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Alja [10]
3 years ago
14

Write the quotient 13/5-i in the form a+bi

Mathematics
1 answer:
vichka [17]3 years ago
3 0

Answer: I found: −7−6i. Explanation: You need to get rid of the i at the denominator first. To do that you can multiply and divide by the complex ...

1 answer

Step-by-step explanation:

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A used car is priced at $2,695. If you borrow the money for the car, your payments will be $122 a month for 30 months. How much
Studentka2010 [4]

Answer: $965

Step-by-step explanation:

The used car is priced at $2,695.

If you borrow the money for the car, your payments will be $122 a month for 30 months. This means that the total amount of money that you would have paid at the end of 30 months at a rate of $122 per month is the amount paid per month multiplied by the total number of months. It becomes

Total payment = 122×30 = $3660

This means that you ended up paying higher than you would have paid if you paid cash.

Amount that you would have saved = amount paid over 30 months - cost of the car

Amount that you would have saved

= 3660 - 2695 = $965

7 0
3 years ago
Help<br> What is the slope of the line on the graph?
Galina-37 [17]
Slope = (Y2-Y1) / (X2-X1)
hope that's helpful
6 0
3 years ago
What is the sum of the two expressions -2.9a+6.8 and 4.4a-7.3
Lady bird [3.3K]

Answer:

1.5a - 0.5

Step-by-step explanation:

Add like terms

4.4 - 2.9 = 1.5

7.3 - 6.8 = 0.5

4 0
3 years ago
Between what two numbers is the exact product of 379 and 8
ANEK [815]
"Product" means to multiply, so you have to multiply 379 by 8 to get 3,032; it's between the numbers 3,031 and 3,033
8 0
3 years ago
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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