All categories

2 years ago

The slope of the line passing through the points (6, 2) and (7, 3) is 3/5 1 -1

2 answers:

7 0

<span>The slope of the line passing through the points (6, 2) and (7, 3) is

$\large\begin{array}{I} \mathtt{ slope= \dfrac{3-2}{7-6}=1} } \end{array}$
</span>

FromTheMoon [43]2 years ago

6 0

Answer:

Step-by-step explanation:

You might be interested in

How much is 7 kilometers

Ira Lisetskai [31]

7 kilometers=7000 meters

6 0

2 years ago

Read 2 more answers

You read 50 pages in an hour. Your sister reads 40 pages in an hour. Explain how to find out how many more pages you read than y

devlian [24]

Since 1/2 is equal to a half, you divide 50 and 40 by 2 and you'll get 25 and 20 and then you subtract 25 and 20 and you get 5 which is the answer!

6 0

3 years ago

Read 2 more answers

Two lines intersect in _____. one plane many planes one point many points

Effectus [21]

Two lines intersect in one point.

5 0

2 years ago

Read 2 more answers

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

Which equation has a slope of -2 and passes through the point (1,-6)?

Naily [24]

Answer:

y=-2x-4

Step-by-step explanation:

Given that,

Slope, m = -2

It passes through the point (1,-6).

We need to find the equation of line that passes throgh the given point. It can be calculated as:

$y-y_1=m(x-x_1)$

Put y₁ = -6, x₁ = 1 and m = -2

$y-(-6)=(-2)(x-1)\\\\y+6=-2x+2\\\\y+2x+6-2=0\\\\y=-2x-4$

Hence, the equation of line is y=-2x-4.

7 0

2 years ago