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Sauron [17]
3 years ago
8

67/20 in simplest form

Mathematics
2 answers:
OLEGan [10]3 years ago
5 0
It does not simplify ...

Decimal version: 3.35 sorry
GuDViN [60]3 years ago
3 0

Answer: it does not simplify

Step-by-step explanation:

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Determine whether the given value is a solution of the equation.<br> 7w = 73; w = 10
kipiarov [429]

Step-by-step explanation:

When w=10, 7w=7×10=70

But the given value of 7w is 73.

So, the given value is not a solution of the equation.

8 0
2 years ago
Which algebraic expression could NOT represent the phrase below?
skad [1K]

Answer:

B

Step-by-step explanation:

because 4 and 3 are grouped together you would add them first. this would change the outcome of the problem.

6 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
I need help to see if their equivalent help fast pls
Taya2010 [7]
They are <em><u>not</u></em><u> </u>equivalent. When you <em>distribute the 1/4</em>, the <em><u>expressions are different</u></em>. 
8 0
3 years ago
Mr.Jacobs has a $200 budget for school supplies. He purchased 12 reams of printer paper for $4 each and six packs of dry-erase m
Nana76 [90]
The answer will be 16
5 0
3 years ago
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