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3 years ago

Please answer as many as you can from problems 2-8

Mathematics

1 answer:

klasskru [66]3 years ago

3 0

W0W alot........
2. 16/5c
3. 3q/2r^2
4. 1
5. yw^2/z^3x^3
6. 20j
7. 0 + m^2-m+m+1/m^2-m
8. 1- 1/b+2
H0P3 It H3LPS :)
#Itwashardwork

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Help me with this 1,2,3. For each sequence, Determine with whether it appears to be arithmetic If it does find a common differen

pshichka [43]

Problem 1

<h3>Answer: Arithmetic, common difference = -4</h3>

-------------------

Explanation:

Pick any term of the sequence, and subtract off the previous term to find that,

term2 - term1 = -12 - (-8) = -12 + 8 = -4
term3 - term2 = -16 - (-12) = -16 + 12 = -4
term4 - term3 = -20 - (-16) = -20 + 16 = -4

Each time we get the same result, so that means we have an arithmetic sequence with common difference -4

This indicates that adding -4 to each term, or subtracting 4 from each term, will generate the next one.

Eg: term2 = term1 - 4 = -8-4 = -12.

==========================================================

Problem 2

<h3>Answer: Arithmetic, common difference = 5</h3>

-------------------

Explanation:

Similar to problem 1, this sequence is also arithmetic because we add on 5 to each term to get the next one

-6+5 = -1
-1+5 = 4
4+5 = 9

Or you could subtract adjacent terms as done in problem 1, to find that the common difference is 5.

==========================================================

Problem 3

<h3>Answer: Not arithmetic</h3>

-------------------

Explanation:

Unlike the previous two problems, this sequence is not arithmetic.

We can see that

term2 - term1 = 12 - 3 = 9
term3 - term2 = 48 - 12 = 36

The gaps of 9 and 36 aren't the same. We need the same common difference between any adjacent terms to have an arithmetic sequence.

This sequence is instead geometric because

term2/term1 = 12/3 = 4
term3/term2 = 48/12 = 4
term4/term3 = 192/48 = 4

Each quotient is 4, showing the common ratio is 4. To find the next term, we multiply the current term by 4. So the next term after 192 would be 4*192 = 768, then 4*768 = 3072 is next, and so on.

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3 years ago

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11 x < - 66

x < - 66 / 11

x < - 6

hope this helps!

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5) You and a friend live 4 miles apart. You hear the sound of thunder 18 seconds before your friend hears it. Assuming sound tra

Leokris [45]

Answer:

see attached

Step-by-step explanation:

At 1100 ft per second for 18 seconds, the sound travels 19,800 ft, or 3.75 miles farther to my friend's house. The set of points that lie 3.75 miles farther from my friend's house than from my house form a hyperbolic curve. This is illustrated by the blue line in the attached graph. (My house is the red dot on the left; my friend's house is the red dot on the right.)

The lightning occurred somewhere on the blue curve.

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If the lightning occurred on the line between our houses, it was 1/8 mile from my house and 3 7/8 mile from my friend's house. (That's close!)

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The formula for the curve in the graph is the distance formula applied to the set of points (x, y). It equates the difference of distance from the two houses to 3.75 miles. If one were to write the equation of the hyperbola in standard form, the equation would look a little different and a restriction would need to be applied so the formula would describe only one branch of the hyperbola.

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3 years ago