-5+d/3=5
d/3-5=5
d-3.5/3=5
d-15/3=5
d-15=15
(d-15)+15=15+15
d-15+15=30
d=30
Hope this helps kiddo
Given:
The function is

To find:
The inverse of the given function.
Solution:
We have,

Substitute m(x)=y.

Interchange x and y.

Add square of half of coefficient of y , i.e.,
on both sides,


![[\because (a-b)^2=a^2-2ab+b^2]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%28a-b%29%5E2%3Da%5E2-2ab%2Bb%5E2%5D)
Taking square root on both sides.

Add
on both sides.

Substitute
.

We know that, negative term inside the root is not real number. So,


Therefore, the restricted domain is
and the inverse function is
.
Hence, option D is correct.
Note: In all the options square of
is missing in restricted domain.
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I think this is what you're looking for?
a) (0.17 ac)×(126 min/ac) = 21.42 min
b) I can mow 1/126 ac/min; you can mow 1/117 ac/min. Together, we can mow at a rate that is the sum of these:
... (1/126 + 1/117) ac/min = (117 + 126)/(126×117) = 243/14,742 ac/min
Then it will take 14,742/243 min/ac × 0.17 ac/lawn ≈ 10.31 min/lawn
1. Round 50.75 to 50 and 0.18 to 0.20.
50 x 0.2 = 10
2) Round 96 to 100 and 0.499 to 0.5
100 divided by 0.5 = 50
3a) Round 8.2 to 8, 6.7 to 7, and 0.46 to 0.50
8 x 7 divided by 0.50 = 112
3b) Round 23.4 to 20, 13.9 to 10, and 0.18 to 0.20
20 x 10 divided by 0.20 = 2,000