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<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:
Step-by-step explanation:
Let A = R−{0}, the set of all nonzero real numbers, and consider the following relations on A × A.
Given that (a,b) R (c,d) if
Or (a,b) R (c,d) if determinant
a) Reflexive:
We have (a,b) R (a,b) because ab-ab =0 Hence reflexive
b) Symmetric
(a,b) R (c,d) gives ad-bc =0
Or da-cb =0 or cb-da =0 Hence (c,d) R(a,b). Hence symmetric