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2 years ago

Which expression is equivalent to (4mn/m^-2n^6)^-2

Mathematics

1 answer:

Tanzania [10]2 years ago

4 0

The expression that is equivalent to (4mn/m^-2n^6)^-2 is $\frac{n^{10}}{16m^{6}}$

<h3>How to determine the equivalent expression?</h3>

The expression is given as:

(4mn/m^-2n^6)^-2

Apply the law of indices

$(4m^{1 + 2}n^{1-6})^{-2$

Evaluate the sum and the differences

$(4m^{3}n^{-5})^{-2$

Apply the negative exponent

$\frac{1}{16m^{2*3}}}n^{5*2}$

Evaluate

$\frac{1}{16m^{6}}n^{10}$

Rewrite as:

$\frac{n^{10}}{16m^{6}}$

Hence, the expression that is equivalent to (4mn/m^-2n^6)^-2 is $\frac{n^{10}}{16m^{6}}$

Read more about equivalent expressions at:

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Answer:

Step-by-step explanation:

Let us examine all options:

(a)

Lets verify for t = 0:

h(t = 0) = 7

This isn't allowable as 14 cm is the lower limit for the ball's motion.

This is rejected.

(b)

Let's verify for t = 0:

h(t = 0) = 14

Since at t = 0 sec, the ball is at its minimum, after 2 seconds, it should be at its maximum.

But h(t = 2) = 14, which doesn't satisfy the condition.

Hence, this is rejected.

(c)

Now, let's see at what time instance, the ball is at minimum in this case:

h(t) = 14 = 18sin(πt) + 32

∴ sin(πt) = -1

∴ πt = 3π/2

∴ t = 3/2 seconds

Hence, after 2 seconds, i.e. at 3.5 seconds, the ball should be at its maximum.

h(t = 3.5) = 18sin(3.5π) + 32 = -18 + 32 = 14, which doesn't satisfy the condition.

Hence, this is rejected,

(d)

Now, let's see at what time instance, the ball is at minimum in this case:

h(t) = 14 = 18sin(πt/2) + 32

∴ sin(πt/2) = -1

∴ πt/2 = 3π/2

∴ t = 3 seconds

Hence, after 2 seconds, i.e. at 5 seconds, the ball should be at its maximum.

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Hence, option (D) is the right answer

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4 years ago

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Nesterboy [21]

Answer:

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Step-by-step explanation:

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