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2 years ago

-2021

Mathematics

1 answer:

sukhopar [10]2 years ago

6 0

Answer:

I'm not too sure what you mean but since there was no image I tried to look up how to tell if angles are corresponding based on angle degree and the most I got was the angles will be equal but only if there is a transversal cur but none of the answers equal each other

Step-by-step explanation:

so what advice I can give you is corresponding angles are usually the same degree but on a different level. like on the first parallel line there is an angle and on the second (or whatever) there is another angle (on the same side of the first) -meaning they will both be on the right or the left- and yeah there you go

if you have any questions comment and I'll try my best to help

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2 years ago

If sum of first 6 digits of AP is 36 and that of the first 16 terms is 255,then find the sum of first ten terms.

Cloud [144]

Answer:

100

Step-by-step explanation:

We have the sum of first n terms of an AP,

Sn = n/2 [2a+(n−1)d]

Given,

36= 6/2 [2a+(6−1)d]

12=2a+5d ---------(1)

256= 16/2 [2a+(16−1)d]

32=2a+15d ---------(2)

Subtracting, (1) from (2)

32−12=2a+15d−(2a+5d)

20=10d ⟹d=2

Substituting for d in (1),

12=2a+5(2)=2(a+5)

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∴ The sum of first 10 terms of an AP,

S10 = 10/2 [2(1)+(10−1)2]

S10 =5[2+18]

S10 =100

This is the sum of the first 10 terms.

Hope it will help.

4 0

2 years ago

Read 2 more answers

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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Agata [3.3K]

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2 years ago

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