The area of a triangle can be calculated as half the product of base length and height. We want the area to be no greater than 10 in².
a) (1/2)(4)(2x-3) ≤ 10
b) 4x -6 ≤ 10 . . . . . simplify
... 4x ≤ 16 . . . . . . . .add 6
... x ≤ 4 . . . . . . . . . divide by the coefficient of x
c) The maximum value of (2x -3) in is (2·4 -3) in = 5 in.
The triangle should be no more than 5 in high to have an area less than 10 in².
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Solve for x over the real numbers:18 ° (x - 1) (x - 10 ° x) = 0
Divide both sides by 18 °:(x - 1) (x - 10 ° x) = 0
Split into two equations:x - 1 = 0 or x - 10 ° x = 0
Add 1 to both sides:x = 1 or x - 10 ° x = 0
Collect in terms of x:x = 1 or (1 - 10 °) x = 0
Divide both sides by 1 - 10 °:Answer: x = 1 or x = 0
4 cos² x - 3 = 0
4 cos² x = 3
cos² x = 3/4
cos x = ±(√3)/2
Fixing the squared cosine doesn't discriminate among quadrants. There's one in every quadrant
cos x = ± cos(π/6)
Let's do plus first. In general, cos x = cos a has solutions x = ±a + 2πk integer k
cos x = cos(π/6)
x = ±π/6 + 2πk
Minus next.
cos x = -cos(π/6)
cos x = cos(π - π/6)
cos x = cos(5π/6)
x = ±5π/6 + 2πk
We'll write all our solutions as
x = { -5π/6, -π/6, π/6, 5π/6 } + 2πk integer k
Answer:
<u>Step-by-Step Explanation:</u>
Pythagorean Theorem is: a² + b² = c² , <em>where "c" is the hypotenuse</em>
Note: (15)² + (3√11)² = hypotenuse² → hypotenuse = 18
Note: 8² + 15² = hypotenuse² → hypotenuse = 17
Note: hypotenuse not needed for tan
Note: 2² + 6² = hypotenuse² → hypotenuse = 2√10