All categories

4 years ago

A line passes through (8,2) and (11,-13) write the equation of thr line in standerd form

Mathematics

1 answer:

docker41 [41]4 years ago

8 0

Answer:

Step-by-step explanation:

(8,2),(11,-13)

slope(m) = (-13 - 2) / (11 - 8) = -15/3 = -5

y = mx + b

slope(m) = -5

(8,2)...x = 8 and y = 2

now we sub and find b, the y int

2 = -5(8) + b

2 = -40 + b

2 + 40 = b

42 = b

so ur equation is : y = -5x + 42...now we need it in standard form

Ax + By = C

y = -5x 42

5x + y = 42 <====

You might be interested in

Use the function below to find F(3) f(x)=1/4^x

Harman [31]

There is no function below.

5 0

3 years ago

Read 2 more answers

In which table does y vary directly with x?

g100num [7]

Answer:

In which table does y vary directly with x? ОА. х у 1 -2 2 -4 3 -16 ОВ. Х у 1 -5 2 18 3 41 Ос. х у 1 26 2 52 3 78 - did not match any documents.

Suggestions:

Make sure all words are spelled correctly.

Try different keywords.

Try more general keywords.

Try fewer keywords.

8 0

3 years ago

Starting at the 6th floor, an elevator went down 1 floor, up 6 floors, and then down 4 floors. Find the number of the floor on w

Ymorist [56]

Answer:

The answer is that the elevator stopped at the 7th floor

Step-by-step explanation:

6 0

3 years ago

In 2012 William bought a new boat for $28,500 if the value of the boat declined by 8% each year find the value of the boat in 20

nikklg [1K]

Answer:

$17,281.12

Step-by-step explanation:

28,500 * (.92)^x

x - amount of years after 2012

.92 - 100-8 = 92 which is the percent of the price the previous year

Calculator needed.

28,500 * .92^6

28,500 * ~.60636

17,281.1175383

Round up to 17,281.12

8 0

4 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0

3 years ago