If A=38x-x^2 then
dA/dx=38-2x
d2A/dx2=-2
Since the acceleration, d2A/dx2 is a constant negative, when velocity, dA/dx=0, it will be an absolute maximum for A(x)
dA/dx=0 only when 38=2x, x=19
A(19)=38(19)-19^2
A(19)=722-361
A(19)=361 ft^2
So the maximum possible area is 361 ft^2
(This will always be true as the maximum possible area enclosed by a given amount of material will always be a perfect square...)
4 + 3(10 - 23)
4 + 3 × (-13)
4 - 39
-35 << answer
hope this helps, God bless!
Answer:
Step-by-step explanation:
Given
sinΘ = 
= 
Then the hypotenuse = 5 and one leg of the right triangle is 3
Using Pythagoras' identity to find the second leg (x)
x² + 3² = 5²
x² + 9 = 25 ( subtract 9 from both sides )
x² = 16 ⇒ x = 4
cosΘ = 
=
Answer:
sln
b3=8(dived by 3 both sides)
b=8/3
I hope this help you
Let <em>a</em> and <em>b</em> be the zeroes of <em>x</em>² + <em>kx</em> + 12 such that |<em>a</em> - <em>b</em>| = 1.
By the factor theorem, we can write the quadratic in terms of its zeroes as
<em>x</em>² + <em>kx</em> + 12 = (<em>x</em> - <em>a</em>) (<em>x</em> - <em>b</em>)
Expand the right side and equate the coefficients:
<em>x</em>² + <em>kx</em> + 12 = <em>x</em>² - (<em>a</em> + <em>b</em>) <em>x</em> + <em>ab</em>
Then
<em>a</em> + <em>b</em> = -<em>k</em>
<em>ab</em> = 12
The condition that |<em>a</em> - <em>b</em>| = 1 has two cases, so without loss of generality assume <em>a</em> > <em>b</em>, so that |<em>a</em> - <em>b</em>| = <em>a</em> - <em>b</em>.
Then if <em>a</em> - <em>b</em> = 1, we get <em>b</em> = <em>a</em> - 1. Substitute this into the equations above and solve for <em>k</em> :
<em>a</em> + (<em>a</em> - 1) = -<em>k</em> → 2<em>a</em> = 1 - <em>k</em> → <em>a</em> = (1 - <em>k</em>)/2
<em>a</em> (<em>a</em> - 1) = 12 → (1 - <em>k</em>)/2 • ((1 - <em>k</em>)/2 - 1) = 12
→ (1 - <em>k</em>)²/4 - (1 - <em>k</em>)/2 = 12
→ (1 - <em>k</em>)² - 2 (1 - <em>k</em>) = 48
→ (1 - 2<em>k</em> + <em>k</em>²) - 2 (1 - <em>k</em>) = 48
→ <em>k</em>² - 1 = 48
→ <em>k</em>² = 49
→ <em>k</em> = ± √(49) = ±7
