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3 years ago

10

I know how to find slope I need help finding out what to do after that

2 answers:

MariettaO [177]3 years ago

7 0

Y = mx + b is the slope intercept form of a linear equation.
m is the slope and b is the y intercept.
To find the y intercept you want to find the point where the graph crosses the y axis. On this graph you can see that it is 1.
The slope is rise/run. This is 2/1, so slope is 2.
Putting these into the equation we get y = 2x + 1
This is not an option because the answers are in standard form.
Changing to standard form we get 2x - y = -1
Letter C

Liula [17]3 years ago

5 0

I think its letter D because x + 2y fits with the equation and there for -3

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PLZ I WILL GIVE BRAILIEST TO THE FIRST ANSWER

natima [27]

Option B is the correct answer.

3 is greater than -5, -9, and -1.

Hint: Remember to distribute the negative when solving. A negative directly outside outside of and to the left the parenthesis times a negative number inside the parenthesis makes that number positive.

So for option B, you can simplify it to -3 +4 +2. And that equals 3.

A) -3+-4-(-2) = -5

B) -3-(-4)-(-2) = 3

C) -3+(-4)-2 = -9

D) -3-(-4)-2 = -1

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3 years ago

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Solve the system of equations using elimination<br> x-y=3<br> 2x+2y=50

pishuonlain [190]

Answer:

X=14

Y=11

Step-by-step explanation:

14-11=3

14x2=28

11x2=22

22+28=50

Hope that works :)

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3 years ago

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I need to know what the ansswer is to this we are doing pemdas

mamaluj [8]

Answer:

18

Step-by-step explanation:

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10-4+6(2)

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8 0

3 years ago

767074 rounded to the nearest hundred thousand

andriy [413]

767,074 rounded to the nearest hundred thousand is 800,000.

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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