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2 years ago

Use the following area model to multiply 29x97

Mathematics

1 answer:

alexandr402 [8]2 years ago

6 0

The answer to 29×97 would be 2813

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Can someone pls help me with this.. Thanks in advance

vfiekz [6]

3,6,8,5, these are your answers

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3 years ago

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The bases of a trapezoid are 16.8 yards and 6.9 yards. what is the average of the two bases?

kow [346]

The average = 1/2 (sum of the two bases) = 1/2 ( 16.8 + 6.9) = 11.85 yards

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2 years ago

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Jake wants to surprise his parents with a small anniversary party at their favorite restaurant.

ch4aika [34]

Answer:

16 people

Step-by-step explanation:

35.75 times 16 is 572, times 17 is 607.75, he only has 600 and he cant have a fraction of a person, so he can have 16 people

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2 years ago

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A collection of nickels and quarters is worth 2.85 . There are 3 more nickels than quarters how many nickels and quarters are th

Zolol [24]

A nickel is equal to 5 cents or 0.05 dollars.

A quarter is equal to 25 cents or 0.25 dollars.

Let number of nickels be = n

Let number of quarters be = q

$0.05n+0.25q=2.85$ ...........(1)

As it is given, there are 3 more nickels than quarters so equation becomes,

$n=q+3$ ................(2)

Plug in the value of 'n' from (2) in (1)

$0.05(q+3)+0.25q=2.85$

= $0.05q+0.15+0.25q=2.85$

$0.30q=2.70$

$q=9$

As $n=q+3$ we get, $n=9+3=12$

Hence, there are 12 nickels and 9 quarters.

3 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).

4 0

2 years ago