1. Find the equation of the line AB. For reference, the answer is y=(-2/3)x+2.
2. Derive a formula for the area of the shaded rectange. It is A=xy (where x is the length and y is the height).
3. Replace "y" in A=xy with the formula for y: y= (-2/3)x+2:
A=x[(-2/3)x+2] This is a formula for Area A in terms of x only.
4. Since we want to maximize the shaded area, we take the derivative with respect to x of A=x[(-2/3)x+2] , or, equivalently, A=(-2/3)x^2 + 2x.
This results in (dA/dx) = (-4/3)x + 2.
5. Set this result = to 0 and solve for the critical value:
(dA/dx) = (-4/3)x + 2=0, or (4/3)x=2 This results in x=(3/4)(2)=3/2
6. Verify that this critical value x=3/2 does indeed maximize the area function.
7. Determine the area of the shaded rectangle for x=3/2, using the previously-derived formula A=(-2/3)x^2 + 2x.
The result is the max. area of the shaded rectangle.
Answer:
Refer to the attachment for the labelling of the triangles.
In △ABC & △PQR,
∠A = ∠R (equal pair of angles)
∠B = ∠Q (equal pair of angles)
AC = PR (equal pair of sides)
•°• △ABC ≅ △RQP (Angle-Angle-Side congruence property → AAS property)
Hope it helps ⚜
Hello!
Answer:
3(4+3g-10h)
Step-by-step explanation:
Hope this helps!
Answer:
x = 1, y = 6
or
x = 5, y = 2.
Step-by-step explanation:
y=x2−7x+12
y=−x+7
Substitute for y in the first equation:
- x + 7 = x^2 - 7x + 12
x^2 - 7x + x + 12 - 7 = 0
x^2 - 6x + 5 = 0
(x - 1)(x - 5) = 0
x = 1, 5.
When x = 1, y = -1 + 7 = 6.
when x = 5, y = -5+7 = 2.
Answer:
560/9 or around 63 cubes
Step-by-step explanation:
2 2/3 -> 8/3
3 1/3 -> 10/3
2 1/3-> 7/3
7/3 x 8/3= 56/9
56/9 x 10/3 = 560/27
560/27 divided by 1/3
= 560/9
