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3 years ago

5

PLease help me out with this...I will mark the first one as the brainiest answer.

2 answers:

Slav-nsk [51]3 years ago

3 0

You cannot add row 2 to column 3 because they have different dimensions. You can do any of the other operations, but the only one that makes any sort of sense is ...
Multiply row 2 by -1 and add it to row 3

_____
It makes no sense to multiply a row by zero. That makes the entire row zero and makes the matrix useless for finding any sort of solution.

You can switch columns, but that doesn't get you any closer to a solution here.

If I were trying to find a solution, I might
switch rows 1 and 2
multiply the new row 1 by -3 and add it to the new row 2
multiply the new row 1 by 2 and add it to row 3
This sequence of operations will make the first column [1 0 0], reducing the problem to 2×2 from 3×3.

noname [10]3 years ago

3 0

Answer:

yoooooooooooooooooo

Step-by-step explanation:

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What is the line slope of (10,5) and (1,9)

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Use the slope formula
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so it would be
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Hope that helped :-)

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3 years ago

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3 years ago

Use slope-intercept method help me

g100num [7]

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Step-by-step explanation:

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7 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

6 0

2 years ago