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Ierofanga [76]
2 years ago
8

3n+18= n² solve for n

Mathematics
1 answer:
Dmitriy789 [7]2 years ago
7 0
It =6
i hope this helped you
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Find an equation of the circle that satisfies the given conditions. (Give your answer in terms of x and y.) Center at the origin
Elanso [62]

Answer:

The equation of circle is x^2+y^2=65.

Step-by-step explanation:

It is given that the circle passes through the point (8,1) and center at the origin.

The distance between any point and the circle and center is called radius. it means radius of the given circle is the distance between (0,0) and (8,1).

Distance formula:

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

Using distance formula the radius of circle is

r=\sqrt{\left(8-0\right)^2+\left(1-0\right)^2}=\sqrt{65}

The standard form of a circle is

(x-h)^2+(y-k)^2=r^2         .... (1)

where, (h,k) is center and r is radius.

The center of the circle is (0,0). So h=0 and k=0.

Substitute h=0, k=0 and r=\sqrt{65} in equation (1).

(x-0)^2+(y-0)^2=(\sqrt{65})^2

x^2+y^2=65

Therefore the equation of circle is x^2+y^2=65.

5 0
3 years ago
Math question:<br> solve for x.<br> -7.8(x + 6.5) = -25.74
BlackZzzverrR [31]
My calculator says bad expression.

7 0
3 years ago
There are 90 girls and 60 boys Participating in a race. The coach separates them into two equal groups made of all girls or or b
Lady_Fox [76]
90/2 is 45.. 60/2 is 30.. 30+45=75... 75 students would be able to be in each of the 2 groups, each containing 45 girls and 30 boys ... Hope I helped (:
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3 years ago
Please help me asap
Nitella [24]

Answer:

noestiendo

Step-by-step explanation:

3 0
2 years ago
Find the area lying outside r=6cos(theta) and inside r=3+3cos(theta)
Sloan [31]
The area bounded by the 2 parabolas is A(θ) = 1/2∫(r₂²- r₁²).dθ between limits θ = a,b... 

<span>the limits are solution to 3cosθ = 1+cosθ the points of intersection of curves. </span>
<span>2cosθ = 1 => θ = ±π/3 </span>

<span>A(θ) = 1/2∫(r₂²- r₁²).dθ = 1/2∫(3cosθ)² - (1+cosθ)².dθ </span>
<span>= 1/2∫(3cosθ)².dθ - 1/2∫(1+cosθ)².dθ </span>
<span>= 9/8[2θ + sin(2θ)] - 1/8[6θ + 8sinθ +sin(2θ)] .. </span>
<span>.............where I have used ∫(cosθ)².dθ=1/4[2θ + sin(2θ)] </span>
<span>= 3θ/2 +sin(2θ) - sin(θ) </span>

<span>Area = A(π/3) - A(-π/3) </span>
<span>= 3π/6 + sin(2π/3) -sin(π/3) - (-3π/6) - sin(-2π/3) + sin(-π/3) </span>
<span>= π.</span>
7 0
3 years ago
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