How to solve the problem, and complete answer

Mathematics

1 answer:

rusak2 [61]2 years ago

5 0

Notice the picture below

thus, solve for "y"

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The distance of a trip

Vladimir [108]

Answer:

T-.5 = x (200)

Step-by-step explanation:

Let T = total time in hours

30 mn = .5 hours 30 /60 = .5

$ 2.00 is the rate per hour.

so T − .5 = the time that will be charged at the rate

Therefore the total charge is then

T − .5 × 2.00

4 0

2 years ago

Factor y2 + 10z - 10y - yz

stira [4]

Group and factor
group y's with y's and z's with z's
(y²-10y)+(10z-yz)
factor
y(y-10)+z(10-y)
force undistribute -1
-y(10-y)+z(10-y)
factor
(10-y)(-y+z)
(z-y)(10-y) or
(y-z)(y-10)

6 0

2 years ago

Read 2 more answers

Distance between parallel lines y=3x+10 and y=3x-20

Alecsey [184]

1. Take an arbitrary point that lies on the first line y=3x+10. Let x=0, then y=10 and point has coordinates (0,10).

2. Use formula $d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$ to find the distance from point $(x_0,y_0)$ to the line Ax+By+C=0.

The second line has equation y=3x-20, that is 3x-y-20=0. By the previous formula the distance from the point (0,10) to the line 3x-y-20=0 is:

$d=\dfrac{|3\cdot 0-10-20|}{\sqrt{3^2+(-1)^2}}=\dfrac{30}{\sqrt{10}}=3\sqrt{10}$ .

3. Since lines y=3x+10 and y=3x-20 are parallel, then the distance between these lines are the same as the distance from an arbitrary point from the first line to the second line.

Answer: $d=3\sqrt{10}$ .