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leva [86]
3 years ago
10

Solve by factoring

Mathematics
1 answer:
True [87]3 years ago
7 0

2×2-13(19÷13)+15=0

So x= 19/13

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Tickets to a baseball game are 20 dollars for an adult and 15 dollars far a student. A school bus tickets for 45 adu6and 600 stu
yulyashka [42]

Answer:The total amount of money that the school will spend for tickets is $9900

Step-by-step explanation:

Tickets to a baseball game are 20 dollars for an adult and 15 dollars for a student. A school bus tickets for 45 adults and 600 students. This means that the total amount spent by the school on adult tickets is 20 × 45 = $900

The total amount spent by the school on students tickets is 15 × 600 = $9000

The total amount of money that the school will spend for tickets will be the sum of the amount spent on adult tickets and student tickets. It becomes

900 + 9000 = $9900

4 0
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).
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3 years ago
I Need Help Asappp! So Pls answerrr!
lisabon 2012 [21]

Answer:

y= -3x+6

Step-by-step explanation:

3 0
2 years ago
If a+b-c=d, if a-b+c=e, then a= ?
yaroslaw [1]
A + b = d+c 
<span>a - b = e - c </span>
The b and c cancel out. 
<span>2a = d+e </span>
<span>a = (d+e)/2 </span>

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3 years ago
Solve the following equation for the value for x.<br> 7(0.3x - 0.9) = 6x-0.6 (18 +4x)
tangare [24]
The answer to the question is: x=3
4 0
3 years ago
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