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3 years ago

Please help me If you give me a good answer I will award brainiest

Mathematics

2 answers:

SVEN [57.7K]3 years ago

5 0

Answer:

integer

Step-by-step explanation:

it is an integer

Yuliya22 [10]3 years ago

4 0

It’s an Integer

Because it’s an integer

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How many times does 6 go into 532

mezya [45]

Answer:

88.67

Step-by-step explanation:

divide 532 from 6

532/6 = 88.67

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3 years ago

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Suppose that the lifetimes of TV tubes are normally distributed with a standard deviation of 1.1 years. Suppose also that exactl

joja [24]

The TV equals 20% because 1.1-4 years equals points

7 0

3 years ago

Please answer correctly !!!! Will mark brainliest !!!!!

7nadin3 [17]

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f(x) = − 4x/5 − 8/5

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3 years ago

A container manufacturer plans to make rectangular boxes whose bottom and top measure 4x by 3x. The container must contain 48in.

alex41 [277]

The height of the container that will be able to minimize the cost will be 3.08cm.

<h3>How to calculate the height?</h3>

The volume of the box will be:

= (3x)(4x)h

= 12x²h

From the information given, we are told that the container must contain 48in³. Therefore,

48 = 12x²h

h = 4/x²

The function cost will be:

= 3.50(2)(12x²) + 4.40(14x)h

= 84x² + 61.6x(4/x²)

= 84x² + 246.4/x

We'll use the first derivative. This will be:

dC/dx = 168x - 246.4/x²

x = 1.14.

Therefore, the height will be:

h = 4/x² = 4/1.14² = 3.08cm

In conclusion, the height is 3.08cm.

Learn more about height on:

brainly.com/question/1557718

4 0

2 years ago

Find the smallest positive $n$ such that \begin{align*} n &\equiv 3 \pmod{4}, \\ n &\equiv 2 \pmod{5}, \\ n &\equiv

Alex777 [14]

4, 5, and 7 are mutually coprime, so you can use the Chinese remainder theorem right away.

We construct a number $x$ such that taking it mod 4, 5, and 7 leaves the desired remainders:

$x=3\cdot5\cdot7+4\cdot2\cdot7+4\cdot5\cdot6$

Taken mod 4, the last two terms vanish and we have

$x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod4$

so we multiply the first term by 3.

Taken mod 5, the first and last terms vanish and we have

$x\equiv4\cdot2\cdot7\equiv51\equiv1\pmod5$

so we multiply the second term by 2.

Taken mod 7, the first two terms vanish and we have

$x\equiv4\cdot5\cdot6\equiv120\equiv1\pmod7$

so we multiply the last term by 7.

Now,

$x=3^2\cdot5\cdot7+4\cdot2^2\cdot7+4\cdot5\cdot6^2=1147$

By the CRT, the system of congruences has a general solution

$n\equiv1147\pmod{4\cdot5\cdot7}\implies\boxed{n\equiv27\pmod{140}}$

or all integers $27+140k$ , $k\in\mathbb Z$ , the least (and positive) of which is 27.

3 0

3 years ago