Answer:
We now want to find the best approximation to a given function. This fundamental problem in Approximation Theory can be stated in very general terms. Let V be a Normed Linear Space and W a finite-dimensional subspace of V , then for a given v ∈ V , find w∗∈ W such that kv −w∗k ≤ kv −wk, for all w ∈ W.
Step-by-step explanation:
Answer:
He needs 5x+56 ft
Step-by-step explanation:
To find how much of fencing he needs , we find the perimeter of the given figure
All sides are equal in a square
To find perimeter of the square we add all the sides
4 sides we have for the square
one side is x, so perimeter of square = x+x+x+x= 4x
Now we find perimeter of rectangle
Opposite sides of rectangle are equal
Here for rectangle we consider only three sides
because fourth side is common for rectangle and square
So perimeter of the rectangle (with 3 sides) = 28 +x+ 28 = 56+x
Total fencing = perimeter of square + perimeter of rectangle
4x + 56 + x= 5x+56
Answer:
I think it's 45.
Step-by-step explanation:
46 + 89 = 135
adding all the angles of a triangle will always equal 180.
180 - 135 = 45
the last angle is 45.
<span>96-2r=6r-112
Add 2r to both sides
96=8r-112
Add 112 to both sides
208=8r
Divide 8 on both sides
Final Answer: 26=r</span>
Answer:
a. 360.323 m
b. 
Step-by-step explanation:
Distance covered from A to B = speed x time
= 110 x 2.8
= 308 m
Distance covered from B to C = speed x time
= 110 x 1.7
= 187 m
The sum of angles at B = 
+ 
= 
a. The distance of the plane from it starting point to C can be determined by applying the cosine rule.
= 
+ 
- 2ac Cos B
Sot hat;
= 
+ 
- 2(187 x 308) Cos
But, Cos = 0
So that,
= +
= 34969 + 94864
= 129833
b = 
= 360.323
The distance from A to C is 360.323 m.
b. Applying the sine rule;
= 
= 
= 
Sin A = 
= 0.5190
⇒ A = 
0.5190
= 
The bearing of the plane from its original location = +
=
