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timurjin [86]
1 year ago
14

Over the summer, Nancy exercised daily and lost from 145 pounds to 132 pouds. What is the percent of change?

Mathematics
1 answer:
jeka57 [31]1 year ago
5 0

Answer:

8.97\text{ \%}

Explanation: Over the summer, Nancy exercised daily, and her weight went 145 pounds to 132 pounds. we need to find the percent change:

Change in weight is:

145p-132p=13p

Change is percent is:

\frac{\text{ Change in Weight}}{\text{Original Weight}}\times100=(\frac{13}{145})\times100=8.97\text{ \%}

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Elden [556K]

Answer:

1) x_1 = \frac{2 - 2\sqrt{13}}{2}= 1-\sqrt{13}

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2) x_1 = \frac{6 - 2\sqrt{10}}{1}= 6-2\sqrt{10}

x_2 = \frac{6 + 2\sqrt{10}}{1}= 6+2\sqrt{10}

3) p_1 = \frac{-8 - 2\sqrt{30}}{4}= -2-\frac{1}{2}\sqrt{30}

p_2 = \frac{-8 + 2\sqrt{30}}{4}= -2+\frac{1}{2}\sqrt{30}

4) y_1 = \frac{-3 - 9}{4}=-3

y_2 = \frac{-3 + 9}{4}= \frac{3}{2}

Step-by-step explanation:

The quadratic formula is given by:

x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}

We can use this formula in order to solve the following equations:

1. x^2 − 2x = 12 → a = 1, b = −2, c = −12

For this case if we apply the quadratic formula we got:

x = \frac{-(-2) \pm \sqrt{(-2)^2 -4(1)(-12)}}{2(1)}

x_1 = \frac{2 - 2\sqrt{13}}{2}= 1-\sqrt{13}

x_2 = \frac{2 + 2\sqrt{13}}{2}= 1+\sqrt{13}

2. 1/2x^2 − 6x = 2 → a = 1 / 2, b = −6, c = −2

For this case if we apply the quadratic formula we got:

x = \frac{-(-6) \pm \sqrt{(-6)^2 -4(1/2)(-2)}}{2(1/2)}

x_1 = \frac{6 - 2\sqrt{10}}{1}= 6-2\sqrt{10}

x_2 = \frac{6 + 2\sqrt{10}}{1}= 6+2\sqrt{10}

3. 2p^2 + 8p = 7 → a = 2, b = 8, c = −7

For this case if we apply the quadratic formula we got:

p = \frac{-(8) \pm \sqrt{(8)^2 -4(2)(-7)}}{2(2)}

p_1 = \frac{-8 - 2\sqrt{30}}{4}= -2-\frac{1}{2}\sqrt{30}

p_2 = \frac{-8 + 2\sqrt{30}}{4}= -2+\frac{1}{2}\sqrt{30}

4. 2y^2 + 3y − 5 = 4 → a = 2, b = 3, c = −9

For this case if we apply the quadratic formula we got:

y = \frac{-(3) \pm \sqrt{(3)^2 -4(2)(-9)}}{2(2)}

y_1 = \frac{-3 - 9}{4}=-3

y_2 = \frac{-3 + 9}{4}= \frac{3}{2}

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4 years ago
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Answer:

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Answer:

0 is the answer of to question

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Answer:

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using the law of logarithms

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John should have found a common denominator.
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