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3 years ago

Please need help thank you

Mathematics

1 answer:

aev [14]3 years ago

4 0

Answer:

12 miles

Step-by-step explanation:

Area=length*width

To find the length using the area, do area/width.

96/8=12

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Point F is at (0, -10) and point H is at (-6, 5). Find the coordinates of point G on line segment FH such that the ratio of FG t

Tanya [424]

Answer:

G(x,y)=(-4,0)

Step-by-step explanation:

We use the section formula:

$(x,y)=\left(\dfrac{mx_2+nx_1}{m+n}, \dfrac{my_2+ny_1}{m+n}\right)$

Given:

$F(x_1,y_1)=(0, -10)\\H(x_2,y_2)=(-6, 5)\\FG:GH=m:n=2:1$

We substitute the values to get:

$G(x,y)=\left(\dfrac{2*-6+1*0}{2+1}, \dfrac{2*5+1*-10}{2+1}\right)\\=\left(\dfrac{-12}{3}, \dfrac{10-10}{3}\right)\\=\left(\dfrac{-12}{3}, \dfrac{0}{3}\right)\\\\G(x,y)=(-4,0)$

5 0

4 years ago

A new company spent $9,399.74 on 24 tablets for their employees. How much did each tablet cost? Round your answer to the nearest

Nitella [24]

$391.655 or $391.66 per table

8 0

4 years ago

Read 2 more answers

There are 7 ushers and 11 technicians helping at the Harper Middle School fall play.

bearhunter [10]

Answer:

<h3>The ratio of technicians to all helpers is 11 : 7, or $\frac{11}{7}$ or 11 to 7.</h3>

Step-by-step explanation:

Given that there are 7 ushers and 11 technicians helping at the Harper Middle School fall play.
Let x be the number of ushers ( or helpers ).
Therefore x=7 helpers.
Let y be the number of technicians.
Therefore y=11 technicians.

<h3>To find the ratio of technicians to all helpers :</h3>

That is to find the ratio of y to x.

We can write the ratio of technicians to all ushers(helpers) as y : x

Which implies that 11 : 7, (since y=11 and x=7)

Or $\frac{11}{7}$ or 11 to 7

<h3>The ratio of technicians to all helpers is 11 : 7, or $\frac{11}{7}$ or 11 to 7</h3>

4 0

4 years ago

Are the answers correct or am I missing some?

AlekseyPX

Well,

Emily received $40.00 and then withdrew $40.00. The final amount would be zero.

Jacoby climbed 782 feet to get to his campsite, and then descended 782 feet to get back to his car. His final change is zero.

The Eagles mad two points, but then the Bulldogs evened it out with another two points. The difference between the two scores would be the same.

Marcus ends up going down more than he goes up, so this is not a correct option.

The Ravens scored two points, but at the end the Raiders scored an additional 5 points, which set both teams to the same score.

Since there are an equal number of protons and electrons, and the protons have the same (but opposite) amount of charge as the electrons, the charges cancel each other out, leaving the atom with a neutral overall charge (0).

8 0

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).

4 0

3 years ago