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3 years ago

What is the equation of a line that is parallel to -2x+3y=-6 and passes through the point (-2,0)

1 answer:

4 0

First, a line that is parallel, means a line that has the same slope as the original. To find the slope of the original equation, we have to solve for y.
-2x+3y=-6
3y=2x-6
y=2/3x-2
From this equation, we can see that the slope of the line is 2/3. For every 2 units you go up, you move three units over.

Now we need to use the point (-2,0) to find the equation of the parallel line.
y-y=m(x-x)

Plug in the point coordinates and the slope, and solve for the final equation of the line.
y-0=2/3(x+2)
y=2/3x+ 4/3

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William has a pair of identical number cubes. the faces of each cube are numbered 1 through 6. william will roll the cubes one t

Angelina_Jolie [31]

The chances will be : 3 and 6 and 4 and 5
The probability of getting a sum of 9 would be 2/9

5 0

3 years ago

Read 2 more answers

What is (5+3)+7=5+(3+7) is it Associative, Identity, Commutative or Substitution?

timama [110]

Answer:

Associative

Step-by-step explanation:

The associative property of addition states that you can add numbers however they are grouped inside parentheses.

It doesn't matter where you put the parentheses. You will always get the same answer.

For example,

$\begin{array}{rcl}(5 + 3) + 7 & = & 5 + (3 + 7)\\8 + 7& = & 5 + 10\\15 & = & 15\\\end{array}$

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3 years ago

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio

JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

$Initial =1\ million$

$6\ years\ later = 500,000$

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

$\frac{dA}{dt} = kA$

Where k represents the constant of proportionality

$\frac{dA}{dt} = kA$

Multiply both sides by dt/A

$\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}$

$\frac{dA}{A} = k\ dt$

Integrate both sides

$\int\ {\frac{dA}{A} = \int\ {k\ dt}$

$ln\ A = kt + lnC$

Make A, the subject

$A = Ce^{kt}$

$t = 0\ when\ A =1\ million$ i.e. At initial

So, we have:

$A = Ce^{kt}$

$1000000 = Ce^{k*0}$

$1000000 = Ce^{0}$

$1000000 = C*1$

$1000000 = C$

$C =1000000$

Substitute $C =1000000$ in $A = Ce^{kt}$

$A = 1000000e^{kt}$

To solve for k;

$6\ years\ later = 500,000$

i.e.

$t = 6\ A = 500000$

So:

$500000= 1000000e^{k*6}$

Divide both sides by 1000000

$0.5= e^{k*6}$

Take natural logarithm (ln) of both sides

$ln(0.5) = ln(e^{k*6})$

$ln(0.5) = k*6$

Solve for k

$k = \frac{ln(0.5)}{6}$

$k = \frac{-0.693}{6}$

$k = -0.1155$

Recall that:

$\frac{dA}{dt} = kA$

Where

$\frac{dA}{dt}$ = Rate

So, when

$A = 600000$

The rate is:

$\frac{dA}{dt} = -0.1155 * 600000$

$\frac{dA}{dt} = -69300$

Hence, the amount of oil was decreasing at 69300 barrels, yearly

7 0

2 years ago

Solve for a = ____. 1. 6 2. 4 3. 9

garik1379 [7]

The answer is 3, or 9

8 0

3 years ago

PLEASE HELP!!!!!!!!!!

Montano1993 [528]

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5 0

2 years ago