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3 years ago

PLEASE HELP! ASAP PLEASE

Mathematics

2 answers:

Ganezh [65]3 years ago

6 0

The answer is x=45*.

djverab [1.8K]3 years ago

5 0

Answer:

x = 45

Step-by-step explanation:

x = (360 - 135 * 2) / 2

x = (360 - 270) / 2

x = 90 / 2

x = 45

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Write the function for the following chart,

timurjin [86]

Dy/dx=1/2, 1/2, 1/2 etc

So this is a linear equation of the form y=mx+b where m=dy/dx=1/2 so

y=x/2 +b, now we can use any point to solve for the y-intercept, "b", I'll use (7,0)

0=7/2 +b

b=-7/2 so

y=x/2-7/2

y=(x-7)/2

7 0

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

2 years ago

Hi there. Pls help me. Also the question is too long to type um so i took screenshot. Ty if u solved it and showed me how to do

Schach [20]

The answer is 6 days

5 0

2 years ago

Read 2 more answers

Please help me!! Will get brainliest!

makvit [3.9K]

Completing the square follows the principle of taking $a x^{2} +bx+c$ and converting it into $(x+ \frac{b}{2})^2+d$ where d is the 'correctional number' as I like to call it - i.e. the number that converts the expanded bracket into the +c, since the expanded bracket will give us $a x^{2} +b$ .

In this case, 2/2=1 so we have the first part: $(w+1)^2$ .
Expanding this gives us $w^2+2w+1$ . We need c to be 9, so we can just add 8.
Putting this together:

$w^2+2w+9=(w+1)^2+8$

Now we can solve it more easily.
Rearranging:
$(w+1)^2=-8 \\ w+1= +-\sqrt{-8} = +-\sqrt{8}i=+-2 \sqrt{2}i \\ \boxed{w=-1+ 2\sqrt{2}i\ or\ -1-2 \sqrt{2}i }$

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3 years ago

What is the value of this expression when d equals three

kotykmax [81]

Answer:

What is the expresion I am confused?

Step-by-step explanation:

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2 years ago