For b: \(\log_5w+\frac{\log_5u}{3}-\frac{\log_5v}{3}\rightarrow \log_5w+\frac{\log_5u/v}{3} \rightarrow \log_5w+\log_5((\frac{u}{v})^{1/3})\rightarrow \log_5(w(\frac{u}{v})^{1/3})\)
or you could express this as \(\frac{\log_5(w^3u/v)}{3}\)
For e you should have 2x, not just 2.
Prove as follows:
(Note: your expressions for a and b are incorrect).
If f(x) = -2x2 + 8x + 10 then f(3x) = -2(3x)2 + 8(3x) + 10 or f(3x) = -18x2 + 24x + 10
An alternative to heureka's method is to notice that after 5 moves the ant can only end up at one of the four dots directly connected to A. There are four of them, and nothing to prefer one over the other, hence the probability of ending up at B is 1/4 or 25%.
f-1 means inverse rather than reciprocal, so if f(x) = a/(x+2) then f-1(x) = a/x - 2
f(0) = a/2
f-1(3a) = 1/3 - 2 = -5/3
So a/2 = -5/3 or a = -10/3
25 = 5*5
35 = 5*7
45 = 3*3*5
55 = 5*11
Angle ABD = 180 - 35 = 145º
Angle ADB = 180 - (145+16) = 19º
Using the sine rule: BD/sin(16) = 950/sin(19) = ...
Then DC = BDsin(35) = ...
I’ll leave you to fill in the blanks.
9x^2+24x+16 can be written as (3x + 4)2
If the side lengths of the square are as ax + b, then a = 3 and b = 4.
Number of positive integers, n, as follows:
Assuming a and b are orthogonal then:
a.b is 0
a.c is positive
a.d is negative
b.c is positive
b.d is positive
c.d is negative