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3 years ago

Jamal has $35 to spend at the mall. If pairs of shorts are $9.99, how many can he buy?

Mathematics

2 answers:

mixas84 [53]3 years ago

6 0

Answer: he can buy 3 pairs

Step-by-step explanation: $9.99 x 3 = $29.97 | $9.99 x 4 = $39.96

Mrac [35]3 years ago

5 0

$35 / $9.99 = 3.5 pairs of shorts

We have to round down because you cannot buy half a pair of shorts

The correct answer is 3 pairs of shorts.

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<h3>Answer: choice B) counterclockwise rotation of 90 degrees around the origin</h3>

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3 years ago

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3 years ago

Evaluate if m=-6 what is m/2 + m =

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Answer:

Step-by-step explanation:

they said that m=6 which you can plug in into the problem, now that you know what 'm' is. It will look like this:

6/2 + 6 = ?

6 divided by 2 = 3 so it'll look like this

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3 years ago

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Natasha2012 [34]

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3 years ago

Read 2 more answers

6. If the net investment function is given by

Pachacha [2.7K]

The capital formation of the investment function over a given period is the

accumulated capital for the period.

(a) The capital formation from the end of the second year to the end of the fifth year is approximately <u>298.87</u>.

(b) The number of years before the capital stock exceeds $100,000 is approximately <u>46.15 years</u>.

Reasons:

(a) The given investment function is presented as follows;

$I(t) = 100 \cdot e^{0.1 \cdot t}$

(a) The capital formation is given as follows;

$\displaystyle Capital = \int\limits {100 \cdot e^{0.1 \cdot t}} \, dt =1000 \cdot e^{0.1 \cdot t}} + C$

From the end of the second year to the end of the fifth year, we have;

The end of the second year can be taken as the beginning of the third year.

Therefore, for the three years; Year 3, year 4, and year 5, we have;

$\displaystyle Capital = \int\limits^5_3 {100 \cdot e^{0.1 \cdot t}} \, dt \approx 298.87$

The capital formation from the end of the second year to the end of the fifth year, C ≈ 298.87

(b) When the capital stock exceeds $100,000, we have;

$\displaystyle \mathbf{\left[1000 \cdot e^{0.1 \cdot t}} + C \right]^t_0} = 100,000$

Which gives;

$\displaystyle 1000 \cdot e^{0.1 \cdot t}} - 1000 = 100,000$

$\displaystyle \mathbf{1000 \cdot e^{0.1 \cdot t}}} = 100,000 + 1000 = 101,000$

$\displaystyle e^{0.1 \cdot t}} = 101$

$\displaystyle t = \frac{ln(101)}{0.1} \approx 46.15$

The number of years before the capital stock exceeds $100,000 ≈ <u>46.15 years</u>.

Learn more investment function here:

brainly.com/question/25300925

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2 years ago