Answer:
Polynomial expression that represents the area of blanket:
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If
:
Step-by-step explanation:
The area of the rectangle can be calculated with the formula:

Being l the lenght of the rectangle and w the width of the rectangle.
In this case, the lenght and the width are represented with:


Substitute them into
:

Then:
Use Distributive property (Remember the Product of powers property:
):

Add like terms:
(Simplied form)
Evaluate
:

Answer:
$400
Step-by-step explanation:
In the table y is increased 70 for every 1 in x
(210/3 = 70)
on the graph for every 1 increase on x y increases by 55
(110/2 = 55)
so for the graph for x = 11, y =55 x 11 = 605
for the table for x = 11, y = 11*70 = 770
the difference is 770-605 = 165
the 2nd answer is the correct one.
You can either use the inverse function theorem or compute the general derivative using implicit differentiation. The first method is slightly faster.
The IFT goes like this: if f(x) is invertible and f(a) = b, then finv(b) = a (where "finv" means "inverse of f").
By definition of inverse functions, we have
f(finv(x)) = finv(f(x)) = x
Differentiating both sides of the second equality with respect to x using the chain rule gives
finv'(f(x)) * f'(x) = 1
When x = a, we get
finv'(b) * f'(a) = 1
or
finv'(b) = 1/f'(a)
Now let f(x) = sin(x), which is invertible over the interval -π/2 ≤ x ≤ π/2. In the interval, we have sin(x) = √3/2 when x = π/3. We also have f'(x) = cos(x).
So we take a = π/3 and b = √3/2. Then
arcsin'(√3/2) = 1/cos(π/3) = 1/(1/2) = 2
Answer:
4
Step-by-step explanation:
WY=WX+XY
28=12+4p
28-12=4p
16=4p
16/4=p
4=p