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4 years ago

Can someone answer these questions ? thank you

Mathematics

1 answer:

Naya [18.7K]4 years ago

5 0

Answer:

Step-by-step explanation:

12a)To rationalize the denominator, multiply the denominator and numerator by √5.

$\frac{15}{\sqrt{5}}=\frac{15*\sqrt{5}}{\sqrt{5}*\sqrt{5}}\\\\=\frac{15\sqrt{5}}{5}\\\\=3\sqrt{5}$

b) (a+b)² = a² + 2ab + b²

(1 +√3)² = 1² + 2*1*√3 + (√3)²

= 1 + 2√3 + 3

= 4 + 2√3

a = 4 ; b =2

13) (a + b)(a - b) = a² - b²

$\frac{(6-\sqrt{5})(6+\sqrt{5})}{\sqrt{31}}=\frac{6^{2}-(\sqrt{5})^{2}}{\sqrt{31}}\\\\ =\frac{36-5}{\sqrt{31}}\\\\=\frac{31}{\sqrt{31}}\\\\=\frac{31*\sqrt{31}}{\sqrt{31}*\sqrt{31}}\\\\=\frac{31\sqrt{31}}{31}\\\\=\sqrt{31}$

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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

F(x)=2x+5 find f(-6)

ser-zykov [4K]

Answer:

Step-by-step explanation:

f(x) = 2x + 5

f(-6)= 2(-6) + 5 = -12 + 5 = -7

4 0

3 years ago

On a coordinate plane, 2 cube root functions are shown. Function f (x) goes through (negative 3, negative 1.5), has an inflectio

Strike441 [17]

Using translation concepts, it is found that the equation that represents the graph of g(x) is:

$g(x) = \sqrt[3]{x + 2}$ .

<h3>What is a translation?</h3>

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.

In this problem, we have that g(x) is a shift left of 2 units of f(x), hence:

$f(x) = \sqrt[3]{x}$

$g(x) = f(x + 2) = \sqrt[3]{x + 2}$

More can be learned about translation concepts at brainly.com/question/4521517

#SPJ1

5 0

2 years ago

I will give brainliest to whoever answers first

mojhsa [17]

2(3x - 4) = 2x + 12

6x - 8 = 2x + 12

4x = 20

x = 5

AB = 3(5) - 4 = 11

AC = 2(5) + 12 = 22

BC = 7(5) - 2 = 35 - 2 = 33

AC:BC = 22:33 = 2:3

4 0

2 years ago

Read 2 more answers

Write two different mixed numbers where the LCD (lowest common denominator) is 12.<br> HELP ASAP

alex41 [277]

Answer:

1 1/2, 3 1/3

<h3>Step-by-step explanation:</h3>

For the denominators (3, 4) the least common multiple (LCM) is 12. Calculations to rewrite the original inputs as equivalent fractions with the LCD:

10/3 = 10/3 × 4/4 = 40/12

6/4 = 6/4 × 3/3 = 18/12

5 0

3 years ago