Answer:
first you find out how much will 1/2 kg of carrot cost, that way you can multiply it by 3 to get what 3/2 cost.
So, 3÷ 5 = 0.6 per 1/2 kg. Then 0.6 × 3 = 1.8 dollars for 3/2 kg
The total money is $2,178
Answer:
l=0.1401P\\
w =0.2801P
where P = perimeter
Step-by-step explanation:
Given that a window is in the form of a rectangle surmounted by a semicircle.
Perimeter of window =2l+\pid/2+w

Or 
To allow maximum light we must have maximum area
Area = area of rectangle + area of semi circle where rectangle width = diameter of semi circle


Hence we get maximum area when i derivative is 0
i.e. 

Dimensions can be

These are the steps, with their explanations and conclusions:
1) Draw two triangles: ΔRSP and ΔQSP.
2) Since PS is perpendicular to the segment RQ, ∠ RSP and ∠ QSP are equal to 90° (congruent).
3) Since S is the midpoint of the segment RQ, the two segments RS and SQ are congruent.
4) The segment SP is common to both ΔRSP and Δ QSP.
5) You have shown that the two triangles have two pair of equal sides and their angles included also equal, which is the postulate SAS: triangles are congruent if any pair of corresponding sides and their included angles are equal in both triangles.
Then, now you conclude that, since the two triangles are congruent, every pair of corresponding sides are congruent, and so the segments RP and PQ are congruent, which means that the distance from P to R is the same distance from P to Q, i.e. P is equidistant from points R and Q
Answer:
The weight of the water in the pool is approximately 60,000 lb·f
Step-by-step explanation:
The details of the swimming pool are;
The dimensions of the rectangular cross-section of the swimming pool = 10 feet × 20 feet
The depth of the pool = 5 feet
The density of the water in the pool = 60 pounds per cubic foot
From the question, we have;
The weight of the water in Pound force = W = The volume of water in the pool given in ft.³ × The density of water in the pool given in lb/ft.³ × Acceleration due to gravity, g
The volume of water in the pool = Cross-sectional area × Depth
∴ The volume of water in the pool = 10 ft. × 20 ft. × 5 ft. = 1,000 ft.³
Acceleration due to gravity, g ≈ 32.09 ft./s²
∴ W = 1,000 ft.³ × 60 lb/ft.³ × 32.09 ft./s² = 266,196.089 N
266,196.089 N ≈ 60,000 lb·f
The weight of the water in the pool ≈ 60,000 lb·f