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3 years ago

What are the intercepts of y = 4x − 2?

Mathematics

1 answer:

maw [93]3 years ago

4 0

Answer: y = −2

Step-by-step explanation:

0 = −4x − 2
x = −1/2
To find the y-intercept, substitute in 0 for x and solve for y. Y = − 4 * 0 −2
Which then gives you y = - 2

* Hopefully this helps:) Mark me the brainliest:)

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I have no clue what to do... please help

strojnjashka [21]

Answer:

Multiply the base by the exponent

Step-by-step explanation:

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3 years ago

Amanda earned a score of 940 on a national achievement test that was normally distributed. The mean test score was 850 with a st

hichkok12 [17]

Answer:

lower than Amanda: 816 students

Step-by-step explanation:

An equivalent way in which to state this problem is: Find the area under the standard normal curve to the left (below) 940.

Most modern calculators have built in distribution functions.

In this case I entered the single command normalcdf(-1000,940, 850, 100)

and obtained 0.816.

In this particular situation, this means that 0.816(1000 students) scored lower than Amanda: 816 students.

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3 years ago

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What is the value of x in the equation

serg [7]

The answer to this question would be 2

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3 years ago

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At 8:00 am, here's what we know about two airplanes: Airplane #1 has an elevation of 80870 ft and is decreasing at the rate of 4

wel

Let's begin by listing out the information given to us:

8 am

airplane #1: x = 80870 ft, v = -450 ft/ min

airplane #2: x = 5000 ft, v = 900ft/min

We must note that the airplanes are moving at a constant speed. The equation for the airplanes is given by:

$\begin{gathered} E=x_1+vt----1 \\ E=x_2+vt----2 \\ where\colon E=elevation,ft;x=InitialElevation,ft; \\ v=velocity,ft\text{/}min;t=time,min \\ x_1=80,870ft,v=-450ft\text{/}min \\ E=80870-450t----1 \\ x_2=5,000ft,v=900ft\text{/}min \\ E=5000+900t----2 \end{gathered}$

We equate equations 1 & 2 to get the time both airlanes will be at the same elevation. We have:

$\begin{gathered} 5000+900t=80870-450t \\ \text{Add 450t to both sides, we have:} \\ 900t+450t+5000=80870-450t+450t \\ 1350t+5000=80870 \\ \text{Subtract 5000 from both sides, we have:} \\ 1350t+5000-5000=80870-5000 \\ 1350t=75870 \\ \text{Divide both sides by 1350, we have:} \\ \frac{1350t}{1350}=\frac{75870}{1350} \\ t=56.2min \\ \\ \text{After }56.2\text{ minutes, both airplanes will be at the same elevation} \end{gathered}$

The elevation at that time (when the elevations of the two airplanes are the same) is given by substituting the value of time into equations 1 & 2. We have:

$\begin{gathered} E_1=80870-450t \\ E_1=80870-450(56.2) \\ E_1=80870-25290 \\ E_1=55580ft \\ \\ E_2=5000+900t \\ E_2=5000+900(56.2) \\ E_2=5000+50580 \\ E_2=55580ft \\ \\ \therefore E_1\equiv E_2=55580ft \end{gathered}$

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1 year ago

Carl is trying to determine the length of segment MN using the Pythagorean theorem.

gavmur [86]

B. Carl's answer is correct but he incorrectly labeled the leg lengths in step 2.

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4 years ago

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