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Andreyy89
4 years ago
7

Simplify (c^2 + 2c- 6)(c+5

Mathematics
1 answer:
Sunny_sXe [5.5K]4 years ago
7 0
Here's your answer
 c^2+7c-6
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3 5/8 +2 2/3 +7 3/4 what is the answer to this question
kumpel [21]
The answer is: 14\frac{1}{24}
5 0
3 years ago
Principal=2000,R=12percent,Time=2years and 6 months.​
almond37 [142]
  • Principal=P=2000
  • Rate of interest=12%
  • Time=2years and 6months=2.5years
  • Interest=I

\boxed{\sf I=\dfrac{PRT}{100}}

\\ \sf\longmapsto I=\dfrac{2000(12)(2.5)}{100}

\\ \sf\longmapsto I=\dfrac{24000(2.5)}{100}

\\ \sf\longmapsto I=\dfrac{60000}{100}

\\ \sf\longmapsto I=600

7 0
3 years ago
Read 2 more answers
Imarindra thers.otes
Wittaler [7]

Answer:

V=176\pi\ in^3

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

we know that

The given figure is in the shape of a cylinder

so

To find out how much grain can this container hold, determine the volume of the cylinder

The volume of the cylinder is given by the formula

V=\pi r^{2} h

we have

r=4\ in\\h=11\ in

substitute

V=\pi (4)^{2} (11)

V=176\pi\ in^3

4 0
3 years ago
Read 2 more answers
Answer the thing on screen. Make sure to use the right &lt; or &gt; thing
BlackZzzverrR [31]

Answer:

12 \leq v

Step-by-step explanation:

first you subtract 7 from 7 and 12

19\leq v+ 7\\-7           -7

then you get 19 \leq v

(btw the variable always goes first with these problems)

so the final answer would be v \geq 12

3 0
3 years ago
The sum of any two consecutive prime numbers is also prime
jek_recluse [69]
Let's consider two prime numbers where each is larger than 2

Say the primes 7 and 11. Adding them gets us 7+11 = 18. This counter example disproves the initial claim since 18 = 9*2 = 6*3 making 18 composite (not prime)

In general, if we let p and q be two primes such that q > p > 2 and q is the next prime after p, then p and q are both odd. If any of them were even then they wouldn't be prime (2 would be a factor)

Adding any two odd numbers together leads to an even number

Proof:
p = 2k+1 where k is some integer
q = 2m+1 where m is some integer
p+q = 2k+1+2m+1 = 2(k+m) + 2 = 2(k+m+1) which is in the form of an even number

That proof above shows us that adding any prime larger than 2 to its next prime up leads to an even number. This further shows us that the claim is false overall. It is only true if you restrict yourself to the primes 2 and 3, which add to 5. Otherwise, the claim is false. 
3 0
4 years ago
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