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Neporo4naja [7]
3 years ago
9

Can someone help me out with these?"write a linear equation for line m,n,l,p"​

Mathematics
2 answers:
Debora [2.8K]3 years ago
8 0

m: y=-2x+4

<u>slope: 2</u>

2/1=2

Its negative because the line is going down from left to right

<u>y-intercept:</u>

it crosses the y axis at (0,4)

n: y=x-1

<u>slope: 1</u>

1/1=1

Its positive because the line is going up from left to right

<u>y-intercept:</u>

it crosses the y axis at (0,-1)

l: y=4

<u>slope: 0</u>

A vertical line's slope is always 0

<u>y-intercept:</u>

it crosses the y axis at (0,4)

p: x=-4

To find the equation for a horizontal line, you need to find what x always is no matter what.

aniked [119]3 years ago
3 0

Answer:

Step-by-step explanation:

m:  y = -2x + 4

n:    y = x - 1

l:      y = 4

p:     x = -4

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\displaystyle\\&#10;\text{Verification}\\\\&#10;3x^3-5x^2-14x-4=\\\\&#10;=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\&#10;=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\&#10;=-\frac{6}{9}+\frac{14}{3}-4=\\\\&#10;=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\&#10; =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\&#10;\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\&#10;\Longrightarrow~~~P(x)~\vdots~(3x+1)


\displaystyle\\&#10;3x^3-5x^2-14x-4=0\\&#10;~~~~~-5x^2 = x^2 - 6x^2\\&#10;~~~~~-14x =-2x-12x \\&#10;3x^3+x^2 - 6x^2-2x-12x-4=0\\&#10;x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\&#10;(3x+1)(x^2-2x -4)=0\\\\&#10;\text{Solve: } x^2-2x -4=0\\\\&#10;x_{12}= \frac{-b\pm  \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm  \sqrt{4+16}}{2}=\frac{2\pm  \sqrt{20}}{2}=\frac{2\pm  2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\&#10;x_1 =1+\sqrt{5}\\&#10;x_2 =1-\sqrt{5}\\&#10;\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}



7 0
3 years ago
Explain how a number and it's reciprocal are related adding 28 ?
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Answer:

A(t) = 300 -260e^(-t/50)

Step-by-step explanation:

The rate of change of A(t) is ...

A'(t) = 6 -6/300·A(t)

Rewriting, we have ...

A'(t) +(1/50)A(t) = 6

This has solution ...

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We need to find the values of p and q. Using the differential equation, we ahve ...

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A(0) = 300 +q = 40

q = -260

So, the complete solution is ...

A(t) = 300 -260e^(-t/50)

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