Answer:
cos(θ) = 3/5
Step-by-step explanation:
We can think of this situation as a triangle rectangle (you can see it in the image below).
Here, we have a triangle rectangle with an angle θ, such that the adjacent cathetus to θ is 3 units long, and the cathetus opposite to θ is 4 units long.
Here we want to find cos(θ).
You should remember:
cos(θ) = (adjacent cathetus)/(hypotenuse)
We already know that the adjacent cathetus is equal to 3.
And for the hypotenuse, we can use the Pythagorean's theorem, which says that the sum of the squares of the cathetus is equal to the square of the hypotenuse, this is:
3^2 + 4^2 = H^2
We can solve this for H, to get:
H = √( 3^2 + 4^2) = √(9 + 16) = √25 = 5
The hypotenuse is 5 units long.
Then we have:
cos(θ) = (adjacent cathetus)/(hypotenuse)
cos(θ) = 3/5
Answer:
630
Step-by-step explanation:

just subsitute
f(4)=4²+3(4)-1
f(4)=16+12-1
f(4)=28-1
f(4)=27
it approaches 27 as x approaches 4
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