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3 years ago

9

What are the solutions to the equation y2 – 1 = 48?

2 answers:

Sati [7]3 years ago

8 0

The answer would 23.5 as the final answer

Fynjy0 [20]3 years ago

3 0

I got +/- 7 for this one.............

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What is the ordered pair for point A in this coordinate plane?

Alecsey [184]

Answer:

the answer is A.

Step-by-step explanation:

left 7 (-7)

down 5 (-5)

the coordinates are(-7,-5)

4 0

4 years ago

(03.06 MC)The volume of two spheres is shown: Sphere A = 40 cubic inches Sphere B = 10 cubic inches How many inches more is the

dem82 [27]

0.8 inches

the volume (V) of a sphere is calculated using the formula

V = $\frac{4}{3}$ πr³

change the subject and express in terms of r

multiply both sides by 3

3V = 4πr³ ( divide both sides by 4π )

r³ = 3V / 4π ( take the cube root of both sides )

r = ∛( 3V /4π )

sphere A with V= 40

r = ∛ (120 / 12.56 ) = 2.122

sphere B with V = 10

r = ∛(30 / 12.56 ) = 1.337

sphere A - sphere B = 2.122 - 1.337 = 0.8 inches ( to nearest tenth )

6 0

3 years ago

Read 2 more answers

What percent of 8x is5x

statuscvo [17]

62.5% what type of math question is this??

6 0

4 years ago

Read 2 more answers

Is <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B%20-%2027%7D%20" id="TexFormula1" title=" \sqrt[3]{ - 27} " alt=" \

Alenkinab [10]

Answer:

$\sqrt[3]{-27}$ =-(3*3*3)

$\sqrt[3]{-27}$ =-3

-3 is a rational number, thus $\sqrt[3]{-27}$ will be a rational number.

8 0

3 years ago

Solve for y where y(2)=2 and y'(2)=0 by representing y as a power series centered at x=a

Crank

I'll assume the ODE is actually

$y''+(x-2)y'+y=0$

Look for a series solution centered at $x=2$ , with

$y=\displaystyle\sum_{n\ge0}c_n(x-2)^n$

$\implies y'=\displaystyle\sum_{n\ge0}(n+1)c_{n+1}(x-2)^n$

$\implies y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n$

with $c_0=y(2)=2$ and $c_1=y'(2)=0$ .

Substituting the series into the ODE gives

$\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge0}(n+1)c_{n+1}(x-2)^{n+1}+\sum_{n\ge0}c_n(x-2)^n=0$

$\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge0}c_n(x-2)^n=0$

$\displaystyle2c_2+c_0+\sum_{n\ge1}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge1}c_n(x-2)^n=0$

$\displaystyle2c_2+c_0+\sum_{n\ge1}\bigg((n+2)(n+1)c_{n+2}+(n+1)c_n\bigg)(x-2)^n=0$

$\implies\begin{cases}c_0=2\\c_1=0\\(n+2)c_{n+2}+c_n=0&\text{for }n>0\end{cases}$

If $n=2k$ for integers $k\ge0$ , then

$k=0\implies n=0\implies c_0=c_0$

$k=1\implies n=2\implies c_2=-\dfrac{c_0}2=(-1)^1\dfrac{c_0}{2^1(1)}$

$k=2\implies n=4\implies c_4=-\dfrac{c_2}4=(-1)^2\dfrac{c_0}{2^2(2\cdot1)}$

$k=3\implies n=6\implies c_6=-\dfrac{c_4}6=(-1)^3\dfrac{c_0}{2^3(3\cdot2\cdot1)}$

and so on, with

$c_{2k}=(-1)^k\dfrac{c_0}{2^kk!}$

If $n=2k+1$ , we have $c_{2k+1}=0$ for all $k\ge0$ because $c_1=0$ causes every odd-indexed coefficient to vanish.

So we have

$y(x)=\displaystyle\sum_{k\ge0}c_{2k}(x-2)^{2k}=\sum_{k\ge0}(-1)^k\frac{(x-2)^{2k}}{2^{k-1}k!}$

Recall that

$e^x=\displaystyle\sum_{n\ge0}\frac{x^k}{k!}$

The solution we found can then be written as

$y(x)=\displaystyle2\sum_{k\ge0}\frac1{k!}\left(-\frac{(x-2)^2}2\right)^k$

$\implies\boxed{y(x)=2e^{-(x-2)^2/2}}$

6 0

4 years ago