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3 years ago

Can someone tell me what 2 to the power of 6 is

Mathematics

2 answers:

Ira Lisetskai [31]3 years ago

8 0

${2}^{6} = 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2$

Fiesta28 [93]3 years ago

8 0

So basically:
2x2x2x2x2x2x=
4x2x2x2x2=
8x2x2x2=
16x2x2=
32x2=
64

The answer would be 64.

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Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

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t ---> the time in seconds

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The x-coordinate of the vertex represent the time when the ball reach the maximum

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Convert the equation in vertex form

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$h(t)=-16(t^{2}-\frac{25}{16}t)+5$

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$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}$

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}$

Rewrite as perfect squares

$h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

The vertex is the point $(\frac{25}{32},\frac{945}{64})$

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

$0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

$16(t-\frac{25}{32})^{2}=\frac{945}{64}$

$(t-\frac{25}{32})^{2}=\frac{945}{1,024}$

square root both sides

$(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}$

$t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}$

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$t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec$

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3 years ago

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