The graphed line shown below is y = 5 x minus 10. On a coordinate plane, a line goes through (2, 0) and (3, 5). Which equation, when graphed with the given equation, will form a system that has no solution? y = negative 5 x + 10 y = 5 (x + 2) y = 5 (x minus 2) y = negative 5 x minus 10 Mark this and return

The correct option is:

y = 5(x-2)

Step-by-step explanation:

The graphed line is;

y = 5x - 10

y = mx + c

Thus, slope of graphed line is equal to m=5.

Those line will have no solution which have are parallel to the graphed line (have a slope of 5) and passes through the given points.

Following lines have the slopes 5:

y= 5(x+2)

y= 5(x-2)

Check which equation satisfied points (2,0) and (3,5)

For (2,0)

y=5(x-2)

y=5(2-2)

y=0

For (3,5)

y=5(x-2)

y=5(3-2)

y=5

Thus, correct option is y = 5(x-2)

the figures below are made out of circles, semicircles, quarter … find the area and the perimeter of each figure and give your answers as a completely simplified exact value in terms of π (no approximations). remember that we'll also have to cut whatever areas we get in half, since we're only dealing

step-by-step explanation:

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}