30 km

Contains the point (-1, 2) and is parallel to<br> x – 2y = -3

ivanzaharov [21]

Answer:

see explanation

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Rearrange x - 2y = - 3 into this form

Subtract x from both sides

- 2y = - x - 3 ( divide all terms by - 2 )

y = $\frac{1}{2}$ x + $\frac{3}{2}$ ← in slope- intercept form

with m = $\frac{1}{2}$

• Parallel lines have equal slopes, thus

y = $\frac{1}{2}$ x + c ← is the partial equation

To find c substitute (- 1, 2) into the partial equation

2 = - $\frac{1}{2}$ + c ⇒ c = 2 + $\frac{1}{2}$ = $\frac{5}{2}$

y = $\frac{1}{2}$ x + $\frac{5}{2}$ ← in slope- intercept form

Multiply through by 2

2y = x + 5 ( subtract 2y from both sides )

0 = x - 2y + 5 ( subtract 5 from both sides )

- 5 = x - 2y, thus

x - 2y = - 5 ← in standard form

4 0

3 years ago

The number of bacteria in a certain population increases according to a continuous exponential growth model, with a growth rate

nikdorinn [45]

Log(2)/log(1.064) ≈ 11.17 . . . . hours

_____
The population can be given by
p(n) = p₀×1.064ⁿ . . . . where n is the number of hours
You want to find n whe p(n) = 2*p₀.
2p₀ = p₀×1.064ⁿ . . . . . . . . . . . . substitute the given information
2 = 1.064ⁿ . . . . . . . . . . . . . . . . . divide by p₀
log(2) = n×log(1.064) . . . . . . . . take logs to make it a linear equation
log(2)/log(1.064) = n . . . . . . . . divide by the coefficient of n

4 0

3 years ago

How can you tell if a system of equations has infinite solutions?

kirill [66]

To tell if an equation has infinite solutions, the equation will be equal to each other.

For example,

x = x
x + 1 = x + 1
x - y = x - y

And so on...

And in special cases,

0x = 0
0x = 0(y + 1)

They have infinite solutions because there's no constant to determine the variable.

4 0

3 years ago

The given line segment has a midpoint at (3, 1).

Katena32 [7]

Answer:

$y=\frac{1}{3}x$

Step-by-step explanation:

The given line segment has a midpoint at (3, 1) and goes through (2, 4), (3, 1), and (4, -2). We can use any two of the three points to calculate the equation of the line. Let us use the points (2, 4) and (4, -2)

Therefore the line goes through (2, 4) and (4, -2). The equation of a line passing through $(x_1,y_1)\ and\ (x_2,y_2)$ is:

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$ .

Therefore the line passing through (2, 4) and (4, -2) has an equation:

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}\\\frac{y-4}{x-2}=\frac{-2-4}{4-2}\\\frac{y-4}{x-2}=\frac{-6}{2}\\y-4=x-2(-3)\\y-4=-3x+6\\y=-3x+10$

Comparing with the general equation of line: y = mx + c, the slope (m) = -3 and the intercept on the y axis (c) = 10

Two lines are said to be perpendicular if the product of their slope is -1. If the slope of line one is m1 and the slope of line 2 = m2, then the two lines are perpendicular if:

$m_1m_2=-1$ .

Therefore The slope (m2) of the perpendicular bisector of y = -3x + 10 is:

$m_1m_2=-1\\-3m_2=-1\\m_2=\frac{1}{3}$

Since it is the perpendicular bisector of the given line segment, it passes through the midpoint (3, 1). The equation of the perpendicular bisector is:

$\frac{y-y_1}{x-x_1}=m\\\frac{y-1}{x-3}=\frac{1}{3}\\ y-1= \frac{1}{3}(x-3)\\ y-1=\frac{1}{3}x-1\\y=\frac{1}{3}x$

the equation, in slope-intercept form, of the perpendicular bisector of the given line segment is $y=\frac{1}{3}x$

7 0

3 years ago

Read 2 more answers

Solve for x: 3+(-5)x=19

malfutka [58]

3+(-5)x=19
Subtract 3 from both sides of the equation.
-5x= 16
Divide by -5 on both sides
X= -3.2

4 0

3 years ago

Read 2 more answers