The 3 goes with the square root you don’t need to multiply or anything like that. If you need to simply for example 3 square root of 20 which is when 3 is outside the square root sign. The 20 will be separated with 4*5 in the square root and then you will square root the 4 leaving the 5 in and bringing 2 out. Now you can’t just bring it out and forget about the 3 outside you need to multiply that 2 and 3 and you get 6! So you final answer is 6 square root 5. Hope this helps reply if you have a question!
The function you seek to minimize is
()=3‾√4(3)2+(13−4)2
f
(
x
)
=
3
4
(
x
3
)
2
+
(
13
−
x
4
)
2
Then
′()=3‾√18−13−8=(3‾√18+18)−138
f
′
(
x
)
=
3
x
18
−
13
−
x
8
=
(
3
18
+
1
8
)
x
−
13
8
Note that ″()>0
f
″
(
x
)
>
0
so that the critical point at ′()=0
f
′
(
x
)
=
0
will be a minimum. The critical point is at
=1179+43‾√≈7.345m
x
=
117
9
+
4
3
≈
7.345
m
So that the amount used for the square will be 13−
13
−
x
, or
13−=524+33‾√≈5.655m
30u^2v^3 and 8u^5v is : 2u^2v
Answer:
7% of 40 is 2.80
Step-by-step explanation:
0.07 x 40 = 2.8
One way is to divide 500 by 0.15
500/0.15=3333.3333333
the answer is about 3333 times
