All categories

3 years ago

Simplify: m^2n^3 aanm^2a^2n^4

Mathematics

1 answer:

dolphi86 [110]3 years ago

3 0

9514 1404 393

Answer:

a^4·m^4·n^8

Step-by-step explanation:

The applicable rule of exponents is ...

(a^b)(a^c) = a^(b+c)

m^2n^3 aanm^2a^2n^4

= a^(1+1+2)·m^(2+2)·n^(3+1+4)

= a^4·m^4·n^8

You might be interested in

A water sprinkler has a range of 5 meters as shown. The jet of water from the sprinkler sweeps out at an angle of 85° as the noz

telo118 [61]

I found the image that accompanied this problem.
We need to solve for the area of the sector.

A = n/360 * π * r²
A = 85/360 * 3.14 * 5²
A = 0.2361 * 3.14 * 25
A = 18.5347

Answer is C. 18.5 square meters.

6 0

3 years ago

Read 2 more answers

You can talk on the phone y minutes on day x of month according to the equation y=2x+1 the cell phone company clams you talked 1

ValentinkaMS [17]

Answer: They are wrong

Step-by-step explanation: x is given a value in the problem. Since x equals 4, We plug x into the equation.

y = 2(4) + 1

Y is actually equal to 9 minutes, not twelve.

4 0

3 years ago

Based on the information below, choose the correct answers.

lozanna [386]

Since there are 120 payment and the total interest paid was $ 14,644.95. so the average monthly interest can be calculated by:

Average monthly interest = $ 14,644.95 / 120

Average monthly interest = $ 122.04

% total interest = ($ 14,644.95) / $ 39,644.95 x 100

% total interest =36.94 %

8 0

3 years ago

On Mars the acceleration due to gravity is 12 ft/sec^2. (On Earth, gravity is much stronger at 32 ft/sec^2.) In the movie, John

insens350 [35]

Solution :

Given initial velocity, v= 48 ft/s

Acceleration due to gravity, g = $12\ ft/s^2$

a). Therefore the maximum height he can jump on Mars is

$H_{max}=\frac{v^2}{2g}$

$H_{max} = \frac{(48)^2}{2 \times 12}$

= 96 ft

b). Time he can stay in the air before hitting the ground is

$T=\frac{2v}{g}$

$T=\frac{2 \times 48}{12}$

= 8 seconds

c). Considering upward motion as positive direction.

v = u + at

We find the time taken to reach the maximum height by taking v = 0.

v = u + at

0 = 16 + (12) t

$t=\frac{16}{12}$

$=\frac{4}{3} \ s$

We know that, $S=ut + \frac{1}{2}at^2$

Taking t = $=\frac{4}{3} \ s$ , we get

$S=16 \times\frac{4}{3} + \frac{1}{2}\times(-12) \times \left(\frac{4}{3}\right)^2$

$S=\frac{32}{3}$ feet

Thus he can't reach to 100 ft as it is shown in the movie.

d). For any jump whose final landing position will be same of the take off level, the final velocity will be the initial velocity.

Therefore final velocity is = -16 ft/s

3 0

3 years ago

A survey of 123 college students was done to find out what elective courses they were taking. Let A = the set of those taking ar

Ilya [14]

Answer:

Step-by-step explanation:

8 0

2 years ago