Three irrational numbers are given below. Select the option that correctly plots the approximate value of each me on live where

Which equation represents the line that is perpendicular to and passes through (-40,20)?

emmasim [6.3K]

The equation of the perpendicular line to the given line is: y = -5/4x - 30.

<h3>What is the Equation of Perpendicular Lines?</h3>

The slope values of two perpendicular lines are negative reciprocal of each other.

Given that the line is perpendicular to y = 4/5x+23, the slope of y = 4/5x+23 is 4/5. Negative reciprocal of 4/5 is -5/4.

Therefore, the line that is perpendicular to it would have a slope (m) of -5/4.

Plug in m = -5/4 and (x, y) = (-40, 20) into y = mx + b to find b:

20 = -5/4(-40) + b

20 = 50 + b

20 - 50 = b

b = -30

Substitute m = -5/4 and b = -30 into y = mx + b:

y = -5/4x - 30

The equation of the perpendicular line is: y = -5/4x - 30.

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8 0

1 year ago

What are the first three common multiples of 8 and 12

Nata [24]

Answer:

24, 48, 72

Step-by-step explanation:

Common Multiples of 8: 8,16,24,32,40,48,56,64,72,80….

Common Multiples of 12: 12,24,36,48,60,72,………..

Hence the First 3 Least common multiple of 8 and 12= 24, 48, 72

7 0

2 years ago

A number increace by a 5 is 12

svetlana [45]

7 is the number increased by 5 = 12

8 0

2 years ago

Movie Ticket Sales : 6x + 9y = 1,500 Suppose on this day all tickets were on sale for the child price of $6. How many tickets we

dedylja [7]

The answer to the question

6 0

2 years ago

Read 2 more answers

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>.

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6 0

2 years ago