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Lubov Fominskaja [6]
3 years ago
14

is 3240 a perfect cube? if not then by which smallest number should 3240 be multiplied so that the product is a perfect cube?​

Mathematics
1 answer:
Debora [2.8K]3 years ago
5 0

Answer:

<h2>3240 it's not a perfect cube.</h2><h2>The smallest number should 3240 be multiplied so that the product is a perfect cube is 15² = 225.</h2>

Step-by-step explanation:

3240 is a perfect cube if 3240 = n³ (n ∈ N).

Use the Prime Factorization:

\begin{array}{c|c}3240&2\\1620&2\\810&2\\405&5\\81&3\\27&3\\9&3\\3&3\\1\end{array}\\\\3240=2\cdot2\cdot2\cdot5\cdot3\cdot3\cdot3\cdot3=2^3\cdot3^3\cdot5\cdot3=(2\cdot3)^3\cdot5\cdot3=6^3\cdot5\cdot3

3240=6^3\cdot15\qquad\text{multiply both sides by}\ 15^2\\\\3240\cdot15^2=6^3\cdot15^3=3240\cdot15^2=(6\cdot15)^3=90^3

Used:

a^n\cdot a^m=a^{n+m}\\\\(ab)^n=a^nb^m

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Three towns divided highway repair costs equally.
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Step-by-step explanation:

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3 years ago
A submarine started at the surface of the water and was moving at -15 meters per minute. The submarine traveled at this rate for
velikii [3]

Given :

A submarine started at the surface of the water and was moving at -15 meters per minute.

The submarine traveled at this rate for 48 minutes before coming to rest on the ocean floor.

To Find :

The change in the submarine position.

Solution :

We need to find the position of submarine position from the surface of water.

Distance travelled =  Speed × Time

Distance travelled = -15 m/min × 48 min

Distance travelled = 720 m

Therefore, the position of submarine is 720 m below the surface of ocean.

5 0
3 years ago
How do i figure out these problems ?
Andreyy89

Given two vectors

v=(a,b),\quad w=(c,d)

they are parallel if they are a multiple of each other:

c=ka,\quad d=kb

You can easily test this by checking if

\dfrac{c}{a}=\dfrac{d}{b}

they are orthogonal if their dot product is null:

v\cdot w=ac+bd=0

For example, in the first case, we have

\dfrac{10}{3}\div 2 \neq \dfrac{4}{3}\div 5

So, they aren't parallel. Similarly, you have

2\cdot \dfrac{10}{3}+5\cdot \dfrac{4}{3}\neq 0

So, they aren't orthogonal.

In the second case, we have

\dfrac{5}{-12}\neq \dfrac{-6}{-10}

So, they aren't parallel, and

5(-12)+(-6)(-10)=-60+60=0

So, they are orthogonal.

Finally, we have

\dfrac{2}{-4}=\dfrac{-7}{14}=-\dfrac{1}{2}

So, they are parallel (and thus can't be orthogonal)

7 0
3 years ago
35.
babunello [35]

Hi there! :)

Answer:

\huge\boxed{y = 4x - 2}

Given line with an equation of y = 4x + 3

Parallel lines contain equivalent slopes, so a parallel line to the given equation would contain a slope of m = 4.

Plug in the coordinates of the point given, along with the slope into the equation y = mx + b where:

m = slope

y = y-coordinate of point

x = x-coordinate of point

Solve for the 'b' value, or y-intercept:

y = mx + b

6 = 4(2) + b

6 = 8 + b

b = -2

Rewrite the equation as slope-intercept form:

y = 4x - 2

4 0
3 years ago
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Answer:

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