Solve for t. 16 - 2t = t + 9 + 4t please help.
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Answer:
<u>Step-by-step explanation:</u>
There are 3 conditions that must be satisfied:
- f(x) is continuous on the given interval
- f(x) is differentiable
- f(a) = f(b)
If ALL of those conditions are satisfied, then there exists a value "c" such that c lies between a and b and f'(c) = 0.
f(x) = x³ - x² - 6x + 2 [0, 3]
1. There are no restrictions on x so the function is continuous
2. f'(x) = 3x² - 2x - 6 so the function is differentiable
3. f(0) = 0³ - 0² - 6(0) + 2 = 2
f(3) = 3³ - 3² - 6(3) + 2 = 2
f(0) = f(3)
f'(x) = 3x² - 2x - 6 = 0
This is not factorable so you need to use the quadratic formula:
Only one of these values (1.8) is between 0 and 3.
QUESTION 1
The dimensions of the rectangular blanket are;
and
The perimeter is given by,
We substitute the dimensions to obtain,
Expand bracket to get,
This simplifies to
QUESTION 2
When
The perimeter becomes
QUESTION 3
Area is given by
We expand to get,
This gives us,
QUESTION 4
If
Then the area becomes,
QUESTION 5.
When the length of the blanket is 5cm longer, then the length of the new blanket becomes
The width is still
The perimeter of the new blanket is
This implies that,
Comparing to the old perimeter which is
,
The perimeter changes by 10 units
The dimensions of the rectangular blanket are;
and
The perimeter is given by,
We substitute the dimensions to obtain,
Expand bracket to get,
This simplifies to
QUESTION 2
When
The perimeter becomes
QUESTION 3
Area is given by
We expand to get,
This gives us,
QUESTION 4
If
Then the area becomes,
QUESTION 5.
When the length of the blanket is 5cm longer, then the length of the new blanket becomes
The width is still
The perimeter of the new blanket is
This implies that,
Comparing to the old perimeter which is
The perimeter changes by 10 units