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In-s [12.5K]
3 years ago
14

Other ways to express 10^4

Mathematics
1 answer:
prohojiy [21]3 years ago
4 0

Answer:

one way is 10000

Step-by-step explanation:

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The equation t^3=a^2 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A
ankoles [38]

Answer:

     2√2

Step-by-step explanation:

We can find the relationship of interest by solving the given equation for A, the mean distance.

<h3>Solve for A</h3>

  T^3=A^2\\\\A=\sqrt{T^3}=T\sqrt{T}\qquad\text{take the square root}

<h3>Substitute values</h3>

The mean distance of planet X is found in terms of its period to be ...

  D_x=T_x\sqrt{T_x}

The mean distance of planet Y can be found using the given relation ...

  T_y=2T_x\\\\D_y=T_y\sqrt{T_y}=2T_x\sqrt{2T_x}=(2\sqrt{2})T_x\sqrt{T_x}\\\\D_y=2\sqrt{2}\cdot D_x

The mean distance of planet Y is increased from that of planet X by the factor ...

  2√2

8 0
2 years ago
Which statements about the local maximums and minimums for the given function are true? Choose three options.
Nostrana [21]

Answer:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.

Step-by-step explanation:

The true statements are:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.  

Lets discuss each option one by one:

Over the interval [1, 3], the local minimum is 0

This is a false statement. Look at the graph. The minimum point given is (3.4,-8). Therefore the local minimum is -8 not 0

Over the interval [2, 4], the local minimum is –8.

This statement is true because the given minimum point is(3.4, -8). Thus the  local minimum is -8 which is true

Over the interval [3, 5], the local minimum is –8.

According to the given minimum point, the local minimum  is -8 which is true

Over the interval [1, 4], the local maximum is 0.

Look at the graph. The maximum point given is (2,0). Thus this statement is true because local maximum is 0.

Over the interval [3, 5], the local maximum is 0.

This is a false statement because there is no maximum point

8 0
3 years ago
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