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3 years ago

Which expression is equivalent to 5^{10} · 5^{5}

Mathematics

2 answers:

klemol [59]3 years ago

7 0

I think it’s a but I don’t know don’t listen to me

Wewaii [24]3 years ago

4 0

Answer:

C should be correct

Step-by-step explanation:

I am not sure if you or anybody else does this another way or maybe the same way, who knows.

Anyways, what I did here was just simply add the powers, and no, not the base, I mean what it is being multiplied by.

When I did this, I got 5^15, but I had to check to make sure, which is what I did

5^10x5^5 is 30517578125

And guess what.

5^15 = 30517578125

So the answer is C, and I hope my explanation helps in a way.

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ha purchased a gym membership a while ago. There was a one time $100 fee to join the gym and she had gone for 7 months. So far,

d1i1m1o1n [39]

It is 59.99 each month

8 0

4 years ago

Read 2 more answers

A particle moves along the x–axis so that at time t its position is given by x(t) = (t+1)(t–3)3. For what values of t is the vel

rosijanka [135]

I'll assume that's an exponent at the end there:

$x(t) = (t+1)(t-3)^3$

The first derivative gives the velocity. The second derivative gives the acceleration; increasing velocity is the same as positive acceleration. So we want to find when the second derivative is positive.

Let's see if we can use $d(uv) = u\, dv + v\, du$ to avoid multiplying this out.

$x'(t) = 3 (t+1)(t-3)^2 + (t-3)^3$

That worked; let's do it again:

$x''(t) = 3( 2(t+1)(t-3) + (t-3)^2) + 3(t - 3)^2$

$x''(t) = 3(t-3) ( 2(t+1) + (t-3) + (t - 3) ) = 3(t-3)(4t-4)=12(t-1)(t-3)$

That's a nice parabola. It's zero or negative for $1 \le t \le 3$ so positive everywhere else:

Answer: Increasing velocity when t < 1 or t > 3

4 0

3 years ago

Amar and his friends went to a movie at 3:55 P.M. The movie ended at 5:30 P.M.

kotegsom [21]

Answer: the movie was 1 hour and 35 minutes and Amar got home at 6:15 P.M.

Step-by-step explanation:

to find out how long the movie was you have to find the distance between the time frames 3:55 and 5:30.

3:55+1 hour= 4:55

4:55+5 minutes= 5:00

5:00+30 minutes= 5:30

Add the times to find how long the movie was.

This is how I found how long the movie was.

To find out when Amar got home you have to do the same thing as well.

5:30+30 minutes= 6:00

6:00+15 minutes= 6:15

Hope this helps!

6 0

3 years ago

4 in. 9 in. 12 in. 3 in. 4 in. 6 in.

IgorC [24]

Answer:

$\tt v=336\:in^3$

Step-by-step explanation:

$\tt v=(4)(12)(4)+(3)(4)(2)$

$\tt v=336\:in^3$

7 0

2 years ago

Simplify. −9x − 3 − (5x + 3) − 7

Ket [755]

Answer: I got -14x-13

But it sounds wrong but I have a picture

Step-by-step explanation:

4 0

4 years ago