Answer:
Rectangular area as a function of x : A(x) = 200*x + 2*x²
A(max) = 5000 m²
Dimensions:
x = 50 m
l = 100 m
Step-by-step explanation:
"x" is the length of the perpendicular side to the wall of the rectangular area to be fenced, and we call "l" the other side (parallel to the wall of the barn) then:
A(r) = x* l and the perimeter of the rectangular shape is
P = 2*x + 2*l but we won´t use any fencing material along the wll of the barn therefore
P = 2*x + l ⇒ 200 = 2*x + l ⇒ l = 200 - 2*x (1)
And the rectangular area as a function of x is:
A(x) = x * ( 200 - 2*x) ⇒ A(x) = 200*x + 2*x²
Taking derivatives on both sides of the equation we get:
A´(x) = 200 - 4*x ⇒ A´= 0
Then 200 - 4*x = 0 ⇒ 4*x = 200 ⇒ x = 50 m
We find the l value, plugging the value of x in equation (1)
l = 200 - 2*x ⇒ l = 200 - 2*50 ⇒ l = 100 m
A(max) = 100*50
A(max) = 5000 m²
Answer:
Step-by-step explanation:
You are to take the square root across the inequality sign to make it plus or minus 3 squared.
The answer will therefore be C
W + 3w = 90 is obviously wrong
Answer:
Solve the equation for x by finding a, b, and c of the quadratic then applying the quadratic formula.
Exact Form:
x = − 70 ± √4210/30
Decimal Form:
x = −0.03
Step-by-step explanation:
5x(6x+28)=−23
Step 1: Simplify both sides of the equation.
30x2+140x=−23
Step 2: Subtract -23 from both sides.
30x2+140x−(−23)=−23−(−23)
30x2+140x+23=0
Step 3: Use quadratic formula with a=30, b=140, c=23.
x=−b±√b2−4ac/2a
x=−(140)±√(140)2−4(30)(23)/2(30)
x=−140±√16840/60
x=−7/3+1/30√4210 or x=−7/3+−1/30√4210
Begin by finding the lowest point the quadratic equation can be, the vertex;
x²-1= is just a translation down of the graph x²
vertex; (0, -1) and since the graph of x² would extend to infinity beyond that point, we can say {x| x≥0} for domain and {y| y≥-1}.
For the linear equation, it is possible to have all x and y values, therefore range and domain belong to all real numbers.
Hope I helped :)
