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Leto [7]
2 years ago
10

The ratio of losses to wins for Kyle's team is 3 to 2. If the team had played the same number of games, but had won twice as man

y of its games, what would the ratio of losses to wins have been? Express your answer as a common fraction.
Mathematics
2 answers:
Fiesta28 [93]2 years ago
4 0

Answer: The new ratio will be 1/4

Explanation: The initial ratio of losses to wins is 3 to 2. If we sum the numer of losses and wins 3 + 2 = 5 games, that means they loss 3 out of 5 games , and they win 2 out of 5 games.

So if they had won twice as many of the games, that is 2*2=4. And since the number of games is the same ( 5 ), then they would have won 4 games and loss only 1.

So the new ratio of losses to wins will be 1 to 4, or expressed in a fraction: 1/4

natali 33 [55]2 years ago
3 0

Answer: the ratio would ben 1 to 4, or (1/4) as a common fraction, this means that the number of games that the team lost is equal to 1/4 of the games that the team won.

Step-by-step explanation:

the ratio is 3 to 2

this means that if they played 5 games (because 3 + 2 = 5), 3 of those games were losses and 2 where wins.

Now, if they played the same number of games (5) and won twice as many of the games (2) them would have wined 2*2 = 4 of 5 games.

then the rate of loses to wins is:

1 to 4

this means that they lost 1 game for every 4 wins.

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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

h ----> the height of a ball in feet

t ---> the time in seconds

we have

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

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

so

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

Find the vertex

Convert the equation in vertex form

Factor -16

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

Complete the square

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}

the positive value is

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

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Answer:

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Step-by-step explanation:

7 0
2 years ago
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We have that
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using a graph tool
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Step-by-step explanation:

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4 0
3 years ago
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Andru [333]
<h3>Answer:   10^4</h3>

This is an exponential expression with 10 as the base and 4 as the exponent.

=============================================

Explanation:

We're dividing 10^8 over 10^4. When dividing like this, we subtract the exponents (numerator minus denominator). The bases must be the same value and they stay at the same value for the final answer as well.

The new exponent is 8-4 = 4 which is how we arrive at the answer 10^4.

Side note: 10^4 = 10,000 = ten thousand

7 0
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