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3 years ago

Maria invested $450 in a savings account and earned $126 in interest at the end of 7 years.

1 answer:

5 0

The interest rate is B. 4%

Given:
Principal = 450
Interest = 126
term = 7
rate = ?

Simple Interest is computed by multiplying the Principal, rate, and term.

Interest = P × r × t

126 = 450 × r × 7
126 = 3150r
r = 126/3150
r = 0.04 x 100% = 4%

To check:

I = 450 × 0.04 x 7
I = 18 × 7
I = 126

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If triangle ABC is dilated by a scale factor of 2 with a dilation center of A, what will be the coordinates of point b’?

mojhsa [17]

Answer:

In the given figure the sides of triangle measures as follows:

AB= 4 units,

BC= 6 units,

Since triangle ABC is right angled triangle, to find AC we will have to use Pythagorean theorem,

AB² + BC² = AC²

Plugging the values of AB and BC to find AC,

4² + 6² = AC²

16+36=AC²

AC = 7.48

Now if the triangle is dilated by a scale factor of 2, each side will be multiplied by 2 to get the new triangle A'B'C'

side A'B' = 2* AB = 2*4= 8 units

side B'C' = 2* BC = 2*6 =12 units

side A'C' = 2*AC = 2*7.48 = 14.96 units

Perimeter of triangle = sum of three sides

Perimeter of triangle A'B'C' = A'B' + B'C' + A'C'= 8+12 + 14.96 = 34.96 units.

Perimeter of triangle ABC = AB+BC + AC = 4+6+7.48 = 17.48 units.

Perimeter of triangle A'B'C' = 2* Perimeter of triangle ABC

The perimeter of new triangle A'B'C' is 34.96 units which is twice that of triangle ABC.

Answer : A) The perimeter of A'B'C' is 2 times the perimeter of ABC.

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y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

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