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erica [24]
3 years ago
12

Write 10³ as a multiplication expression having repeated factors.

Mathematics
2 answers:
Gelneren [198K]3 years ago
8 0

Answer:

10 x 10 x 10

Step-by-step explanation:

Since 10^3 is raised to the power of 3, 10 is multiplied by itself 3 times.

10 x 10 x 10 = 10^3 = 1,000

Setler79 [48]3 years ago
6 0

Answer: 10 * 10*10

Step-by-step explanation:

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Y
alina1380 [7]

Answer:

the coordinates of the point of x and y is -1÷ 2 and -√3÷ 2

Step-by-step explanation:

The computation of the coordinates of the point is shown below:

The angle is

= 4π ÷ 3

The radius is 1 unit

Now

x = rcos\theta, and y = rsin\theta

Now

x = 1cos(4π ÷ 3) = -1÷ 2

y = 1sin(4π ÷ 3) = -√3÷ 2

hence, the corordinates of the point of x and y is -1÷ 2 and -√3÷ 2

8 0
3 years ago
Read 2 more answers
What value of x makes 2x-1=-7 true?
forsale [732]

Answer:x=-3

Step-by-step explanation:

2x-1=-7

Add 1 to each side

2x=-6

Divide each side by 2

x=-3

7 0
3 years ago
The half-life of cesium-137 is 30 years. Suppose we have a 180-mg sample. (a) Find the mass that remains after t years. (b) How
DedPeter [7]

Answer:

a) Q(t) = 180e^{-0.023t}

b) 11.4mg of cesium-137 remains after 120 years.

c) 225.8 years.

Step-by-step explanation:

The following equation is used to calculate the amount of cesium-137:

Q(t) = Q(0)e^{-rt}

In which Q(t) is the amount after t years, Q(0) is the initial amount, and r is the rate at which the amount decreses.

(a) Find the mass that remains after t years.

The half-life of cesium-137 is 30 years.

This means that Q(30) = 0.5Q(0). We apply this information to the equation to find the value of r.

Q(t) = Q(0)e^{-rt}

0.5Q(0) = Q(0)e^{-30r}

e^{-30r} = 0.5

Applying ln to both sides of the equality.

\ln{e^{-30r}} = \ln{0.5}

-30r = \ln{0.5}

r = \frac{\ln{0.5}}{-30}

r = 0.023

So

Q(t) = Q(0)e^{-0.023t}

180-mg sample, so Q(0) = 180

Q(t) = 180e^{-0.023t}

(b) How much of the sample remains after 120 years?

This is Q(120).

Q(t) = 180e^{-0.023t}

Q(120) = 180e^{-0.023*120}

Q(120) = 11.4

11.4mg of cesium-137 remains after 120 years.

(c) After how long will only 1 mg remain?

This is t when Q(t) = 1. So

Q(t) = 180e^{-0.023t}

1 = 180e^{-0.023t}

e^{-0.023t} = \frac{1}{180}

e^{-0.023t} = 0.00556

Applying ln to both sides

\ln{e^{-0.023t}} = \ln{0.00556}

-0.023t = \ln{0.00556}

t = \frac{\ln{0.00556}}{-0.023}

t = 225.8

225.8 years.

8 0
3 years ago
Read 2 more answers
You estimate there are 45 marbles in a jar. There are actually 50. What is your percent error?
Katen [24]

Answer:

10%

Step-by-step explanation:

8 0
3 years ago
Pi times 7/2 squared
Goshia [24]

5.49778714378 (Hope I helped!)

5 0
3 years ago
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