My son is 12 years old and my father is 39 years old. When was my father four times as old as my son

My son is 12 years old and my father is 39 years old. When was my father four times as old as my son

When my father was 36, he was four times as old as his son when he was nine
Set X years ago
39-x=4(12-x)
X=3
3 years
（39-12）/3=9
12-9=3
So three years ago, the father was four times as old as his son
The answer should be that x years ago, the father was four times as old as his son
39 - x = 4 (12 -x )
X=3
Three years ago, when my son was nine years old
Hope to help you?
There is no solution to this problem~~~
Suppose that x years ago, the father was four times as old as his son
39 - x = 4 (12 -x )
X=3
Three years ago, when my son was nine years old
Thank you for adopting me!
Suppose that x years ago, the father was four times as old as his son
There are: 39 -- X = 4 (12-x)
We get x = 3
Three years ago, my father was four times as old as my son
How many meters is 9 decimeters
 
Nine tenths and 0.9
May I ask you what is the value of '(- AC) ^ 2
a^2c^2
Equal to (AC) ^ 2 or a ^ 2C ^ 2
If f (SiNx + cosx) = SiNx * cosx, find the value of F (COS π / 6) and the result is 1 / 8. Who knows the idea and the process? Thank you
a=sinx+cosx
a²=sin²x+cos²x+2sinxcosx=1+2sinxcosx
sinxcosx=(a²-1)/2
So f (a) = (A & sup2; - 1) / 2
cosπ/6=√3/2
So the original formula = [(√ 3 / 2) & sup2; - 1) / 2 = - 1 / 8
Let SiNx + cosx = t, then SiNx * cosx = (T ^ 2-1) / 2
f(t)=(t^2-1)/2
f(cosπ/6)=[(cosπ/6)^2-1]/2=-1/8
it's easy
a=sinx+cosx
a²=sin²x+cos²x+2sinxcosx=1+2sinxcosx
sinxcosx=(a²-1)/2
So f (a) = (A & sup2; - 1) / 2
cosπ/6=√3/2
So the original formula = [(√ 3 / 2) & sup2; - 1) / 2 = - 1 / 8
78 cm is equal to a few parts of a meter
78 / 100 = 39 / 50m
How many hours is 1264 days? Thank you, God
30336 hours
Find the maximum value of the function y = SiNx multiplied by cosx + SiNx + cosx, X ∈ [0, π / 2]
Y = SiNx times the maximum of cosx + SiNx + cosx, X ∈ [0, π / 2]
y=sinx*cosx+sinx+cosx+1-1=sinx(cosx+1)+(cosx+1)-1=(sinx+1)(cosx+1)-1
See two integral multiplication of the form of the maximum, think of
Using the mean value theorem: one positive (determined by the domain of definition), two definite, three-phase, etc
That is y ≤ ((SiNx + 1) ^ 2 + (cosx + 1) ^ 2) / 2-1
≤sinx+cosx+1/2
So... If and only if SiNx + 1 = cosx + 1, that is, SiNx = cosx = root 2 / 2, the maximum value of Y is (root 2 + 1 / 2)
3 / 4 = 0.9 = 0.100 = 0.03 = 8 / 25 = 0.78
3 / 4 = 0.75
0.9 = 9 / 10
7 out of 100 = 0.07
0.03 = 3 / 100
8 out of 25 = 0.32
78 = 39 / 50
Sincostan2 minus 60
SDU
Root 2 minus half root 3
Find y = (SiNx) ^ 2 + 3sinxcosx + 4 (cosx) ^ 2 (0)
Y = (SiNx) ^ 2 + 3sinxcosx + 4 (cosx) ^ 2 = 1 + 3 (cosx) ^ 2 + 3sinxcosx = 1 + 3 / 2 * [2 (cosx) ^ 2-1 + 1] + 3 / 2 * 2sinxcosx = 1 + 3 / 2cos2x + 3 / 2 + 3 / 2sin2x = 5 / 2 + 3 / 2 (sin2x + cos2x) = 5 / 2 + 3 / 2 * radical 2 * sin (2x + pi / 4) from 0