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4 years ago

What is the least common denominator for the fractions 5/6 and 4/5 ?

Mathematics

1 answer:

laila [671]4 years ago

5 0

C.30 because if u multiply the denominators you get 30 not the other numbers

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Which one? 10 points

Butoxors [25]

I believe it’s bbbbb

7 0

2 years ago

The diagram shows an empty cylindrical container. Ana puts a solid cube of side length 8 cm into the container. She then pours 1

marshall27 [118]

Answer:

<h3><u>Cylinder</u></h3>

$\textsf{Volume of a cylinder}=\sf \pi r^2 h \quad\textsf{(where r is the radius and h is the height)}$

Given:

r = 6 cm

h = 18 cm

$\begin{aligned}\implies \sf V &= \sf \pi (6)^2(18)\\& = \sf 648 \pi \: cm^3\end{aligned}$

$\textsf{Volume of a cube}=\sf x^3\quad\textsf{(where x is the side length)}$

Given:

x = 8 cm

$\begin{aligned}\implies \sf V &= 8^3\\& = \sf 512 \: cm^3\end{aligned}$

<h3><u>Volume available to be filled with water</u></h3>

Volume of cylinder - volume of cube

= 684π - 512

= 1532.75204 cm³

1 litre = 1000 cm³

⇒ 1.5 litres = 1000 × 1.5 = 1500 cm³

As 1500 < 1532.75204, the volume of water poured into the container is smaller than the empty space available in the cylinder. Therefore, the water will <u>not</u> come over the top of the container.

4 0

2 years ago

36 of 80 marbles are blue. What percent of the marbles are blue

Masja [62]

If 36 of 80 marbles are blue, then the percent of marbles that are blue is 44% i think

5 0

3 years ago

Read 2 more answers

A local Chinese restaurant sells combination meals for $9.50 a piece. If the cost of an order is proportional to the number of m

lakkis [162]

Answer:

Step-by-step explanation:

bghgytythghgf66ef

4 0

3 years ago

Can someone explain to me why are we adding 2kpi when we are doing zeros for sin and cos, but adding kpi when doing zeros for tg

makvit [3.9K]

on the first exercise, you got a solution angle of π/18, that's a good solution for the I Quadrant only, however, on a circle, we have angles that go from 0 to 2π, however we can always keep on going around and continute to 2π + π/2 or 3π or 4π, or 115π/3 or 1,000,000π/18 and so on, and we're really just going around the circle many times over, getting a larger and larger angle, same circular motion.

π/18 on that exercise works for the I Quadrant, however if we continue and go around say 2π, we'll find that 2π/3 + π/18 is a coterminal angle with π/18, and thus that angle has also the same sine value.

π/18 + 2kπ/3 , where k = integer, is a way to say, all angles around the circle that look like this have the same sine, namely

π/18 + 2(1)π/3

π/18 + 2(2)π/3

π/18 + 2(3)π/3

π/18 + 2(5)π/3

π/18 + 2(99999999)π/3

.....

so using the "k" as some sequence multiplier, is a generic notational way to say, "all these angles".

you'll also find that "n" is used as well for the same notation, say for example

2π/3 + 2πn.

8 0

3 years ago