Answer:
Explanation:
<u>1) Calculate the volume of the water in the tank.</u>
- Area of the base of the tank: B = 7 ft × 4 ft = 28 ft²
- Height of the tank: H = 9 in = 9 in × 1 ft / 12 in = 0.75 ft
- Volume of the tank: V = area of the base × height = B × H = 28 ft² × 0.75 ft = 21 ft³.
<u>2) Calculate the weight of 21 ft³ of water.</u>
Since this is not a chemistry question but a math question, I will not use the fomula of density but set a proportion with one unknown:
- 62.4 lb / 1 ft³ = x / 21 ft³
Solve for x:
- x = 21 ft³ × 64 lb / ft³ = 1,310.4 lb.
So, rounding to the next integer, the water in the tank weighs 1,310 pounds, when it is full.
Answer: x < 2
<u>Step-by-step explanation:</u>
2x + 1 < 5
Subtract 1 from both sides:
2x < 4
Divide both sides by 2:
x < 2 <em>(x is less than 2)</em>
Graph: ←-------o -2
Interval Notation: (-∞, -2)
1)
LHS = cot(a/2) - tan(a/2)
= (1 - tan^2(a/2))/tan(a/2)
= (2-sec^2(a/2))/tan(a/2)
= 2cot(a/2) - cosec(a/2)sec(a/2)
= 2(1+cos(a))/sin(a) - 1/(cos(a/2)sin(a/2))
= 2 (1+cos(a))/sin(a) - 2/sin(a)) (product to sums)
= 2[(1+cos(a) -1)/sin(a)]
=2cot a
= RHS
2.
LHS = cot(b/2) + tan(b/2)
= [1 + tan^2(b/2)]/tan(b/2)
= sec^2(b/2)/tan(b/2)
= 1/sin(b/2)cos(b/2)
using product to sums
= 2/sin(b)
= 2cosec(b)
= RHS
-5 = 3/4x - 2
-5 + 2 = 3/4x
-3 = 3/4x
-3 / (3/4) = x
-3 * 4/3 = x
-12/3 = x
-4 = x <==
Answer:
5
Step-by-step explanation:
