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alexdok [17]
3 years ago
11

How to complete the table

Mathematics
2 answers:
Pie3 years ago
8 0
The answer will be 21
Luba_88 [7]3 years ago
3 0
2- 21 Bc 7 times 3 equal 21
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Jim makes the chart shown below to prove that Triangle APD is congruent to triangle BPC:
Oliga [24]

Answer:

Given a square ABCD and an equilateral triangle DPC and given a chart with which Jim is using to prove that triangle APD is congruent to triangle BPC.

From the chart, it can be seen that Jim proved that two corresponding sides of both triangles are congruent and that the angle between those two sides for both triangles are also congruent.


Therefore, the justification to complete Jim's proof is "SAS postulate"



Step-by-step explanation:


6 0
3 years ago
Write an expression for the sequence of operations described below.
Murrr4er [49]

Answer:

<em>8</em><em>+</em><em>s</em><em>+</em><em>n</em>

Step-by-step explanation:

Add s to 8=8+s

and add 3 to result=8+s+3

3 0
3 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
3 years ago
Can someone help me slove this ?
Veronika [31]

Answer:

x<-1

Step-by-step explanation:

I think and hope it helps

6 0
3 years ago
Read 2 more answers
Help please. 10 points to who help me​
Lelechka [254]

9514 1404 393

Answer:

  64/109 ≈ 0.5872

Step-by-step explanation:

41 + 23 = 64 of the events have outcome E.

An additional 35 + 10 = 45 do not.

The probability of an outcome of E is 64 out of a total of 64+45 = 109 events.

  P(E) = 64/109 ≈ 0.5872

4 0
3 years ago
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