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crimeas [40]
3 years ago
8

PLEASE HELPPPPPPLPPPPPPPP

Mathematics
1 answer:
solong [7]3 years ago
3 0
<h3>Answer:  36%</h3>

============================================

Explanation:

There are 20 blue cards out of 10+20+25 = 55 cards total.

Divide the 20 over 55 to get

20/55 = 0.363636...

where the "36"s go on forever

So 20/55 = 0.3636 approximately

Move the decimal point over 2 spots to go from 0.3637 to 36.37%

Then round to the nearest percent: 36.37% ---> 36%

The probability of getting a blue card is approximately 36%

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what is the equation of the line in standard form? please help me I am really having a hard time with this
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Answer:

      -4 - 2

m= --------- (over)

      -1 - 2

Step-by-step explanation:


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Simplify each expression using the order of operations.
meriva

Answer:

Step-by-step explanation:

1. a

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3 years ago
D. (30 * 4) + (8 * 2)
RideAnS [48]
30•4=120, 8•2=16. 120+16=136
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We would like to use the power series method to find the general solution to the differential equation d 2y dx2 − 4x dy dx + 12y
Feliz [49]

y=\displaystyle\sum_{n\ge0}a_nx^n

\dfrac{\mathrm dy}{\mathrm dx}=\displaystyle\sum_{n\ge1}na_nx^{n-1}\implies4x\dfrac{\mathrm dy}{\mathrm dx}=4\sum_{n\ge1}na_nx^n=4\sum_{n\ge0}na_nx^n

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}=\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n

Substituting into the ODE

\dfrac{\mathrm d^2y}{\mathrm dx^2}-4x\dfrac{\mathrm dy}{\mathrm dx}+12y=0

gives

\displaystyle\sum_{n\ge0}\bigg((n+2)(n+1)a_{n+2}-4na_n+12a_n\bigg)x^n=0

so that the coefficients of the series are given according to

\begin{cases}a_0=y(0)\\a_1=y'(0)\\a_{n+2}=\dfrac{4(n-3)a_n}{(n+2)(n+1)}&\text{for }n\ge0\end{cases}

We can shift the index in the recursive part of this definition to get

a_n=\dfrac{4(n-5)a_{n-2}}{n(n-1)}

for n\ge2. There's dependency between coefficients that are 2 indices apart, so we can consider 2 cases:

  • If n=2k, where k\ge0 is an integer, then

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

but since y(0)=0, we have a_0=0 and a_{2k}=0 for all k\ge0.

  • If n=2k+1, then

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

k=1\implies n=3\implies a_3=\dfrac{4(-2)a_1}{3\cdot2}=-\dfrac43a_1

k=2\implies n=5\implies a_5=0

and so a_{2k+1}=0 for all k\ge2. If y'(0)=1, we then have a_1=1 and a_3=-\dfrac43.

So the ODE has solution

y(x)=\displaystyle\sum_{k\ge0}(a_{2k}x^{2k}+a_{2k+1}x^{2k+1})\implies\boxed{y(x)=x-\dfrac43x^3}

8 0
3 years ago
4y + 4x = -28 and<br> x = 2y - 4
Rzqust [24]
4y+4(2y-4)=-28
4y+8y-16=-28
12y-16=-28
12y=-12
y=-1
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2 years ago
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