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N76 [4]
4 years ago
14

Identify the volume and surface area of a sphere in terms of π with a great circle area of 144π ft2.

Mathematics
1 answer:
algol134 years ago
8 0

we have given area of circle=\pi r^2=144\pi

r=12ft

we know that surface area of sphere is 4\pi r^2=4*\pi *(144)=576\pi ft^2

volume of the sphere=\frac{4}{3} \pi r^3=\frac{4}{3} *\pi (12)^3=2304\pi ft^3.

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In a direct variation, y = 18 when x = 6. Write a direct variation equation that
inysia [295]

Step-by-step explanation:

y =kx(where k is constant)

18 = 6k

divide both side by 6

k = 6

The relationship between x and y

y = kx

y =3x

8 0
3 years ago
Rose and Dennis each open a savings account at the same time. Rose invests $2,600 in an account yielding 4.1% simple interest, a
vodka [1.7K]
The formula is
A=p (1+rt)
A future value
P present value
R interest rate
T time in years

Rose investment
A=2,600×(1+0.041×9)
A=3,559.4

Dennis investment
A=2,200×(1+0.057×9)
A=3,328.6

So Rose investment is greater than Dennis investment by
3,559.4−3,328.6=230.8

Hope it helps!
3 0
3 years ago
What is the center of the hyperbola whose equation is (y+3)^2/81-(x-6)^2/89=1?
xxMikexx [17]

We have been given an equation of hyperbola \frac{(y+3)^2}{81}-\frac{(x-6)^2}{89}=1. We are asked to find the center of hyperbola.  

We know that standard equation of a vertical hyperbola is in form \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1, where point (h,k) represents center of hyperbola.

Upon comparing our given equation with standard vertical hyperbola, we can see that the value of h is 6.

To find the value of k, we need to rewrite our equation as:

\frac{(y-(-3))^2}{81}-\frac{(x-6)^2}{89}=1

Now we can see that value of k is -3. Therefore, the vertex of given hyperbola will be at point (6,-3) and option D is the correct choice.

6 0
3 years ago
Find an explicit solution to the Bernoulli equation. y'-1/3 y = 1/3 xe^xln(x)y^-2
NNADVOKAT [17]

y'-\dfrac13y=\dfrac13xe^x\ln x\,y^{-2}

Divide both sides by \dfrac13y^{-2}(x):

3y^2y'-y^3=xe^x\ln x

Substitute v(x)=y(x)^3, so that v'(x)=3y(x)^2y'(x).

v'-v=xe^x\ln x

Multiply both sides by e^{-x}:

e^{-x}v'-e^{-x}v=x\ln x

The left side can be condensed into the derivative of a product.

(e^{-x}v)'=x\ln x

Integrate both sides to get

e^{-x}v=\dfrac12x^2\ln x-\dfrac14x^2+C

Solve for v(x):

v=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x

Solve for y(x):

y^3=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x

\implies\boxed{y(x)=\sqrt[3]{\dfrac14x^2e^x(2\ln x-1)+Ce^x}}

4 0
3 years ago
5x + y = 6 5x + 3y = -4 The y-coordinate of the solution to the system shown is _____. -5 -1 1
anzhelika [568]
5x +y = 6
5x + 3y = -4 
multiply top equation by -1 and bottom by 1
-5x -y = -6
5x+3y = -4
solve and get 
2y = -10
y = -10/2
y = -5 
answer is -5
8 0
3 years ago
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