1) Given ![f(x) = x^{3}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20x%5E%7B3%7D)
![f(bx) = (bx)^{3}= b^{3} x^{3}](https://tex.z-dn.net/?f=f%28bx%29%20%3D%20%28bx%29%5E%7B3%7D%3D%20b%5E%7B3%7D%20%20x%5E%7B3%7D)
Since (0.5,8) lies on f(bx) , ![8=b^{3} 0.5^{3}](https://tex.z-dn.net/?f=8%3Db%5E%7B3%7D%200.5%5E%7B3%7D)
Divide with
on both sides
![b^{3} = \frac{8}{0.5^{3} }](https://tex.z-dn.net/?f=b%5E%7B3%7D%20%3D%20%5Cfrac%7B8%7D%7B0.5%5E%7B3%7D%20%7D)
= 64
![b=64^{\frac{1}{3} } = 4](https://tex.z-dn.net/?f=b%3D64%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D%20%3D%204)
Hence b=4.
2) Given (1,3) lies on y=f(x) plugin x=1 and y=3
that is f(1) = 3
We can find corresponding point for that in y=-2f(x) by plugging in x=1.
That is y= -2f(1) = -2*3 = -6
Hence point is (1,-6)
Answer:
3x-2
Step-by-step explanation:
Answer:
x=2
Step-by-step explanation:
Answer: x = 2921, y = 579
Step-by-Step Explanation:
I am assuming that we just have to Solve for ‘x’ and ‘y’.
‘x’ = No. Of Contemporary Titles
‘y’ = No. Of Classic Titles
=> x + y = 3500 (Eq. 1)
=> x - y = 2342 (Eq. 2)
Adding Eq. 1 and Eq. 2, we get :-
2x = 3500 + 2342
2x = 5842
x = 5842/2
=> x = 2921
Therefore, x = 2921
Substitute value of ‘x’ in Eq. 1 :-
x + y = 3500
(2921) + y = 3500
y = 3500 - 2921
=> y = 579
Therefore, y = 579
Hence,
No. Of Contemporary Titles = 2921
No. Of Classic Titles = 579