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ExtremeBDS [4]
3 years ago
15

Can someone tell me how to find my weighed grade. I have a 97% for homework which is 15% of my grade and 84% which is 35% of gra

de. How do I find my weighed grade for this. Thanks
Mathematics
2 answers:
dlinn [17]3 years ago
6 0
I think your weighted grade is 50%
Nuetrik [128]3 years ago
4 0
Homework= 97%
To work this out you divide 97 by 100 then multiply it by 15
Then you divide 84 by 100 then multiply it by 35
Lastly you add the two totals together
97/100 x 15= 14.55
84/100 x 35= 29.4
14.55 + 29.4= 43.95
You have 43.95 out of 50 at the moment.
Keep up the good work!
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MrRa [10]

Answer:

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3) 21

4) 7

Step-by-step explanation:

2) (-6) -(-17) = 17 - 6 = 11

3) 19 - (-2) = 19 + 2 = 21

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2 years ago
-x-2y-4z=-6<br> X-5y-2z=-2<br> 2x-3y+2z=2
Svetlanka [38]

Answer:

\sqrt{2} or (1/2)

Step-by-step explanation:

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3 years ago
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
3 years ago
Is this right??<br><br> 313.65 x 0.9 = 282.285x
yan [13]
Yes Your Answer Is In Fact 282.285  :)
5 0
3 years ago
What is the meaning of the point with an x-coordinate of 3?
Gnoma [55]

Answer:

answer

Step-by-step explanation:

This means that the point 3 is placed in quarters 3.

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