PLZ HELP ME 1. Find the equation of the line that goes through the point (0,-3) and is perpendicular to the line y= -2/5x+6.

2. Find the equation of the line that is parallel to the line -3x+2y=10 and goes through the point (0,7) .

1.  y = (-5x/2) + -3

2. y = (3x/2) + 7

Step-by-step explanation:

For both these problems, we simply need to apply point-slope form.

1.  The point (0, -3) and the slope that is perpendicular to the line is the reciprocal of the coefficient, slope = -5/2.

From here, we simply apply the formula:

y - y0 = m ( x - x0)

y - -3 = -5/2 ( x - 0 )

y + 3 = -5/2 x

y = (-5x / 2) + -3

2.  The point (0, 7) and the slope that is parallel to the line must maintain the same slope.  First, we need to rewrite the equation into slope-intercept:

-3x + 2y = 10

2y = 3x + 10

y = 3x/2 + 5

Now we want to apply the point-slope form

y - y0 = m ( x - x0 )

y - 7 = (3/2) (x - 0)

y - 7 = 3x / 2

y = 3x/2 + 7

Cheers.

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation: