The question is worded incorrectly. It should say the DENSITY of water is 62 1/2 lbs per cubic foot. Divide the weight by the density to get volume.
(150 lbs) / (62.5 lbs/ft3) = ? ft3 (the two 3's are exponents. so it'd be ft then an exponent of 3.)
Answer:
Height of the pyramid = 3v/y² units
Step-by-step explanation:
We are given;
- The volume of a solid right pyramid with a square base = v units³
- The length of the base edge = y units
The formula for volume of a pyramid is given as;
V = ⅓ x base area x height.
Since the base is square, we will use the formula for square area which is A = side × side.
Thus, v = ⅓ × (y × y) × height
Making height the subject, we have;
Height = 3v/y²
height of the pyramid would be given by the expression 3v/y² units
The domain (input values) of the cosine function is all negative and positive angle measures.
Let the function be f(x) = cos(x)
The domain of cos(x) is -∞ < x < ∞
The range is -1 ≤ f(x) ≤ 1
Hence, domain of cos(x) is all (+) and (-) angle measures.
Same goes with sine function as well
For function f(x) = sin(x)
The domain is -∞ < x < ∞ and range -1 ≤ f(x) ≤ 1
However for f(x) = tan(x) the same is not applicable.
Answer:
x = 2
Step-by-step explanation:
First, distribute 6 to all terms within the parenthesis.
6(2x + 3) = 12x + 18
Isolate the variable x. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.
12x + 18 = 42
First, subtract 18 from both sides.
12x + 18 (-18) = 42 (-18)
12x = 24
Isolate the variable x. Divide 12 from both sides.
(12x)/12 = (24)/12
x = 24/12
x = 2
2 is your answer for x.
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Given:
The dimensions of the given cuboid are 8 in, 6 in and 7 in.
To find:
The volume and the surface area of the given cuboid.
Solution:
The volume of cuboid is:
Where, l is length, b is breadth and h is height.
Therefore, the volume of the cuboid is 336 cubic inches.
The surface area of the cuboid is:
Where, l is length, b is breadth and h is height.
Therefore, the surface area of the cuboid is 292 square inches.
