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guapka [62]
3 years ago
15

Need help! Will mark Brainiest!

Mathematics
1 answer:
kicyunya [14]3 years ago
5 0

Option B: $y=-\sqrt[3]{x-4}+7$ is the new equation

Explanation:

The given equation is $y=-\sqrt[3]{x}$

We need to find the new graph which is shifted 7 units up and 4 units right.

First, we shall shift the graph 7 units up.

The general formula to shift the graph b units up is given by

y=f(x)+b

Thus, to shift the graph 7 units up, let us substitute b=7 and f(x)=-\sqrt[3]{x} in the general formula, we have,

y=$-\sqrt[3]{x}$+7

Now, we shall shift the graph 4 units right.

The general formula to shift the graph b units right is given by

y=f(x-b)

Thus, to shift the graph 4 units right, let us substitute b=4 and f(x)=$-\sqrt[3]{x}+7$ in the above equation, we have,

$y=-\sqrt[3]{x-4}+7$

Therefore, the new equation is $y=-\sqrt[3]{x-4}+7$

Therefore, Option B is the correct answer.

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Answer:

Lola was in lead for \frac{6}{25} of a mile.

Step-by-step explanation:

We have been given that Lola and Brandon had a running race. They ran 3/10 of a mile. Lola was in the lead for 4/5 of the distance.

To find the fraction of a mile for which Lola was in the lead, we need to find 4/5 of 3/10 as:

\text{Fraction for which Lola was in lead}=\frac{3}{10}\times\frac{4}{5}

\text{Fraction for which Lola was in lead}=\frac{3}{5}\times\frac{2}{5}

\text{Fraction for which Lola was in lead}=\frac{3\times2}{5\times5}

\text{Fraction for which Lola was in lead}=\frac{6}{25}

Therefore, Lola was in lead for \frac{6}{25} of a mile.

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Answer:

  5 hours

Step-by-step explanation:

A quick way to look at this is to compare the difference in hourly charge to the difference in 0-hour charge.

The first day, the charge is $3 more than $12 per hour.

The second day, the charge is $12 less than $15 per hour.

The difference in 0-hour charges is 3 -(-12) = 15. The difference in per-hour charges is 15 -12 = 3. The ratio of these is ...

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The charges are the same after 5 hours.

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If you write equations for the charges, they will look like ...

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Equating these charges, we have ...

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You might notice that the math here is very similar to that described in words, above.

The charges are the same after 5 hours.

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