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tatiyna
3 years ago
10

Please help with this math question

Mathematics
1 answer:
Igoryamba3 years ago
5 0

Answer:

243 x^{20} y^{10}

Step-by-step explanation:

Since they are all inside the bracket, they all get the exponent distributed.

This means that (3^5) * (x^4)^5 * (y^2)^5

The exponents get multiplied.

If for example it was x^5 * x^6 the exponent will be added, but here they are multiplied so we add them.

In our case they are exponent, so we multiply them.

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Read 2 more answers
What is the volume of a cone with a height of 6m and a diameter of 12m? Nearest meter.
AlexFokin [52]

Answer:

0.0005m^3

Step-by-step explanation:

V=1/3hπr²

h=6m

d=12m

r=12÷2=6m

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V=0.0005m^3

3 0
3 years ago
Can you please help me out
Taya2010 [7]

The bag contains,

Red (R) marbles is 9, Green (G) marbles is 7 and Blue (B) marbles is 4,

Total marbles (possible outcome) is,

\text{Total marbles = (R) + (G) +(B) = 9 + 7 + 4 = 20 marbles}

Let P(R) represent the probablity of picking a red marble,

P(G) represent the probability of picking a green marble and,

P(B) represent the probability of picking a blue marble.

Probability , P, is,

\text{Prob, P =}\frac{required\text{ outcome}}{possible\text{ outcome}}\begin{gathered} P(R)=\frac{9}{20} \\ P(G)=\frac{7}{20} \\ P(B)=\frac{4}{20} \end{gathered}

Probablity of drawing a Red marble (R) and then a blue marble (B) without being replaced,

That means once a marble is drawn, the total marbles (possible outcome) reduces as well,

\begin{gathered} \text{Prob of a red marble P(R) =}\frac{9}{20} \\ \text{Prob of }a\text{ blue marble =}\frac{4}{19} \\ \text{After a marble is selected without replacement, marbles left is 19} \\ \text{Prob of red marble + prob of blue marble = P(R) + P(B) = }\frac{9}{20}+\frac{4}{19}=\frac{251}{380} \\ \text{Hence, the probability is }\frac{251}{380} \end{gathered}

Hence, the best option is G.

5 0
2 years ago
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