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3 years ago

10. Write an equation in slope-intercept form for a line that is parallel to the line y= 2/5x +1

Mathematics

1 answer:

harina [27]3 years ago

8 0

If two lines are parallel to each other, then they have the same slope, but different y-intercepts.

So any y-intercept works basically

You could do y=2/5x+2 if you wanted or y=2/5x-7 or something like that

Hope this helped.

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|x|+5=18 a. 5 or -5 b.13 or -13 c.18 or -18 d.23 -23

dezoksy [38]

X+5=18
first get rid of the 5
so X+5-5=18-5
which makes X=13
if any questions please ask
$b. 13$

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3 years ago

Which of the following are binomials? Check all that apply.

Nezavi [6.7K]

Binomial is the sum of two monomials.

$\text{binominals:}\\\\C.\ x^2+18\\D.\ x+2$

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3 years ago

In a test, John was able to answer 15 correct out of 40 questions. Beth was able to answer 24 questions correctly out of 50. Mic

Helen [10]

Answer:

that is wrong beth has a better score

Step-by-step explanation:

14 out of 40 is 37.5%

and 24 out of 50 is 48%

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2 years ago

Write 1,340,000,000,000 in expanded form with exponents

Kazeer [188]

1 × 1010+3 × 109+4 × 108+0 × 107+0 × 106+0 × 105+0 × 104+0 × 103+0 × 102+0 × 101+0 × 100

3 0

3 years ago

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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9 months ago