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3 years ago

Determine the slope of the line that passes through the points:

Mathematics

1 answer:

miskamm [114]3 years ago

4 0

Answer:

The slope of the line is 4.

Step-by-step explanation:

Recall that slope = rise over run: rise / run.

From the table we see that if x changes from 0 to 4, a run of 4, y changes from -1 to 15, a rise of 16, the slope of this line must be

m = rise / run = 16/4 = 4

The slope of the line is 4.

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After a storm damage is the community center sent Kiya and her friends hold fundraising events to help pay for repairs after the

andrew-mc [135]

Answer: the total amount of money that they wanted to raise is $2400

Step-by-step explanation:

Let x represent the total amount of money that Keelan and her friends want to raise.

After the first event they raise $240 which is 10% of the total amount they want to raise. The amount of money that raised after the first event is expressed as

10/100 × x = 0.1 × x = 0.1x

Therefore,

0.1x = 240

Dividing both sides of the equation by 0.1, it becomes

x = 240/0.1

x = $2400

7 0

3 years ago

A new bagel store opened. The first day 15 customers entered the store. The number of customers that enter the store triples eac

Lelu [443]

Answer: it triples by 3 so 15,45, 135

Step-by-step explanation: this is by 3 because it goes on by 15,45,135

3 0

3 years ago

Solve xy − 2 = k for x.

MA_775_DIABLO [31]

Isolate x
So xy - 2 = k
Then
xy = k + 2

x = (k+2)/y

6 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

3 years ago

There are 84 boys in a senior class of 146 students. find the ratio of boys to girls. (note: if you know the total number of stu

vladimir1956 [14]

84 to 62. You have to subtract 146-84 to get the number of girls from the class.

4 0

3 years ago