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4 years ago

What is the least common multiples of 5 and 15

Mathematics

1 answer:

Vinil7 [7]4 years ago

6 0

The answer would be 15. Hope this helped

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HELP! My teacher didn't teach half the material! How do I do this!?!?!?!

Sladkaya [172]

Answer:

Step-by-step explanation:

1-(-7.2x+3.5)=5+3(7.4x-1) first parenthesis

1+7.2x-3.5=5+22.2x-3

7.2x-2.5=2+22.2x (x on one side and numbers on other)

22.2x-7.2x= -2-2.5

15x=4.5 divide by 15

x=-4.5/15

5 0

3 years ago

If she was vaping every 15 minutes, assuming she was awake from 7 am- 11 pm (16 hours) how many times in a day would she vape?

Ray Of Light [21]

Answer:

64 times

Step-by-step explanation:

16 hours to minutes: 16*60 = 960 minutes

vaping every 15 minutes: 960/15 = 64

8 0

2 years ago

Hello people ~

Luden [163]

Cone details:

height: h cm

radius: r cm

Sphere details:

radius: 10 cm

================

From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

<u>Using Pythagoras Theorem</u>

(a)

TO² + TU² = OU²

(h-10)² + r² = 10² [insert values]

r² = 10² - (h-10)² [change sides]

r² = 100 - (h² -20h + 100) [expand]

r² = 100 - h² + 20h -100 [simplify]

r² = 20h - h² [shown]

r = √20h - h² ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

===========================

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (\sqrt{20h - h^2})^2 \ ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20h - h^2) (h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20 - h) (h) ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)$

To find maximum/minimum, we have to find first derivative.

(c)

<u>First derivative</u>

$\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )$

<u>apply chain rule</u>

$\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}$

<u>Equate the first derivative to zero, that is V'(x) = 0</u>

$\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0$

$\Longrightarrow \sf 40h-3h^2=0$

$\Longrightarrow \sf h(40-3h)=0$

$\Longrightarrow \sf h=0, \ 40-3h=0$

$\Longrightarrow \sf h=0,\:h=\dfrac{40}{3}$ <u />

<u>maximum volume:</u> <u>when h = 40/3</u>

$\sf \Longrightarrow max= \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )$

$\sf \Longrightarrow maximum= 1241.123 \ cm^3$

<u>minimum volume:</u> <u>when h = 0</u>

$\sf \Longrightarrow min= \dfrac{1}{3} \pi (0)^2(20-0)$

$\sf \Longrightarrow minimum=0 \ cm^3$

6 0

2 years ago

Read 2 more answers

Help needed! How do I solve this?<br>Excuse the answers that are already in there!

anastassius [24]

1) 108
2) 72
3) 36
4) 72
5) 72
6) 144
7) 72
8) 72
9) 108
10) 108
I hope this helped

8 0

3 years ago

(5x10)+(4x1)+(6x1/10)+(2x1/1000

Furkat [3]

This is the answer ,you should first do the multiplication then the sum.

5 0

3 years ago

Read 2 more answers