Answer:
ight bet
Step-by-step explanation:
idek
I'm pretty sure its the first choice. Similar means same ratio and congruent means equal. But this was 9th grade for me :)
<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-a) \sqrt{(b+1-b)}](https://tex.z-dn.net/?f=0%3D%28b%2B1-a%29%20%5Csqrt%7B%28b%2B1-b%29%7D)
0=b+1-c --- equation 2
According to multiple rule of Differentiation,
![y^{\prime}=u^{\prime} y+y^{\prime} u](https://tex.z-dn.net/?f=y%5E%7B%5Cprime%7D%3Du%5E%7B%5Cprime%7D%20y%2By%5E%7B%5Cprime%7D%20u)
so, we get
![{u}^{\prime}=1](https://tex.z-dn.net/?f=%7Bu%7D%5E%7B%5Cprime%7D%3D1)
![v^{\prime}=\frac{1}{2} \sqrt{(x-b)}](https://tex.z-dn.net/?f=v%5E%7B%5Cprime%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%5Csqrt%7B%28x-b%29%7D)
![y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}](https://tex.z-dn.net/?f=y%5E%7B%5Cprime%7D%3D1%20%5Ctimes%20%5Csqrt%7B%28x-b%29%7D%2B%28x-a%29%20%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%20%5Csqrt%7B%28x-b%29%7D)
By putting value of point A and putting value of eq 2 we get
![y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}](https://tex.z-dn.net/?f=y%5E%7B%5Cprime%7D%3D%5Csqrt%7B%28b%2B1-b%29%7D%2B%28b%2B1-a%29%20%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%20%5Csqrt%7B%28b%2B1-b%29%7D)
![y^{\prime}=\frac{d y}{d x}=1](https://tex.z-dn.net/?f=y%5E%7B%5Cprime%7D%3D%5Cfrac%7Bd%20y%7D%7Bd%20x%7D%3D1)
Hence proved that the gradient of the curve at A is 1.
Answer:
B
Step-by-step explanation:
∠1 = ∠ 4
corresponding angles are congruent
The answer is 5 ok.................................