V= x*(10-2x)(8-2x)
for greatest volume, differentiate v w.r.t x and equate to 0,
dv/dx= (10-2x)(8-2x) + x(-2)(8-2x)+ x(10-2x)(-2)=0
you get x from solving the quadratic equation. volume will be max at that x
Easy
y=a(x-h)^2+k
vertex is (h,k)
we know that vertex is (4,0)
input that point for (h,k)
y=a(x-4)^2+0
y=a(x-4)^2
passes thorugh the point (6,1)
input that point to find a
1=a(6-4)^2
1=a(2)^2
1=a(4)
divide both sides by 4
1/4=a
thefor the equation is
y=(1/4)(x-4)^2
or
y=(1/4)x^2-2x+4
Answer:
d. x = -3
Step-by-step explanation:
<u />
÷ (x+8)(x-2) = 0
f(x) / g(x) = 0
Variable x cannot be equal to any of the values −8,2 since division by zero is not defined. Multiply both sides of the equation by (x−2)(x+8).
<u /> = 0 ------ square both sides
x + 3 = 0
solve for x:
x = -3
Answer:
18
Step-by-step explanation:
Remark
This is one of those questions that can throw you. The problem is that do you include the original rectangle or not. The way it is written it sounds like you shouldn't
However if you don't the question gives you 2 complex answers. (answers with the sqrt( - 1) in them.
Solution
Let the width = x
Let the length = x + 5
Area of the rectangle: L * w = x * (x + 5)
Area of the smaller squares (there are 2)
Area = 2*s^2
x = s
Area = 2 * x^2
Area of the larger squares = 2 * (x+5)^2
Total Area
x*(x + 5) + 2x^2 + 2(x + 5)^2 = 120 Expand
x^2 + 5x + 2x^2 + 2(x^2 + 10x + 25) = 120 Remove the brackets
x^2 + 5x + 2x^2 + 2x^2 + 20x + 50 = 120 collect the like terms on the left
5x^2 + 25x + 50 = 120 Subtract 120 from both sides.
5x^2 + 25x - 70 = 0 Divide through by 5
x^2 + 5x - 14 = 0 Factor
(x + 7)(x - 2) = 0 x + 7 has no meaning
x - 2 = 0
x = 2
Perimeter
P = 2*w + 2*L
w = 2
L = 2 + 5
L = 7
P = 2*2 + 2 * 7
P = 4 + 14
P = 18
Answer:
b
Step-by-step explanation:
the graph gets translated 5 units above its parent graph of y = x
