It wants it to be in slope-intercept form.
y=mx+b
We have to first find the slope and plug it into point-slope form.
y-y1=m(x-x1)
Find the slope of the second line. (I did this one first on accident)
Rise/run= 3/1= 3 The slope is 3. Plug that in along with the point (0,3)
y-3=3(x-0)
y-3=3x
Add 3 to the other side.
y= 3x +3 <- <em>for the second line</em><em>
</em>
Now, the second.
rise/run= 1/2= .5 Use point (6,0)
y-0=.5(x-6)
y= .5x-3
y=.5x-3 <- for the first line
I hope this helps!
~kaiker
4b + 22 = 5 (b + 4) - 1 (remove the parantheses)
4b + 22 = 5b + 20 - 1 (calculate)
4b + 22 = 5b + 19 (move the terms)
4b - 5b = 19 - 22 (collect like terms and calculate)
-b = -3 (change the signs)
<em>b = 3</em>
The point at which the lines k(x) = 5x - 1 and h(x) = -3x - 1 meet is (0, -1)
Given: k(x) = 5x - 1, h(x) = -3x - 1
We need to find the point(if any) at which these two lines k and h meets.
To find point of intersection(if any), we need to set the functions equal as at the point of intersection the (x, y) value will be same for both of the lines.
Therefore, k(x) = h(x)
=> 5x - 1 = -3x - 1
=> 8x = 0
=> x = 0
k(x=0) = 5 * 0 - 1 = -1
Hence the point at which the lines k(x) = 5x - 1 and h(x) = -3x - 1 meet is (0, -1)
Know more about "point of intersection" problems here: brainly.com/question/16929168
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Answer:
1 hour
Step-by-step explanation:
Hello, let's say that her departure trip takes t in minutes, as her return speed is 3 times her departure speed, she took t/3 for the return and we know that this 40 minutes less, so we can write.
t/3=t-40
We can multiply by 3
t = 3t -40*3 = 3t - 120
This is equivalent to
3t -120 = t
We subtract t
2t-120 = 0
2t = 120
We divide by 2
t = 120/2 = 60
So this is 60 minutes = 1 hour.
Thank you.