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LenaWriter [7]
2 years ago
5

A. y = 8 − 7x B. y = 7x + 8 C. y = 7x − 8 D. y = 8x − 7

Mathematics
1 answer:
Anna71 [15]2 years ago
7 0

Answer:

I'm getting

y=-7x+8

The 7 is negative for me..You should maybe try yourself too and see what you get I might have made a mistake

Step-by-step explanation:

I'm guessing you live somewhere in the U.S. so I'll use the equation you use there

Equation:

y = mx + b

m =  \frac{y1 - y2}{x1 - x2}

first pick two points

I'll pick (0,8) and (1,1)

m =  \frac{8 - 1}{0 - 1} =  \frac{7}{ - 1 } =  - 7

this means that the slope is -7

So far the equation will look like this

y =  - 7x + b

to get b pick one point, I'll pick (0,8) and replace y and x

8 =  - 7 \times 0 + b

when you solve this you'll get b

b = 8

So the equation will look this

y =  - 7x + 8

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Can someone please help me solve 1-|1/3q-5|=-6
FrozenT [24]
Hello,

2 cases:

if 1/3*q-5>0 then

 |1/3*q-5|=1/3*q-5
1-|1/3*q-5|=-6
1-(1/3*q-5)=-6
1-1/3*q+5=-6
6-1/3*q=-6
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else

 |1/3*q-5|=-(1/3*q-5)
1-|1/3*q-5|=-6
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4 0
3 years ago
Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i
jenyasd209 [6]

Answer:

If the limit that you want to find is \lim_{x\to \infty}\dfrac{P(x)}{Q(x)} then you can use the following proof.

Step-by-step explanation:

Let P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} and Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0} be the given polinomials. Then

\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}

Observe that

\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}

and

\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}

Then

\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}

3 0
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Step-by-step explanation:

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