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4 years ago

A⋅a/\2⋅a/\3=?? I need help ASAP

Mathematics

1 answer:

wolverine [178]4 years ago

7 0

Answer:

a^6

Step-by-step explanation:

a^1 * a^2 * a^3 = 1+2+3(add all power when same variable)

a^6

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rosijanka [135]

19. -4x(3x-5y)

20. 6x+ 20

21. $\frac{a+1}{2a}$

22. $\frac{5x-23}{4}$

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3 years ago

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A box contains 15 resistors. twelve of them are labelled 50ω and the other three are labeled 100ω. what is the probability that

Igoryamba

Answer : P(second resistor is 100ω , given that the first resistor is 50ω) is given by

$\frac{1}{5}$

Explanation :

Since we have given that

Total number of resistors =15

Number of resistors labelled with 50ω = 12

Number of resistors labelled with 100ω =3

Let A: Event getting resistor with 50ω

B: Event getting resistor with 100ω

Since A and B are independent events .

So,

$P(A\cap B)=P(A).P(B)$

Now, According to question , we can get that

$P(A)= \frac{12}{15}=\frac{4}{5}\\\\P(B)=\frac{3}{15}=\frac{1}{5}$

So,

$P(A\cap B)=P(A).P(B)\\\\P(A\cap B)=\frac{4}{5}\times \frac{1}{5}\\\\P(A\cap B)=\frac{4}{25}$

So, by using the conditional probability , which state that

$P(B\mid A)=\frac{P(A\cap B)}{P(A)}$

$P(B\mid A)=\frac{\frac{4}{25}}{\frac{4}{5}}\\\\P(B\mid A)=\frac{5}{25}\\\\P(B\mid A)=\frac{1}{5}$

So, P(second resistor is 100ω , given that the first resistor is 50ω) is given by

$\frac{1}{5}$

4 0

4 years ago

What are the prime factorization of 2100 in exponential form

Mama L [17]

2100|2
1050|2
525|3
175|5
    35|5
   7|7
   1|

2100 = 2 × 2 × 3 × 5 × 5 × 7 = 2² × 3 × 5² × 7

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4 years ago

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YALL REALLY NEED TO

Allisa [31]

Answer:

Not similar

Step-by-step explanation:

In triangle MLN & WXY,

Right angles - > 1

There is nothing common in these 2 triangles other than (1)

Therefore,

Not similar

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3 years ago

PLEASE HELP 50 POINT AND BRAINLIEST

shepuryov [24]

Hello!

You can't use the product of powers rule because the numbers do not have exponents

If they did have exponents you still won't be able to use it because the bases have to be the same

Hope this helps!

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3 years ago

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