The triangles are similar so ∆JKL is just an enlargement of ∆WVU. Therefore the ratios of the side lengths are the same. 10:20 = 1:2 so the side lengths for ∆JKL must be in the ratio 1:2. 15 × 2 = 6x so 30 = 6x so x = 5
Answer:
48/7
Step-by-step explanation:
3 1/5 . 2 1/7
= 16/5 * 15/7
= 16/1 * 3/7
= 48/7
Answer: 31/20
Step-by-step explanation:
To add fractions, find the LCD and then combine.
9514 1404 393
Answer:
1) f⁻¹(x) = 6 ± 2√(x -1)
3) y = (x +4)² -2
5) y = (x -4)³ -4
Step-by-step explanation:
In general, swap x and y, then solve for y. Quadratics, as in the first problem, do not have an inverse function: the inverse relation is double-valued, unless the domain is restricted. Here, we're just going to consider these to be "solve for ..." problems, without too much concern for domain or range.
__
1) x = f(y)
x = (1/4)(y -6)² +1
4(x -1) = (y-6)² . . . . . . subtract 1, multiply by 4
±2√(x -1) = y -6 . . . . square root
y = 6 ± 2√(x -1) . . . . inverse relation
f⁻¹(x) = 6 ± 2√(x -1) . . . . in functional form
__
3) x = √(y +2) -4
x +4 = √(y +2) . . . . add 4
(x +4)² = y +2 . . . . square both sides
y = (x +4)² -2 . . . . . subtract 2
__
5) x = ∛(y +4) +4
x -4 = ∛(y +4) . . . . . subtract 4
(x -4)³ = y +4 . . . . . cube both sides
y = (x -4)³ -4 . . . . . . subtract 4
The greatest common factor of 36x3 and 8x is simply 4X
