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Step2247 [10]
3 years ago
14

N is an integer. Write the values of n such that -15 < 3n <6 I need know please

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
3 0

Answer:

-5<n≤2

Step-by-step explanation:

Divide 3 by -15 nad 6 since we need to find the "n" we need to divide on all sides, so -15 divided by 3 is -5 so replace -15 with -5 and 6 divided by 3 is 2 so replace 6 with 2

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Ymorist [56]
Okay!

1. (12,12)
2. (10,14)
3. (8,16)
4. (6,18)
5. (4,20)
6. (2,22)

Hope this helped!

:)
6 0
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What is the answer for number 14
andrew-mc [135]
40% of 200 is 80 so 120 were sold at the door
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Domingo earns $13 dollars per hour and worked 47.5 hours that week. He is paid
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3 years ago
Find the area of the following regions, expressing your results in terms of the positive integer n ≥ 2. The region bounded by
cluponka [151]

Answer:

The area of the searched region is A= a+b+ \frac{2n}{n+1}- \frac{n(a^{\frac{n+1}{n} }+b^{\frac{n+1}{n} }) }{n+1}-2

Step-by-step explanation:

If you want to find the area of a region bounded by functions f(x) and G(x) between two limits (a,b), you have to do a double integral. you must first know which of the functions is greater than the other for the entire domain.

In this case, for 0<x<1, f(x)<g(x)

while for 1<x, g(x)<f(x).

Therefore if our domain is all real numbers superior to 0 (where the limit 0<a<1 and 1<b), we have to do 2 integrals:

A=A(a<x<1)+A(1<x<b)

A(a

A(1

A=a-1 +\frac{n}{n+1} - \frac{na^{\frac{n+1}{n} } }{n+1} +  b-1 + \frac{n}{n+1} - \frac{nb^{\frac{n+1}{n} } }{n+1} = a+b+ \frac{2n}{n+1}  - \frac{n(a^{\frac{n+1}{n} }+b^{\frac{n+1}{n} }) }{n+1}   -2

3 0
3 years ago
Sqrt45a^5 simplified
Anestetic [448]

\sqrt{45a^5}

[scomposition of 45]

\sqrt{9*5*a^5}

[9 = 3^2, must use this notation (not 3*3)]

\sqrt{3^2*5*a^5}

[We appy one of the proprieties of square roots]

\sqrt{3^2}* \sqrt{5}*\sqrt{a^5}

[now we semplify: we must take out as much as possible all the elements under roots]

[to do that, we must divide the esponent of each element with the index of square roots (2)]

so

\sqrt{3^2}, 2/2 = 1

\sqrt{5}, 1/2 = 0 with 1 of rest

\sqrt{a^5}, 5/2 = 2 with 1 rest

[well, after do that, we can take out the elements under tbhe square roots!]

The quotient of each division is the esponent of the element out of the root

The rest of each division is the esponent of the element under the root

so:

3^{1} (quotient = 1, see the first operation) * \sqrt{5} (rest = 1, see the second operation) * a^{2} (quotient = 2, see the third operation) * \sqrt{a^1} (rest = 1, see the third operation)

The final result is:

3 (=3^1) * a² * √5 * √a

3a²√5a

It's more intuitive and easy, but the explanation (necessary) is very long. If you have other questions, ask me here in the comments! Also sorry for my english, not so good!

8 0
4 years ago
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