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2 years ago

Line m passes through (-5,0) and (-3,2)

Mathematics

1 answer:

nata0808 [166]2 years ago

5 0

Answer:

Yes, they are parallel

Step-by-step explanation:

Find the slopes of the lines using rise over run (y2 - y1) / (x2 - x1)

Find the slope of line M:

(y2 - y1) / (x2 - x1)

(2 - 0) / (-3 + 5)

2/2

= 1

Find the slope of line N:

(y2 - y1) / (x2 - x1)

(4 - 2) / (0 + 2)

2/2

= 1

So, the two lines have the same slope of 1, meaning that they are parallel because parallel lines have the same slope.

They are parallel.

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Find the slope of the line that passes through the following points. Write "undefined" if the slope is undefined.

Katyanochek1 [597]

The slope of the line that passes through the following points is undefined

Solution:

Given that, We have to find the slope of line passing through two points

Given points are (-1, 5) and (-1, 6)

The slope of line is given by formula:

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

From given,

$(x_1, y_1) = (-1, 5)\\\\(x_2, y_2) = (-1, 6)$

Substituting the values we get,

$m = \frac{6-5}{-1-(-1)}\\\\m = \frac{1}{-1+1}\\\\m = \frac{1}{0}$

Thus the slope is undefined

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2 years ago

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Answer:

There is no way to know.

Step-by-step explanation:

You need to know how much either burritos cost or how much tacos cost to answer that, let your teacher know.

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2 years ago

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Both millimeters and quarts are units for volume.

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2 years ago

How to solve this and find x and y variables.

Colt1911 [192]

2x=60 and (y-8)+124=180

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2 years ago

Use the exponential decay model, Upper A equals Upper A 0 e Superscript kt, to solve the following. The half-life of a certai

Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

$A=A_0 e^{kt}$

Where,

$A_0$ = Initial amount of the substance,

If the half life of the substance is 19 years,

Then if t = 19, amount of the substance = $\frac{A_0}{2}$ ,

i.e.

$\frac{A_0}{2}=A_0 e^{19k}$

$\frac{1}{2} = e^{19k}$

$0.5 = e^{19k}$

Taking ln both sides,

$\ln(0.5) = \ln(e^{19k})$

$\ln(0.5) = 19k$

$\implies k = \frac{\ln(0.5)}{19}\approx -0.03648$

Now, if the substance to decay to 78% of its original amount,

Then $A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0$

$0.78 A_0=A_0 e^{-0.03648t}$

$0.78 = e^{-0.03648t}$

Again taking ln both sides,

$\ln(0.78) = -0.03648t$

$-0.24846=-0.03648t$

$\implies t = \frac{0.24846}{0.03648}=6.81085\approx 7$

Hence, approximately the substance would be 78% of its initial value after 7 years.

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2 years ago