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kkurt [141]
3 years ago
6

How do you add a fraction with different denomirers

Mathematics
1 answer:
e-lub [12.9K]3 years ago
5 0
Hey there!
When we're adding with different denominators, our goal is to keep the equivalent fraction, but create like denominators.
Let's think of an easier situation. If we have the number 5 and we want an equivalent number, we multiply by one. It's no different with fractions. We want to multiply by some version of one, like 2/2 or 4/4
For example, if we have:
2/8 + 4/6

Our LCM is 24. Therefore, we multiply 2/8 by 3/3:
2/8(3/3) = 6/24

And 4/6 by 4/4:
4/6(4/4) = 16/24

As you can see, we multiplied by versions of 1, so they're still the same fraction.

We have:
16/24 + 6/24 = 22/24 = 11/12

Hope this helps!
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Use synthetic division to solve (3x^4+6x^3+2x^2+9x+10)/ (x+2)
zhenek [66]

Hello!

I believe the answer is (3x^2 - 3x + 5) • (x + 1)

I hope it helps!

7 0
3 years ago
Read 2 more answers
Here is an expression: 3+ t2 1 Evaluate the expression when t is 2​
Zanzabum

Step-by-step explanation:

it would be 45 if you add it as 21 but you didn't put another symbol hope I can help :)

4 0
3 years ago
In an effort to improve scores on college-entrance exams, the principal at a large high school chooses 35 students
denpristay [2]

Using the t-distribution, it is found that the test statistic for the hypotheses test is given by:

t = \frac{-1.75 - 0}{\frac{7.12}{\sqrt{35}}}

<h3>What are the hypotheses tested?</h3>

At the null hypotheses, it i tested if the mean difference remains the same, that is:

H_0: \mu  = 0

At the alternative hypotheses, it is tested if it has decreased, hence:

H_a: \mu < 0.

<h3>What is the test statistic?</h3>

The test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

The parameters are given as follows:

\overline{x} = -1.75, s = 7.12, n = 35.

Hence the test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{-1.75 - 0}{\frac{7.12}{\sqrt{35}}}

More can be learned about the t-distribution at brainly.com/question/16162795

#SPJ1

5 0
2 years ago
Please help<br><br> find the value of x in the given ratio.<br><br> 2/3=x/15
RoseWind [281]
The answer to this is x=3
4 0
3 years ago
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
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