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Mashcka [7]
3 years ago
11

How do you get the diameter of a circle from the circumference?

Mathematics
1 answer:
stiv31 [10]3 years ago
5 0
Because the circumference of a circle is the diameter times 3.14, all you have to do to get the diamater is reverse it and divide the circumference by 3.14
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arah mailed her computer to Store B for repairs. Her bill was: Total Cost of Computer Repair = $1,444.50 Shipping = 2 percent of
nataly862011 [7]
1415.61 IS HER TOTAL BILL BEFORE SHIPPING. 1415.61X .15=212.34 

1415.61 REPAIR
28.89 SHIPPING
212.34 TIP
1656.84 TOTAL
7 0
3 years ago
Read 2 more answers
What is 4 over 8+ 2 over 8+ 2 over 4?
Lostsunrise [7]
You need LCD or the least common denominator. For 4 over 8 and 2 over 8 have the same denominators except for 2 over 4. You need to ask yourself what times 4 equals to 8? 4 times 2 equals to 8, so multiply 4x2 for the denominator and for the numerator multiply 2x2. Whatever you do to the denominator you have to do the same thing to the numerator. You multiplied 4x2 for the denominator and 2x2 for the numerator and now your fraction is 4 over 8. Now add 4 over 8+ 4 over 8 which is 8 over 8 or 1. Now add 1+ 2 over 8 which is 1 2/8. If you simplify 1 2/8 it will become 1 1/4.

Hope that helped:D
3 0
3 years ago
Read 2 more answers
Calculate the area of a rectangle if its dimensions are listed below:
Marizza181 [45]

Answer:

Length=10x

Breadth=7x

Area=10x×7x

=70x²

3 0
2 years ago
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On a family trip, natalie counted 3 groups of 5 motocycles and additonal 7 solo motocycles. write an expression for the number o
timama [110]
3×5=× would be the answer
4 0
2 years ago
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Solve the given initial-value problem. x^2y'' + xy' + y = 0, y(1) = 1, y'(1) = 8
Kitty [74]
Substitute z=\ln x, so that

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dz}\cdot\dfrac{\mathrm dz}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)

Then the ODE becomes


x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}+x\dfrac{\mathrm dy}{\mathrm dx}+y=0\implies\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)+\dfrac{\mathrm dy}{\mathrm dz}+y=0
\implies\dfrac{\mathrm d^2y}{\mathrm dz^2}+y=0

which has the characteristic equation r^2+1=0 with roots at r=\pm i. This means the characteristic solution for y(z) is

y_C(z)=C_1\cos z+C_2\sin z

and in terms of y(x), this is

y_C(x)=C_1\cos(\ln x)+C_2\sin(\ln x)

From the given initial conditions, we find

y(1)=1\implies 1=C_1\cos0+C_2\sin0\implies C_1=1
y'(1)=8\implies 8=-C_1\dfrac{\sin0}1+C_2\dfrac{\cos0}1\implies C_2=8

so the particular solution to the IVP is

y(x)=\cos(\ln x)+8\sin(\ln x)
4 0
3 years ago
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