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3 years ago

What is an equation of the line that passes through the points (-6, -3) and (-3, -5)?

Mathematics

1 answer:

RUDIKE [14]3 years ago

6 0

Answer:

m= -2/3

Step-by-step explanation:

m= y2-y1/ x2-x1

m= -5-(-3)/ -3-(-6)

m= -5+3/ -3+6

m= -2/3

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Which of these number words represents the numeral 4,056? four thousand, five hundred six four thousand, five hundred fifty-six

hjlf

Solution:

we have been asked to find

Which of the given number words represents the numeral 4,056?

Its about the counting.

On unit place we have 6.

on Tenth place we have 5.

on hundredth place we have 0.

On Thousand place we have 4.

so in our number word representation there will be no "Hundreds".

So it will make

Hence the given number words representing the numeral 4,056 is Four thousands and fifty six

4 0

4 years ago

Read 2 more answers

EASY 5POINT MATH QUESTION PLEASE SOLVE

nika2105 [10]

The equation x = 180 - (31 + 40) can be used. Because the angles are in a straight line ad straight line equals 180 degrees

3 0

4 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 298.87.

(b) The number of years before the capital stock exceeds $100,000 is approximately 46.15 years.

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 ≈ 46.15 years.

Learn more investment function here:

brainly.com/question/25300925

6 0

3 years ago

What is the solution to the equation e^3x=12 Round your answer to the nearest hundredth

Lunna [17]

Keywords:

equation, variable, clear, round, centesima, neperian logarithm, exponential

For this case we have the following equation $e ^ {3x} = 12$ , from which we must clear the value of the variable "x" and round to the nearest hundredth. To do this, we must apply properties of neperian and exponential logarithms. By definition:

$ln (e ^ x) = x$

So:

$e ^ {3x} = 12$ We apply Neperian logarithm to both sides:

$ln (e^{3x}) = ln (12)\\3x = ln (12)$

We divide between "3" both sides of the equation:

$\frac {3x} {3} = \frac {ln (12)} {3}\\x = \frac {ln (12)} {3}\\x = 0.828302217$

Rounding out the nearest hundredth we have:

$x = 0.83$

Answer:

$x = 0.83$

3 0

3 years ago

Read 2 more answers

Can someone help and explain and do the solution pls :(

brilliants [131]

Consider ∆JWZ and ∆JKZ

WZ~KJ (given)

/ WZJ~/ KJZ (given)

JZ~JZ (common)

Therefore,

∆JWZ~∆JKZ by SAS congruence rule.

JW~ZK by CPCT.

6 0

3 years ago