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pishuonlain [190]
3 years ago
11

Simplify this, include instructions

Mathematics
2 answers:
monitta3 years ago
7 0
3n-2n = 1n
1+8 = 9
1n + 9
Cerrena [4.2K]3 years ago
5 0
<span><span><span>3n</span>+1</span>−<span>2n</span></span>+<span>8

</span><span>Combine Like Terms:
</span>
<span><span><span>3n</span>+1</span>+<span>−<span>2n</span></span></span>+<span>8
</span>(3n+−2n)+(1+8)
=<span>n+<span>9
</span></span>
Answer is n+9


Hope I helped

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You estimate that a baby pig weighs 21 lbs but it weighs 28 lbs what is the percent of error
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Your estimate of the baby pig's weight is 2 pounds. However the actual weight of it is only 16 pounds.
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A survey was conducted to determine the amount of time, on average, during a given week SCAD students spend outside of class on
fiasKO [112]

Answer:

Standard Deviation = 5.928

Step-by-step explanation:

a) Data:

Days  Hours spent  (Mean - Hour)²

1              5                61.356

2             7                34.024

3            11                  3.360

4           14                   1.362

5           18               26.698

6          22               84.034

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mean

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Therefore, the square root of 35.139 = 5.928

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5 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

1.

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}

2.

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}

3.

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}

6 0
9 months ago
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