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Sauron [17]
3 years ago
5

How do you divide fractions?

Mathematics
1 answer:
Kaylis [27]3 years ago
3 0
First flip the 2nd number. (denominator becomes numerator and vice versa) then multiply the numerators and multiply the denominators. that's the answer.
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Find the missing angle.<br> 100<br> 60<br> 68<br> 45<br> Will mark brain list !
Temka [501]
I think 45 because if you look at the top of the other 45 the graphs look the same lengths except this one stretches out so 45°
5 0
3 years ago
What is the median. mode. &amp; mean? ty!
Vikki [24]

Answer:

The median is: 9

The mode is: 9,10

The mean is: 8

Step-by-step explanation:

Hope this helps!

5 0
2 years ago
Read 2 more answers
I will give brainiest to whoever answers correctly !!
Kisachek [45]
I think the answer is $166,742.24
4 0
2 years ago
Given a right triangle with legs a, b and hypotenuse c, solve for b if a= 72 and c= 75
kotykmax [81]

Answer:

The answer is a. 21.

Step-by-step explanation:

We solve this question by using Pythagoras theorem to relate the sides of a right angled triangle.

Here,

Hypotenuse of the given triangle(c)=75

Two sides of right angled triangle are:

a=72

b=?

Then,

for a given right angled triangle abc,

Using Pythagoras theorem,

c=\sqrt{a^{2} +b^{2} }

Squaring on both sides,

c^{2} =a^{2}+b^{2}

or,75^{2}=72^{2}+b^{2}

or, b^{2}=75^{2}-72^{2}

or,b^{2}=5625- 5184

or,b^{2} =441

∴b=21

So, the value of b is obtained to 21 by the use of Pythagoras theorem.

7 0
3 years ago
What is the sum of the first 70 consecutive odd numbers? Explain.
expeople1 [14]

The sum of the first n odd numbers is n squared! So, the short answer is that the sum of the first 70 odd numbers is 70 squared, i.e. 4900.

Allow me to prove the result: odd numbers come in the form 2n-1, because 2n is always even, and the number immediately before an even number is always odd.

So, if we sum the first N odd numbers, we have

\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1

The first sum is the sum of all integers from 1 to N, which is N(N+1)/2. We want twice this sum, so we have

\displaystyle 2\sum_{i=1}^N i = 2\cdot\dfrac{N(N+1)}{2}=N(N+1)

The second sum is simply the sum of N ones:

\underbrace{1+1+1\ldots+1}_{N\text{ times}}=N

So, the final result is

\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1 = N(N+1)-N = N^2+N-N = N^2

which ends the proof.

5 0
2 years ago
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