Answer:
= 330/1 ÷ 3/4
= (330 × 4) / (3 × 1)
= 1320/3
= 440/1
<span>= 440
Hope this helps! </span>
Answer:
Dimensions of the rectangular plot will be 500 ft by 750 ft.
Step-by-step explanation:
Let the length of the rectangular plot = x ft.
and the width of the plot = y ft.
Cost to fence the length at the cost $3.00 per feet = 3x
Cost to fence the width of the cost $2.00 per feet = 2y
Total cost to fence all sides of rectangular plot = 2(3x + 2y)
2(3x + 2y) = 6,000
3x + 2y = 3,000 ----------(1)
3x + 2y = 3000
2y = 3000 - 3x
y = ![\frac{1}{2}[3000-3x]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B3000-3x%5D)
y = 1500 - 
Now area of the rectangle A = xy square feet
A = x[
]
For maximum area 
A' =
= 0
1500 - 3x = 0
3x = 1500
x = 500 ft
From equation (1),
y = 1500 - 
y = 1500 - 750
y = 750 ft
Therefore, for the maximum area of the rectangular plot will be 500 ft × 750 ft.
two fencing 3(500+500) = $3000
other two fencing 2(750+750) = $3000
Step-by-step explanation:
3y +4 =5y -10
2y = 14
y = 7
DEF = (3Y+4)× 2 OR (5Y-10) ×2 OR (5Y-10+ 3Y+4)
Def= (3 ×7+4)×2
=50 degree
.5 -(-7) would be just 7 because of the imaginary -1 it’s being multiplied 7+1+3-7=11 then divide
Volume of a cone = π r² h/3
I envisioned a cone inside a cube. I only identified the necessary value to compute for the exact volume of the largest cone.
Edges of the cube is 10 inches. When the circle part of the cone is placed in the circle, the diameter would be 10 inches also. I divided it by 2 to get the value of the radius, resulting to 5 inches.
The height of the cone is also the height of the cube. Thus, it is 10 inches.
Volume of the cone = 3.14 * (5in)² * 10in/3
V = 3.14 * 25in² * 3.33 in
V = 261.405 in³ or 261.41 in³ VOLUME OF THE LARGEST CONE THAT CAN FIT IN A 10 INCH CUBE