Answer:
y = - 2x + 1
Step-by-step explanation:
Parallels line have the same slope, when an equation is in the form y= mx + b, m is the slope. In this problem slope = - 2
Now with the slope what is missing is the y-intercept, the problem says that the line contains the point (-2, 5), replacing that point in the equation you can solve it to find the y-intercept
y = mx + b
5 = (-2)(-2) + b
5 = 4 + b
1 + b
y = - 2x +1
Answer:
Step-by-step explanation:
What is the full question?
<u>Answer:</u>
A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.
<u>Solution:</u>
We need to show that the gradient of the curve at A is 1
Here given that ,
--- equation 1
Also, according to question at point A (b+1,0)
So curve at point A will, put the value of x and y

0=b+1-c --- equation 2
According to multiple rule of Differentiation,

so, we get



By putting value of point A and putting value of eq 2 we get


Hence proved that the gradient of the curve at A is 1.
Answer:
x = 3
Step-by-step explanation:
5x - 8 + 8 = 7 + 8
5x = 15
5x/5 = 15/5
15 divided by 5 equals 3
Hope this helped you out :)