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Reika [66]
4 years ago
8

GEOMETRY HELP 2 questions?!!?

Mathematics
1 answer:
anzhelika [568]4 years ago
8 0
The surface area is 72ft^2

hope this helped!
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what's the error? MAX said that 36,594,145 is less than 5,980,251 because 3 is less than 5. Describe max error and give the corr
forsale [732]
Just because the 3 is less than 5 doesn't mean it's less. What max did wrong was automatically see the first number and say it was greater/least because it's just the first digit. 36,594,145 is greater than 5,980,251 because the last 6 digits are invisible right now so thing about the numbers before the first comma- 36 and 5. Which number is greater? 36 so just because it's the first digit in the number doesn't make it automatically greater or less. *-sorry for such long paragraph*
7 0
3 years ago
Divide and simplify.
balandron [24]
The last one is the correct one.


8 0
4 years ago
How much longer can s 3 meters than 3 yards
sergey [27]

0.91 meters = 1 yard

0.91 * 3 = 2.73

3 - 2.73 = 0.27

Your answer is 0.27

Hope this helps you!

4 0
3 years ago
Please hurry and answer this for me. willing to give 15 points.
arlik [135]

Answer: Range - 12 Median - 59

Step-by-step explanation: To find the range order your numbers from least to greatest then subtract the smallest from the highest 61-49 = 12 then for the median do the same in ordering the numbers then count how many you have. The number in the middle is your median. If there is two add them and divided by two. 59. Hope this helped!

8 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
1 year ago
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