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likoan [24]
2 years ago
14

How many lines of symmetry does the picture have? A. 3 B. 2 C. 1 D. 0

Mathematics
1 answer:
brilliants [131]2 years ago
3 0

the answers is D cuz there is no line in the picture

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21) Rearrange the formula c2 = a2 + b2 for b.
Anvisha [2.4K]
C2 + a2 +2 = b
Would be the new formula
8 0
3 years ago
4380 rounded to the nearest 100​
maks197457 [2]

4,400

The number 3 is in the hundredths place the 8 is behind the 3 so it’s larger than 5 so you turn the 3 into a 4(4,400) and turn all the numbers behind zero (4,400).

4 0
3 years ago
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2 times a number that is at least 42
xxMikexx [17]

Answer:

21

Step-by-step explanation:

42 divided by 2 = 21. or 2×21= 42

Hope I helped!!✌

3 0
3 years ago
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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
Rearrange the equation so n is the independent variable.<br> m+1=-2(n+6)<br> m=
S_A_V [24]

Answer:

m=-2n-13

Step-by-step explanation:

m+1=-2(n+6)

m+1=-2n-12

   -1.        -1

--------------------

m=-2n-13

4 0
3 years ago
Read 2 more answers
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