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2 years ago

Please help me with this surds question. Explain how you got your answer.

Mathematics

1 answer:

andre [41]2 years ago

7 0

Answer:

1/5

Step-by-step explanation:

b and c = $\sqrt{50}$ because 10 x 5 = 50

therefore the fraction is √2/ √50

to simplify you need to simplify the √50 into a number of √2 as the top is √2

√50 = √25 x √2

= 5√2

√2/5√2 = 1/5 if you divide both top and bottom by √2

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What is the ratio a:b if 8a=3b?

Digiron [165]

The ratio a:b is 3:8

Step-by-step explanation:

In simple words, the ratio a:b means the fraction a/b.

So we will have to solve the given equation for a/b

Given

$8a=3b$

Dividing both sides by 8

$\frac{8a}{8}=\frac{3b}{8}\\a=\frac{3b}{8}$

Dividing both sides by b

$\frac{a}{b}=\frac{3b}{8*b}\\\frac{a}{b} =\frac{3}{8}$

The ratio a:b is 3:8

Keywords: Ratios, Fractions

Learn more about ratios at:

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2 years ago

Y and x have a proportional relationship and y=9 when x=2 what is the value of y when x=3

Paladinen [302]

If they are proportional, then y/x should always be the same.

So 9/2= 4.5

We need to find the y value at x=3 to make y/x equal 4.5

4.5 = y/3

y = 4.5*3 = 13.5

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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).

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2 years ago

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Nadusha1986 [10]

Answer:

tan 11 pi dived by 6 is equal to -118.307586549

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2 years ago

From a barrel of colored marbles, you randomly select 3 blue, 2 yellow, 7 red, 8 green, and 2 purple marbles. Find the experimen

svet-max [94.6K]

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3 years ago

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