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bazaltina [42]
3 years ago
10

Divide f(x) = | x + 20| by g(x) = x + 1.

Mathematics
1 answer:
Rudik [331]3 years ago
3 0

Answer:

ii am not sure sooo ueah lol

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How can you use similar triangles to solve problems?
mars1129 [50]

Answer:

you can use similar triangle to make known degrees in problems to make then easier to solve. With similar triangle, the angles are the same, but the scale is different. So by using this, one can solve both at the same time, and just just scale up the smaller one or scale down the larger, by the given/found scale.

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Find the sum or difference. a. -121 2 + 41 2 b. -0.35 - (-0.25)
s344n2d4d5 [400]

Answer:

2

Step-by-step explanation:

The reason an infinite sum like 1 + 1/2 + 1/4 + · · · can have a definite value is that one is really looking at the sequence of numbers

1

1 + 1/2 = 3/2

1 + 1/2 + 1/4 = 7/4

1 + 1/2 + 1/4 + 1/8 = 15/8

etc.,

and this sequence of numbers (1, 3/2, 7/4, 15/8, . . . ) is converging to a limit. It is this limit which we call the "value" of the infinite sum.

How do we find this value?

If we assume it exists and just want to find what it is, let's call it S. Now

S = 1 + 1/2 + 1/4 + 1/8 + · · ·

so, if we multiply it by 1/2, we get

(1/2) S = 1/2 + 1/4 + 1/8 + 1/16 + · · ·

Now, if we subtract the second equation from the first, the 1/2, 1/4, 1/8, etc. all cancel, and we get S - (1/2)S = 1 which means S/2 = 1 and so S = 2.

This same technique can be used to find the sum of any "geometric series", that it, a series where each term is some number r times the previous term. If the first term is a, then the series is

S = a + a r + a r^2 + a r^3 + · · ·

so, multiplying both sides by r,

r S = a r + a r^2 + a r^3 + a r^4 + · · ·

and, subtracting the second equation from the first, you get S - r S = a which you can solve to get S = a/(1-r). Your example was the case a = 1, r = 1/2.

In using this technique, we have assumed that the infinite sum exists, then found the value. But we can also use it to tell whether the sum exists or not: if you look at the finite sum

S = a + a r + a r^2 + a r^3 + · · · + a r^n

then multiply by r to get

rS = a r + a r^2 + a r^3 + a r^4 + · · · + a r^(n+1)

and subtract the second from the first, the terms a r, a r^2, . . . , a r^n all cancel and you are left with S - r S = a - a r^(n+1), so

(IMAGE)

As long as |r| < 1, the term r^(n+1) will go to zero as n goes to infinity, so the finite sum S will approach a / (1-r) as n goes to infinity. Thus the value of the infinite sum is a / (1-r), and this also proves that the infinite sum exists, as long as |r| < 1.

In your example, the finite sums were

1 = 2 - 1/1

3/2 = 2 - 1/2

7/4 = 2 - 1/4

15/8 = 2 - 1/8

and so on; the nth finite sum is 2 - 1/2^n. This converges to 2 as n goes to infinity, so 2 is the value of the infinite sum.

8 0
3 years ago
Eva wants to make two pieces of pottery. She needs 3/5 pound of clay for one piece and 7/10 pound of clay for the other piece. S
kow [346]
The answer is 1.1 pounds left.
4 0
3 years ago
What's the value of 8 in the number 56, 782, 010, 000
Doss [256]

okay so 56,782,010,000 and you are just looking at the 8 so  you could shorten it to 82,000,000 what is the value of the 8?

<span>82000000, eighty-two million
</span>
3 0
3 years ago
Read 2 more answers
The area of a square is 625 cm^2. What is the measure of the diagonal to the nearest tenth?
anyanavicka [17]
Area = 625 cm^2


Area = s^2
Diagonal = s*sqrt(2)



625 = s^2
s = sqrt(625)
s = 25 cm.


Diagonal = s*sqrt(2)
= 25sqrt(2)
=35.36 cm.
=35.4 cm.
4 0
3 years ago
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