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2 years ago

write the equation of the line, in standard form, that is perpendicular to y=3x-2 and passes through (-7,2)

Mathematics

1 answer:

34kurt2 years ago

8 0

Answer:

x+3y=-1

Step-by-step explanation:

y=3x-2 so m=3 old slope

the new slope must be -1/3 (opposite reciprocal of the old slope).

y-y0=m*(x-x0)

y-2=-1/3*(x-(-7))

y-2=-1/3x-1/3*7

y-2=-1/3x-7/3

3y-6=-x-7

The standard form for linear equations in two variables is Ax+By=C.

x+3y=6-7

x+3y=-1

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A rectangular poster is 3 times as long as it is wide. A rectangular banner is 5 times as it is wide. Both the banner and the po

Svetradugi [14.3K]

The length and width of poster are 9 inches and 3 inches respectively and length and width of banner are 10 inches and 2 inches respectively

<u> Solution:</u>

Given, A rectangular poster is 3 times as long as it is wide.

Rectangular banner is 5 times as long as it is wide.

Both the banner and the poster have perimeters of 24 inches

Let, the width of poster be p, then its length will be 3p

And, the width of banner be b, then its length will be 5b

Now, according to given information,

Perimeter of poster = 24 inches and perimeter of banner = 24 inches

2(length of poster + width of poster) = 24

2(3p + p) = 24

2(4p) = 24

8p = 24

p = 3

And 2(length of banner + width of banner) = 24

2(5b + b) = 24

2(6b) = 24

12b = 24

b = 2

So, width of poster is 3 inches, then length of poster is 3p = 3 x 3 = 9 inches

And, width of banner is 2 inches, then length of banner is 5b = 5 x 2 = 10 inches

8 0

3 years ago

What is the size of angle x !!!

klemol [59]

Answer:

∠x = 67°

Step-by-step explanation:

∠x = 67° because they both are inscribed angles of the same arc

7 0

3 years ago

Read 2 more answers

4. Which value for x makes the sentence true?<br> 3x - 8 = 16<br> x=

zzz [600]

Answer: x = 8

Step-by-step explanation:

This is a very, very, very simple algebra problem. I assume you know the basics of algebra.

First, move 8 to the other side:

3x = 24

divide by 3 on both sides

x = 8

There! Would it kill you to give me a quartic question to spice my life up?

4 0

1 year 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

3 years ago

Can someone please do 1-9

andriy [413]

1-9=-8 is your answer

5 0

3 years ago