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3 years ago

What is an written expression for 20+5×9

Mathematics

1 answer:

BartSMP [9]3 years ago

7 0

Answer:

Five times nine plus twenty

Step-by-step explanation:

Just remember the order of operations... multiplication and division are done first before adding and subtracting.

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26(92' please someone tell me what this means

Sphinxa [80]

It could mean something with measurements

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2 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

Please help picture below

Gekata [30.6K]

A. The sculptor will have to sculpt away 1M on each side.

Explanation: 3 • 1_1/2 • 1_1/2
To get to 2 • 1/2 • 1/2, subtract 1 meter from each side of the rectangular prism and you’ll get that! :)

7 0

3 years ago

1. MONDAY (5/18): Which of the following functions matches the graph to the right

Strike441 [17]

Answer:

a.) f(x) = -⅙(x+3)²+6

Step-by-step explanation:

The maximum value, our vertex, is at point (-3,6).

We can insert this value into the vertex form of a quadratic function and then solve for a as follows...

$4.5 = a(0 +3)^2 +6\\4.5=a(9)+6\\-1.5 = 9a \\-.17=a\\a = -1/6$

a equals -1/6... We can input this into the original equation we used...

f(x) = -1/6(x+3)^2+6

Good luck on the bellwork ;)

7 0

3 years ago

Solve the problem. Use 3.14 for 4072-07-02-02-00_files/i0170000.jpg, and round your answer to the nearest tenth.

nalin [4]

Volume= (3.14)r^2h
Volume= 3.14 (9.5)^2 5
Volume= 1416.9 ft^3

7 0

3 years ago