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3 years ago

Which is the best estimate for the mass of a rose

Mathematics

1 answer:

andre [41]3 years ago

4 0

Answer:

Step-by-step explanation:

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What’s the trinomial Mila (x-6)^2

MatroZZZ [7]

Hello

(x-6)² = x² -6*2*x + 6²

= x² -12x + 36

8 0

3 years ago

Ms. Nina has 10 lbs of 25% sugar syrup. How much water does she need to add to make 10% sugar syrup?

ryzh [129]

She has 10lbs of 25% syrup... so, in the 10lbs, 25% of that is syrup, the rest, namely the 75% remaining is water or other substances.

let's say she adds "x" lbs of water, to get "y" lbs for the 10% mixture.

how much is 25% of 10lbs? well, (25/100) * 10, or 2.5.

the water has no sugar syrup in it, so is just pure water, so the amoun of syrup in it is 0%, how much is 0% of "x" lbs? well, (0/100) * x, or 0.00x, which is just 0.

how much is 10% of "y" lbs? well (10/100) * y, or 0.10y.

whatever "x" and "y" are, we know that 10 + x = y.

we also know that the syrup amount in that is also 2.5 + 0.00x = 0.10y, thus

$\bf \begin{array}{lccclll}
&\stackrel{lbs}{syrup}&\stackrel{concentration~\%}{syrup}&\stackrel{concentration}{amount}\\
&------&------&------\\
\textit{25\% syrup}&10&0.25&2.5\\
\textit{pur water}&x&0.00&0.00x\\
------&------&------&------\\
\textit{10\% mixture}&y&0.10&0.10y
\end{array}
\\\\\\
\begin{cases}
10+x=\boxed{y}\\
2.5+0.00x=0.10y\\
----------\\
2.5 = 0.10\left( \boxed{10+x} \right)
\end{cases}
\\\\\\
2.5=1 + 0.10x\implies 1.5=0.10x
\\\\\\
\cfrac{1.5}{0.10}=x\implies 15=x$

8 0

3 years ago

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1.2%of 45 what is the answer

drek231 [11]

Answer:

0.54

Step-by-step explanation:

1.2/100*45

6 0

3 years ago

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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

a stock that was selling for $48 a share split 2-for-1. Before the split, the company had 3.4 million shares of stock outstandin

nalin [4]

Answer:

666

Step-by-step explanation:

4 0

3 years ago

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