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fomenos
2 years ago
8

What is 2/3 divided by 3/4

Mathematics
1 answer:
ad-work [718]2 years ago
8 0

Answer:

8/9

Step-by-step explanation:

It can sometimes be difficult to divide fractions, such as 2/3 divided by 3/4! When we divide two fractions, such as 2/3 ÷ 3/4, we flip the second fraction and then we simply multiply the numerators with each other and the denominator with each other. We also simplify the answers to fraction problems whenever possible.

Hope This Helps You!

-TheAnimeCatUwU

#LearnWithBrainly

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What is the factored form of x^3 + 125?​
klemol [59]

Answer:

(X+5)(x^2-5x+25)

Step-by-step explanation:

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3 years ago
How do you find surface area ? i don't remember. also new friends :) ?
Montano1993 [528]

Answer:

be my bestie

Step-by-step: The surface area is the number of square units that fit into the square. As shown in the picture, the surface area of this square is 16 total square units. With a rectangle and square we can also get the surface area by multiplying width (W) x length (L).tep explanation:

4 0
3 years ago
What is the value of the expression below after it has been simplified completely? (2)(3)-(6)(4) A. –24 B. –18 C. 18 D. 24
AlexFokin [52]

Answer:

B

Step-by-step explanation:

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4 0
3 years ago
1. Let a; b; c; d; n belong to Z with n > 0. Suppose a congruent b (mod n) and c congruent d (mod n). Use the definition
lukranit [14]

Answer:

Proofs are in the explantion.

Step-by-step explanation:

We are given the following:

1) a \equi b (mod n) \rightarrow a-b=kn for integer k.

1) c \equi  d (mod n) \rightarrow c-d=mn for integer m.

a)

Proof:

We want to show a+c \equiv b+d (mod n).

So we have the two equations:

a-b=kn and c-d=mn and we want to show for some integer r that we have

(a+c)-(b+d)=rn. If we do that we would have shown that a+c \equiv b+d (mod n).

kn+mn   =  (a-b)+(c-d)

(k+m)n   =   a-b+ c-d

(k+m)n   =   (a+c)+(-b-d)

(k+m)n  =    (a+c)-(b+d)

k+m is is just an integer

So we found integer r such that (a+c)-(b+d)=rn.

Therefore, a+c \equiv b+d (mod n).

//

b) Proof:

We want to show ac \equiv bd (mod n).

So we have the two equations:

a-b=kn and c-d=mn and we want to show for some integer r that we have

(ac)-(bd)=tn. If we do that we would have shown that ac \equiv bd (mod n).

If a-b=kn, then a=b+kn.

If c-d=mn, then c=d+mn.

ac-bd  =  (b+kn)(d+mn)-bd

          =    bd+bmn+dkn+kmn^2-bd

          =           bmn+dkn+kmn^2

          =            n(bm+dk+kmn)

So the integer t such that (ac)-(bd)=tn is bm+dk+kmn.  

Therefore, ac \equiv bd (mod n).

//

3 0
3 years ago
2k+10 show all the steps
12345 [234]

Hello!

Explanation:

↓↓↓↓↓↓↓↓↓↓

First, rewrite 10 as 2*5

2k+2*5

Then, factor out common term 2.

=2(k+5)

<em><u>Answer⇒⇒⇒⇒2(k+5)</u></em>

Hope this helps!

Thank you for posting your question at here on Brainly.

-Charlie

8 0
3 years ago
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