A diameter of a circle has endpoints p(-10,-2) and Q(4,6)

Mathematics

1 answer:

kvasek [131]2 years ago

6 0

Answer:

a) center: (-3, 2)

b) radius: √65

c) equation: (x +3)² +(y -2)² = 65

Step-by-step explanation:

a) The center (point A) is the midpoint of the diameter, so its coordinates are the average of the endpoint coordinates:

A = (P +Q)/2 = ((-10, -2) +(4, 6))/2

= (-10+4, -2+6)/2 = (-6, 4)/2

A = (-3, 2)

b) The radius is the distance from the center to one end of the diameter. The distance formula can be used to find that.

r = √((x2 -x1)² +(y2 -y1)²) = √((4-(-3))² +(6 -2)²) = √(49+16)

r = √65

c) The circle centered at (h, k) with radius r has formula ...

(x -h)² +(y -k)² = r²

So the formula for this circle is ...

(x +3)² +(y -2)² = 65

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20 Answer: a₁ = 4

Step-by-step explanation:

$a_n=324,\ r=3,\ n=5\\\\a_n=a_1 \cdot r^{n-1}\\\\324=a_1\cdot 3^{5-1}\\\\\dfrac{324}{3^4}=a_1\\\\\dfrac{324}{81}=a_1\\\\\large\boxed{4}=a_1$

21 Answer: n = 13

Step-by-step explanation:

$a_n=\dfrac{1}{64},\ a_1=64,\ r=\dfrac{1}{2}\\\\a_n=a_1 \cdot r^{n-1}\\\\\dfrac{1}{64}=64\cdot \bigg(\dfrac{1}{2}\bigg)^{n-1}\\\\\dfrac{1}{64\cdot 64}=\bigg(\dfrac{1}{2}\bigg)^{n-1}\\\\\dfrac{1}{2^6\cdot 2^6}=\dfrac{1}{2^{n-1}}\\\\6+6=n-1\\\\\large\boxed{13}=n$

22 Answer: n = 5

Step-by-step explanation:

$a_n=48,\ a_1=1875\ r=\dfrac{2}{5}\\\\a_n=a_1 \cdot r^{n-1}\\\\48=1875\cdot \bigg(\dfrac{2}{5}\bigg)^{n-1}\\\\\dfrac{48}{1875}=\bigg(\dfrac{2}{5}\bigg)^{n-1}\\\\\dfrac{16}{625}=\bigg(\dfrac{2}{5}\bigg)^{n-1}}\\\\\bigg(\dfrac{2}{5}\bigg)^4=\bigg(\dfrac{2}{5}\bigg)^{n-1}}\\\\4=n-1\\\\\large\boxed{5}=n$