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.
Answer:
B- One-Quarter
Step-by-step explanation:
To solve this problem you must apply the proccedure shown below:
1. You have that the following information given in the problem above: The<span> photograph is reduced by a scale factor of 3/8 and the original had a length of 20 inches,
2- Therefore you have:
length=20 inches-</span>(20 inches x 3/8)<span>
length=20 inches-7.5 inches
length=12.5 inches
3- Then, as you can see, the answer is: 12.5 inches.</span>
![\sqrt[4]{ x^{3} } = x^{3/4}](https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B%20x%5E%7B3%7D%20%7D%20%3D%20x%5E%7B3%2F4%7D%20)
![\sqrt[10]{ x^{5} * x^{4}* x^{2} } = \sqrt[10]{ x^{11} }= x^{11/10}](https://tex.z-dn.net/?f=%20%5Csqrt%5B10%5D%7B%20x%5E%7B5%7D%20%2A%20x%5E%7B4%7D%2A%20x%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%5B10%5D%7B%20x%5E%7B11%7D%20%7D%3D%20x%5E%7B11%2F10%7D%20%20)
