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3 years ago

Write a number in which the value of the 3 is 10 times the value of the digit 3 in 156.32

Mathematics

2 answers:

Alex3 years ago

7 0

Answer: The number in which the value of the 3 is 10 times the value of the digit 3 in 156.32 = 3

Step-by-step explanation:

The given number = 156.32

The place value of 3 in 156.32 is 3 tenth [since it the left most number after the decimal].

3 tenth can be written as $\frac{3}{10}$

The number 10 times $\frac{3}{10}$ = $10\times\frac{3}{10}$

Therefore, The number 10 times $\frac{3}{10}$ =3

Hence, the number in which the value of the 3 is 10 times the value of the digit 3 in 156.32 = 3

Marat540 [252]3 years ago

5 0

Answer:

1563.2

Step-by-step explanation:

If you just multiply 0.3 by 10 you will get 3

Now all you need to do is write 3 or have a number which has unit place value of 3

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A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

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$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

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$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

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$t = 2.60s$

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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