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vichka [17]
3 years ago
13

the number or years that brad has been in the soccer team is 2 les than 5 times the number of years that scoth has . in total th

e boys has being in the soccer team fo 10 years . how long has brad bieng on the soccer team
Mathematics
1 answer:
vaieri [72.5K]3 years ago
3 0

Answer:

x+5x-2 = 10

6x = 12

x=2

Brad was on the team for 2 years

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Find the equation of a line that passes through the points (-4, -2) and (6, 3).
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Step-by-step explanation:

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2 years ago
Find the average of 62 ℓ, 28 ℓ, 45 ℓ and 33 ℓ.
artcher [175]

Answer:

the average is 168

Step-by-step explanation:

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3 years ago
Binomial Expansion/Pascal's triangle. Please help with all of number 5.
Mandarinka [93]
\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}

The rows add up to 1,2,4,8,16, respectively. (Notice they're all powers of 2)

The sum of the numbers in row n is 2^{n-1}.

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When n=1,

(1+x)^1=1+x=\dbinom10+\dbinom11x

so the base case holds. Assume the claim holds for n=k, so that

(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k

Use this to show that it holds for n=k+1.

(1+x)^{k+1}=(1+x)(1+x)^k
(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)
(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}

Notice that

\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}
\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}

So you can write the expansion for n=k+1 as

(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}

and since \dbinom{k+1}0=\dbinom{k+1}{k+1}=1, you have

(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}

and so the claim holds for n=k+1, thus proving the claim overall that

(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n

Setting x=1 gives

(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n

which agrees with the result obtained for part (c).
4 0
3 years ago
11,20,29, Find the 30th term.
Mekhanik [1.2K]

Answer:

372

There are already 3 terms. So, now you have to find the 30th term.

The difference of 30 - 3 is 27. Now, since you add 9 for every term, multiply it by 27.

27 x 9 = 243.

The third term is 29. So you add 29.

Your answer will be 372. Hope it helps.

4 0
3 years ago
What is the equation of the line in slope-intercept form?
belka [17]

Answer:

y = 3/5 x+3

Step-by-step explanation:

two points on the graph

(-5,0) (0,3)

the y intercept is 3  (this is where it crosses the y axis)

the slope is

change in y     0 to 3            up 3

------------------  = -------------- =  -------        =   3/5

change in x      -5 to 0             right 5


we can tell the slope is positive because it goes from bottom left to top right

a negative slope goes from top left to bottom right

slope intercept form is

y=mx+b

y = 3/5 x+3


3 0
4 years ago
Read 2 more answers
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