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Kipish [7]
3 years ago
15

Who wants to do my math work for me for $20 for a month?

Mathematics
2 answers:
TEA [102]3 years ago
8 0

Answer:

what type of math is it?

nadezda [96]3 years ago
3 0

Answer:

i will do your math for a brainliest for every question i answer and for no pay  

Step-by-step explanation:

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Uhh can someone tell me how brainly works
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Convert: 100 yards to meters. <br><br> A)9.14<br> B)91.4<br> C)914<br> D)9140
astra-53 [7]
The answer is B. 91.44
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Find the sum of the first 17 terms of the arithmetic sequence 10, 14, 18, 22, 26...
kari74 [83]
10, 14, 18 ....

notice, we get the next term by simply adding 4 to the current term, thus "4" is the "common difference, and we know that 10 is the first term.

\bf n^{th}\textit{ term of an arithmetic sequence}&#10;\\\\&#10;a_n=a_1+(n-1)d\qquad &#10;\begin{cases}&#10;n=n^{th}\ term\\&#10;a_1=\textit{first term's value}\\&#10;d=\textit{common difference}\\&#10;----------\\&#10;a_1=10\\&#10;d=4\\&#10;n=17&#10;\end{cases}&#10;\\\\\\&#10;a_{17}=10+(17-1)(4)\implies a_{17}=10+(16)(4)&#10;\\\\\\&#10;a_{17}=10+64\implies a_{17}=74\\\\&#10;-------------------------------

\bf \textit{ sum of a finite arithmetic sequence}&#10;\\\\&#10;S_n=\cfrac{n(a_1+a_n)}{2}\qquad &#10;\begin{cases}&#10;n=n^{th}\ term\\&#10;a_1=\textit{first term's value}\\&#10;----------\\&#10;a_1=10\\&#10;a_{17}=74\\&#10;n=17&#10;\end{cases}&#10;\\\\\\&#10;S_{17}=\cfrac{17(10+74)}{2}\implies S_{17}=\cfrac{17(84)}{2}\implies S_{17}=714
6 0
2 years ago
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Whats the value for 24/3*4/2
bogdanovich [222]

Answer: 16

Step-by-step explanation:

Multiple the numerators

24 x 4 = 96

Then multiple the denominators

3 x 2 = 6

So 96/6 equals 16

4 0
3 years ago
Prove by mathematical induction that
postnew [5]

For n=1, on the left we have \cos\theta, and on the right,

\dfrac{\sin2\theta}{2\sin\theta}=\dfrac{2\sin\theta\cos\theta}{2\sin\theta}=\cos\theta

(where we use the double angle identity: \sin2\theta=2\sin\theta\cos\theta)

Suppose the relation holds for n=k:

\displaystyle\sum_{n=1}^k\cos(2n-1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}

Then for n=k+1, the left side is

\displaystyle\sum_{n=1}^{k+1}\cos(2n-1)\theta=\sum_{n=1}^k\cos(2n-1)\theta+\cos(2k+1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta

So we want to show that

\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta=\dfrac{\sin(2k+2)\theta}{2\sin\theta}

On the left side, we can combine the fractions:

\dfrac{\sin2k\theta+2\sin\theta\cos(2k+1)\theta}{2\sin\theta}

Recall that

\cos(x+y)=\cos x\cos y-\sin x\sin y

so that we can write

\dfrac{\sin2k\theta+2\sin\theta(\cos2k\theta\cos\theta-\sin2k\theta\sin\theta)}{2\sin\theta}

=\dfrac{\sin2k\theta+\sin2\theta\cos2k\theta-2\sin2k\theta\sin^2\theta}{2\sin\theta}

=\dfrac{\sin2k\theta(1-2\sin^2\theta)+\sin2\theta\cos2k\theta}{2\sin\theta}

=\dfrac{\sin2k\theta\cos2\theta+\sin2\theta\cos2k\theta}{2\sin\theta}

(another double angle identity: \cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta)

Then recall that

\sin(x+y)=\sin x\cos y+\sin y\cos x

which lets us consolidate the numerator to get what we wanted:

=\dfrac{\sin(2k+2)\theta}{2\sin\theta}

and the identity is established.

8 0
2 years ago
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