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Setler79 [48]
3 years ago
11

I NEED THIS ASAP- Point A is located at (-5, -8). Point A is reflected across the X-axis to create point B. What is point B?

Mathematics
1 answer:
Rasek [7]3 years ago
6 0

Answer:

(5,-8)

Step-by-step explanation:

if it is reflected across the X axis and this is in quadrant 3 that means if it was reflected it would fall in quadrant 2

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Find principal of a loan at 11% for 5 years with $426.25 simple interest
Ahat [919]

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P=775

Step-by-step explanation:

The Simple Interest Equation is A = P(1 + rt)

where

A = Total Accrued Amount (principal + interest)

P = Principal Amount

I = Interest Amount

t = Time Period involved in months or years

In this case, we do not know the values ​​of the equation (A and P), but we know the amount of interest accrued

If we define our principal whit this formula, we are able to know the rest of the values:

A-P= interest

clearing

A=interest+P

replacing

426.25 + P = P (1+(0.11(5))

Solving

426.25+P=1.55P

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-0.55P=-426.65

P=\frac{-426.25}{-0.55}

P=775

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If a point is inside a circle, the distance from the center of the circle to that point ____.
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A. is less than the radius.

Step-by-step explanation:

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Surface area of storage box that measures 15 centimeters by 12 centimeters by 10 centimeters?
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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)!}}

4 0
2 years ago
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