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marishachu [46]
2 years ago
13

Naomi‘s fish is 40 mm long. Her guinea pig is 25 cm long. How much longer is your guinea pig than her fish?

Mathematics
1 answer:
hjlf2 years ago
3 0

Answer:

21 cm

Step-by-step explanation:

25-4=21

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What is the sign of the product (–7)(–2)(–5)(1)?
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The product would be negative. Since there are three negative integers, the product would be negative. Two negatives make a positive, but another negative makes the end result negative.
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A man travelled 3/8 of his journey by a rail 1/4 by a taxi 1/8 by and the remaining 2km on foot what is the length of his total
Burka [1]
The man travelled in different ways: by rail, by taxi, by ___ and by foot. I placed a blank there because there seems to be a missing word in the given problem above. For sample purposes, let's just assume that is travel by bus.

Since all of these travels are equal to 1 whole journey, you can express each travel as a fraction. When you add them up, the answer would be 1. So,

3/8 + 1/4 + 1/8 + x = 1

The variable x here denotes the fraction of his travel by foot. We are only given the exact distance travelled on foot which is 2 km. We have to find the fraction of the travel by foot to determine the length of the total distance travelled. Solving for x,

x = 1 - 3/8 - 1/4 - 1/8
x = 1/4

That means that the travel by foot comprises 1/4 of the whole journey. Thus,

Let total distance be D.

1.4*D = 2 km
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4 0
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

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The answer would be C. both plane A and B because it's on line s which lies on both planes.
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Answer:

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Step-by-step explanation:

-1/2(-3y+10)

Expand the brackets.

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Multiply.

3/2y - 5

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2 years ago
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