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Ahat [919]
4 years ago
14

a 2,000 dollar loan has an annual interest rate of 4.5% on the amount borrowed. How much time has elapsed if the interest is now

270 dollars.
Mathematics
1 answer:
lesya [120]4 years ago
6 0
Love to help but dont understand
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Find a third degree polynomial function of the lowest degree that has the zeros below and whose leading coefficient is one.
sineoko [7]
Zeros are the x-values that make the function zero.
0 = (x-0)(x- -4)(x-3)

f(x) = x(x+4)(x-3)
5 0
4 years ago
Which graph represents the linear function y=-2x+3
klasskru [66]
..............
.........

5 0
4 years ago
Scenario:
KiRa [710]

Answer:

do you still need help with this?

Step-by-step explanation:

5 0
3 years ago
Two lines in a coordinate plane have no points of intersection. Which pair could be the equations of the lines? A. 3x-2y=6, 6x-4
dem82 [27]

Answer:

A.

Step-by-step explanation:

if to re-write all the equations:

A. y=3/2 x-3; y=3/2 x-3; B. y=-7x+10; y=3/2 x+3/2; C. y=x+5; y=-3/4 x+2; D. y=x+1; y= -3/4 x+8.

According to the re-written view the correct answer is 'A'.

4 0
3 years ago
Solve this differential equation using power series and indicial roots about (0,0):
Ivanshal [37]
Let y=\displaystyle\sum_{k\ge0}a_kx^k, so that

y'=\displaystyle\sum_{k\ge1}ka_kx^{k-1}
y''=\displaystyle\sum_{k\ge2}k(k-1)a_kx^{k-2}

Substituting into the ODE gives

\displaystyle3x\sum_{k\ge2}k(k-1)a_kx^{k-2}+6\sum_{k\ge1}ka_kx^{k-1}+\sum_{k\ge0}a_kx^k=0
\displaystyle3\sum_{k\ge2}k(k-1)a_kx^{k-1}+6\sum_{k\ge1}ka_kx^{k-1}+\sum_{k\ge0}a_kx^k=0

The first series starts with a linear term, while the other two start with a constant. Extract the first term from each of the latter two series:

\displaystyle6\sum_{k\ge1}ka_kx^{k-1}=6\sum_{k\ge2}ka_kx^{k-1}+6a_1
\displaystyle\sum_{k\ge0}a_kx^k=\sum_{k\ge1}a_kx^k+a_0

Finally, to get the series to start at the same index, shift the index of the first two series by replacing k with k+1. Then the ODE becomes

\displaystyle3\sum_{k\ge1}k(k+1)a_{k+1}x^k+6\sum_{k\ge1}(k+1)a_{k+1}x^k+\sum_{k\ge1}a_kx^k+6a_1+a_0=0

which can be consolidated to get

\displaystyle\sum_{k\ge1}\bigg[(3k(k+1)+6(k+1))a_{k+1}+a_k\bigg]x^k+6a_1+a_0=0
\displaystyle\sum_{k\ge1}\bigg[3(k+1)(k+2)a_{k+1}+a_k\bigg]x^k+6a_1+a_0=0

You're fixing the solution so that it contains the origin, which means

y(0)=\displaystyle\sum_{k\ge0}a_kx^k=a_0=0

which in turn means a_1=0. With the given recurrence, it follows that a_k=0 for all k\ge2, so the solution would be y=0. This is to be expected, since x=0 is clearly a singular point for the ODE.
8 0
4 years ago
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