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1 year ago

9

Use mathematical induction to prove that, for n ≥ 1, n³ – n is divisible by 3.

1 answer:

ICE Princess25 [194]1 year ago

5 0

Answer:

Step-by-step explanation:

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What is the gcf of 48m^5n and 81m^2n^2

JulijaS [17]

Answer:

gcf is $3m^2n$

Step-by-step explanation:

we are given

$48m^5n$ and $81m^2n^2$

Firstly, we will find factors of both expression

$48m^5n=2\times 2\times2\times 2\times 3\times m^2\times m^3\times n$

$81m^2n^2=3\times 3\times 3\times 3\times m^2\times n\times n$

now, we will find common factors among them

we can see that

For finding gcf, we always find common factors

so,

common terms are 3,m^2 and n

so, gcf is

$=3\times m^2\times n$

$=3m^2n$

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2 years ago

you need to mix 42 pounds of mortar. each pound of mortar mix requires 0.03 liters of water. how many liters of water do you nee

mart [117]

You need to mix 1.26 liters of water

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3 years ago

Letters in the word mathematics

liubo4ka [24]

Answer:

<em>Answer is</em><em> </em><em>given below</em><em> </em><em>.</em>

Step-by-step explanation:

Letters in the word "<em>MATHEMATICS</em><em> </em><em>"</em>

<em>M</em><em>,</em><em>A</em><em>,</em><em>T</em><em>,</em><em>H</em><em>,</em><em>E</em><em>,</em><em>I</em><em>,</em><em>C</em><em>,</em><em>S</em><em>.</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>HAVE A NICE DAY</em><em>!</em>

<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>. </em>

3 0

3 years ago

Please solve with explanation I’ve been asking all day (this is not a multiple choice question)

Anastaziya [24]

Answer:

a) $SA = 522.9~cm^2$

b) $V_{cone} = 670.2~cm^3$

c) $V_{empty} = 1340.4~cm^3$

Step-by-step explanation:

a)

For a cone,

$SA = \pi r (L + r)$

where L = slant height

$L = \sqrt{r^2 + h^2}$

We have r = 8 cm; h = 10 cm

$L = \sqrt{(8~cm)^2 + (10~cm)^2}$

$L = \sqrt{164~cm^2}$

$SA = (\pi)(8~cm)(\sqrt{164~cm^2} + 8~cm)$

$SA = 522.9~cm^2$

b)

$V_{cone} = \dfrac{1}{3}\pi r^2 h$

$V_{cone} = \dfrac{1}{3}(\pi)(8~cm)^2(10~cm)$

$V_{cone} = 670.2~cm^3$

c)

$V_{cylinder} = \pi r^2 h$

empty space = volume of cylinder - volume of cone

$V_{empty} = V_{cylinder} - V_{cone}$

$V_{empty} = \pi r^2 h - \dfrac{1}{3}\pi r^2 h$

$V_{empty} = (\pi)(8~cm)^2(10~cm) - \dfrac{1}{3}(\pi)(8~cm)^2(10~cm)$

$V_{empty} = 1340.4~cm^3$

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3 years ago

Help,, i'll mark as brainliest

PtichkaEL [24]

The first question would be distance from the start, as it is steadily going up as time goes on.

The second question would be distance from the end, as it is steadily going down as time goes on.

The third question would be speed, as the speed is staying stable as shown by the straight lines seen within the distance from start/end graphs being linear lines.

7 0

2 years ago