Answer:
Solving the given formula for v2 gives us:
Step-by-step explanation:
Solving an equation for a particular variable means that the variable has to be isolated on one side of the equation.
Given equation is:
Multiplying both sides by t2-t1
Adding v1 to both sides of the equation
Hence,
Solving the given formula for v2 gives us:
The question is somewhat poorly posed because the equation doesn't involve <em>θ</em> at all. I assume the author meant to use <em>x</em>.
sec(<em>x</em>) = csc(<em>x</em>)
By definition of secant and cosecant,
1/cos(<em>x</em>) = 1/sin(<em>x</em>)
Multiply both sides by sin(<em>x</em>) :
sin(<em>x</em>)/cos(<em>x</em>) = sin(<em>x</em>)/sin(<em>x</em>)
As long as sin(<em>x</em>) ≠ 0, this reduces to
sin(<em>x</em>)/cos(<em>x</em>) = 1
By definition of tangent,
tan(<em>x</em>) = 1
Solve for <em>x</em> :
<em>x</em> = arctan(1) + <em>nπ</em>
<em>x</em> = <em>π</em>/4 + <em>nπ</em>
(where <em>n</em> is any integer)
In the interval 0 ≤ <em>x</em> ≤ 2<em>π</em>, you get 2 solutions when <em>n</em> = 0 and <em>n</em> = 1 of
<em>x</em> = <em>π</em>/4 <u>or</u> <em>x</em> = 5<em>π</em>/4
Answer:
The answer is B. 600
Step-by-step explanation:
Hope that helps!
Hello!
1 foot = 12 inches
12 × 6 = 72
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ANSWER:
Since there are 72 inches in 6 feet, 6 feet are greater than 7 inches.
<h2>
Answer and Explanation:</h2><h2 />
The density of a substance is defined as its mass per unit volume. In a mathematical language, this can be written as follows:
From the problem, a student is trying to calculate the density of a ball. She already knows the mass, but is facing a problem because she also needs to find the volume. From geometry, we know that the volume of a sphere is:
Since a ball is a sphere, then the student can use this formula. She just need to measure the radius of the ball and compute the Volumen, then she have to plug both the value of the mass and the value of the Volume in the equation 
. An easy way to find the radius is to take the measure of the diameter of the ball and divide that value by 2.
