it in the form asked in the question, we get

$(x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}$

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3 years ago

write each of the following in the form of a x + B Y + C is equal to zero and find the values of a b and c ​

write each of the following in the form of a x + B Y + C is equal to zero and find the values of a b and c