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notsponge [240]
3 years ago
6

Factorise 3x^2 +37x+70

Mathematics
1 answer:
erik [133]3 years ago
5 0

Answer:

If it needs to be factored by grouping the answer would be (3x+7) and (x+10).

Step-by-step explanation: I don't know how to explain but I hope this helped.

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Plz help! read the instructions and write a proper answer and if you don't i will report you! And the first one to answer right
ahrayia [7]

Answer:

1 = 0.61, 2 = 1.13

Step-by-step explanation:

The first is 25 + 25 + 10 + 1. This is 61 cents which is 0.61 of 1 dollar. The second is 100 + 10 + 1 + 1 + 1. This is equal to 113, 0r 1.13

BRAINLIEST PLEASE

3 0
2 years ago
What is the sum of 2x2 + 6x + 5 and 3x2 – 2x – 1?
Kipish [7]
5x2 + 4x + 4

Hope this helps!!!!!

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3 0
3 years ago
Find The Sum. 6 2/3 + 1 5/6=
patriot [66]
6 2/3 + 1 5/6 = 8 1/2

i am a mathematics teacher. if anything to ask please pm me
5 0
2 years ago
Read 2 more answers
A certain positive integer has exactly 20 positive divisors. What is the smallest number of primes that could divide the integer
Alika [10]

As per my explanation to (b) above, the largest number of primes that could factor such a number is 4.

Note that  2,3,5 and 7   are the smallest primes, then use the reasoning from

(b) above. we are looking for four exponents, that, when 1 is added to each and all are multiplied together, would equal 20.

But no such integers  k, l, m, and n  exist such that (k + 1)(l + 1)(m + 1) (n + 1)  = 20 where k, l,m, and n ≥ 1 so this number, whatever it is, can't have 4 prime factors

Let's drop 7 out of the mix and suppose it has just 3 prime factors 2, 3, and 5 again we are looking for three exponents,  that, when 1 is added to each and all are multiplied together, would equal 20.  Put another way, we are looking for k, l and m ≥ 1   such that (k + 1)(l + 1)(m + 1) = 20

Note that the  only possibility here  is when we have 2 *2 *5  = 20 and the smallest possible product would be 2^(4 )* 3^(1) * 5^(1) =  2^4 * 3 * 5 = 240

Now......the only remaining possibility is  that this number is composed of the two smallest primes, 2 and 3,  and we are looking for  some k  and l ≥ 1 such that (k + 1)(l + 1) = 20 clearly, the only possibilities  are when k = 4 and l = 5, or vice-versa

So this number would factor as either 2^3 * 3^4   = 648  or 2^4 * 3^3 = 432 and both are > 240.

Learn more about positive integers at

brainly.com/question/1367050

#SPJ4

8 0
1 year ago
PLEASE HELP ME WITH THIS
valentinak56 [21]

Answer:

Ionic Compound

Step-by-step explanation:

:D

3 0
2 years ago
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