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2 years ago

8

Which pair of undefined terms is used to define the term parallel lines?

2 answers:

LuckyWell [14K]2 years ago

8 0

Answer:

Point and line

belka [17]2 years ago

6 0

Answer:

The answer is A: point and line : )

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Write an equation for the function

Mila [183]

Answer:

m(u) = -0.25(u +2)² +1 or -0.25u² -u

Step-by-step explanation:

The equation is fairly easily written in vertex form, as the vertex point is on a grid line intersection at (-2, 1). The parabola opens downward, so the scale factor is negative.

The vertical change from the vertex is only a fraction of a unit when u differs from the vertex by 1. It is 1 unit when u differs from the vertex by 2, so the magnitude of the vertical scale factor is 1/2² = 1/4.

Our equation will be of the form ...

m(u) = (vertical scale factor)(u - (horizontal vertex location))² + (vertical vertex location)

For this graph, the equation is ...

m(u) = -0.25(u +2)² +1

or, simplifying, we get ...

m(u) = -0.25u² -u

8 0

2 years ago

Pls help with qn 13!!!!

balandron [24]

4243.6-4000=243.6/4000= 0.06

Interest rate 0.06%

4 0

2 years ago

Which of these equations have no solution? Check all that apply.

ollegr [7]

Answer:

the first one

Step-by-step explanation:

6 0

2 years ago

Read 3 more answers

If f (x) = x + 7 and g (x) = StartFraction 1 Over x minus 13 EndFraction, what is the domain of (f circle g) (x)?

Rashid [163]

Answer:

R-{13}

Step-by-step explanation:

We are given that

$f(x)=x+7$

$g(x)=\frac{1}{x-13}$

We have to find the domain of fog(x).

$fog(x)=f(g(x))$

$fog(x)=f(\frac{1}{x-13})$

$fog(x)=\frac{1}{x-13}+7=\frac{1+7x-91}{x-13}=\frac{7x-90}{x-13}$

Domain of f(x)=R

Because it is linear function.

Domain of g(x)=R-{13}

Because the g(x) is not defined at x=13

fog(x) is not defined at x=13

Therefore, domain of fog(x)=R-{13}

6 0

2 years ago

Read 2 more answers

Find the volume of the following solid. The solid between the cylinder f(x,y)equals=e Superscript negative xe−x and the region

PilotLPTM [1.2K]

It is hard to comprehend your question. As far as I understand:

f(x,y) = e^(-x)

Find the volume over region R = {(x,y): 0<=x<=ln(6), -6<=y <= 6}.

That is all I understood. It would be easier to understand with a picture or some kind of visual aid.

Anyways, to find the volume between the surface and your rectangular region R, we must evaluate a double integral of f on the region R.

$\iint_{R}^{ } e^{-x}dA=\int_{-6}^{6} \int_{0}^{ln(6)} e^{-x}dx dy$

Now evaluate,

$\int_{0}^{ln(6)}e^{-x}dx$

which evaluates to, 5/6 if I did the math correct. Correct me if I am wrong.

Now integrate this w.r.t. y:

$\int_{-6}^{6}\frac{5}{6}dy = 10$

So,

$\iint_{R}^{ } e^{-x}dA=\int_{-6}^{6} \int_{0}^{ln(6)} e^{-x}dx dy = 10$

7 0

2 years ago