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3 years ago

6

6 + (-4) = 6 + (-6) = ???

2 answers:

Mkey [24]3 years ago

8 0

Answer:

a}2

b}0

Step-by-step explanation:

=6+(-4)

=6-4

=2

=6+(-6)

=6-6

=0

adoni [48]3 years ago

4 0

6+(-4) is almost like saying 6-4 which is 2
Same for 6+(-6) is just like 6-6 which is 0

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The figures abov are similar find the missing length show your work 5in by 3in 3in by x

IgorLugansk [536]

Make each a ratio, and set them equal to each other.

Larger rectangle = 3/5
smaller rectangle = x/3

3/5 = x/3

cross multiply

3/5(5)(3) = x/3(3)(5)

3(3) = x(5)

9 = 5x

Isolate the x, divide 5 from both sides

9/5 = 5x/5

x = 9/5

x = 1.8

1.8 in. is your answer

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3 years ago

A storage bin is 12ft long 4ft wide 2ft high what is the volume in cubic ft

MariettaO [177]

Hello!

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3 years ago

Draw and lable a pentagon and a quadrilateral

Drupady [299]

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3 years ago

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vfiekz [6]

Answer:

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2 years ago

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

6 0

2 years ago