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3 years ago

Please help!

Mathematics

2 answers:

Musya8 [376]3 years ago

5 0

Answer:

C. $-6x-2y=12$

Step-by-step explanation:

We are given that a graph.

We have to find the equation of given graph.

The line passes through the point (-2,0) and (0,-6).

Slope:m $=\frac{y_2-y_1}{x_2-x_1}$

By using the formula

$m=\frac{-6-0}{0+2}=-3$

Slope-intercept form: $y=mx+C$

Where m=Slope of line

C=y-intercept

y-intercept: It is that value of y for which x=0

We have y- intercept=-6

Using the formula

Equation of line

$y=-3x-6$

$3x+y=-6$

Multiply by -2 on both sides then we get

$-6x-2y=12$

dezoksy [38]3 years ago

3 0

Answer:

Answer 3 (-6x-2y=12)

Step-by-step explanation: Turn to slope form to check

First- Add the 6x from both sides

-2y=6x+12

Second- Divide by two on both sides

y= -3x-6

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The answer would be $15

All you need to do is multiply each one by .4 and see which one equals 6.

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2 years ago

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Which statement correctly describes this expression?<br> 2m^3 - 11

Nady [450]

Answer:

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3 years ago

Find the standard form polynomial that represents this product. (6 - y - 4y^2)(-5 + 7y^2)

Tema [17]

Answer:

-28y^4 - 7y^3 + 62y^2 + 5y - 30

Step-by-step explanation:

<u>Opening the parenthesis and simplifying by combining like terms:</u>

(6 - y - 4y^2)(-5 + 7y^2) =
6(-5) +6(7y^2) - y(-5) -y(7y^2) -4y^2(-5) -4y^2(7y^2) =
-30 + 42y^2 + 5y - 7y^3 +20y^2 -28y^4 =
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4 years ago

At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a

morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = $\dfrac{15}{120}=0.125$

As we know the poisson process, we get that

$P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda$

So, for exactly one car would be

P(n=1) is given by

$=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599$

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

$0.2599\times 18=4.6798$

We will find the traffic flow q such that

$P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr$