Answer:
Now, 3x 2+x+5⩾0
This is because b² −4ac=1−4×5×3
=−59 (roots are imaginary)
3x²+x+5=(x−3)² =x²+9−6x and x−3⩾0
2x²+7x−4=0
2x²+8x−x−4=0
2x(x+4)−1(x+4)=0
(2x−1)(x+4)=0
x= 1/2 and−4 but x≥3
∴ No solution.
lol hehehe
Yes, because of the AAA criteria.
Two lines that intersect and form right angles are called perpendicular lines. The symbol ⊥ is used to denote perpendicular lines.
Given,
Diameter of the can = 3"
Height of the can = 7"
Looking at how the cans are arranged in the box, that is 4 x 5 (4 rows of cans [width] with 5 cans in each row [length])
The length of the box (L) = 5 cans multiplied by each can's diameter = 5 × 3" = 15"
The width of the box (W) = 4 cans multiplied by each can's diameter) = 4 × 3" = 12"
The height of the box (H) = 2 layers of cans = 2 cans multiplied by each can's height = 2 × 7" = 14"
Therefore, the volume of the box = Length (L) × Width (W) × Height (H) = 15" × 12" × 14" = 2520 inches³
Volume of the box = 2520 inches³
=====================================================
There is also an alternative method to calculate the volume of the box:
Consider each can. Although the can is cylindrical, each can would occupy the space required by a cuboid.
So, for each Cuboid space, the diameter of the can will be the length and width of the cuboid and the height of the can will be the height of the cuboid.
Therefore, for each can,
Length (L) = 3"
Width (W) = 3"
Height (H) = 7"
Volume occupied by one can (that is a cuboid) = L × W × H = 3" × 3" ×7" = 63 inches³
There are 40 such cans in total inside the box; therefore,
Volume of the box = 40 × 63 inches³ = 2520 inches³
