What is the area of a right triangle with a base of 4 feet and a height of 8 feet?
1 answer:
Answer:
16 feet squared
Step-by-step explanation:
We'll use the formula for finding the area of a triangle:
area = of the base times the height
First, we'll find of the base. Our base is 4 feet, and half of 4 feet is 2 feet. Now, we have to multiply that by the height. The height is 8 feet, so we have to multiply 8 feet by 2 feet. 8 times 2 is 16, and feet times feet gives us feet squared. So we have 16 feet squared as our final answer.
Hopefully that was helpful! Let me know if you have any more questions.
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Just replace 
with its given value:
Answer:
x = 4
Step-by-step explanation:
For an equation of this sort, my first step is to rewrite it so the solutions are the x-intercepts of the graph:
(x +5)^(3/2) -(x +1)^3 = 0
The graph is attached. The one real solution is x=4.
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The solution method for something like this is necessarily ad hoc. Here, it appears that progress can be made by raising both sides of the equation to the 2/3 power:
((x +5)^(3/2))^(2/3) = ((x -1)^3)^(2/3)
x +5 = (x -1)^2
Now, you have an ordinary quadratic that can be solved using any of the usual methods.
x^2 -2x +1 = x +5 . . . . swap sides, expand the square
x^2 -3x -4 = 0 . . . . . . put in standard form
(x -4)(x +1) = 0 . . . . . . factor
The solutions to this are ...
x = 4, x = -1 . . . . . . . values of x that make the factors zero
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To see if either of these solutions is extraneous, we can check them:
(4 +5)^(3/2) = (4 -1)^3 . . . . . substitute 4 for x
9^(3/2) = 3^3 . . . . . . . . . true
(-1 +5)^(3/2) = (-1 -1)^3 . . . . . substitute -1 for x
4^(3/2) = (-2)^3
8 = -8 . . . . . . . . . . . false; x = -1 is extraneous
The solution is x = 4.
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<em>Additional comment</em>
You can eliminate the x=-1 "solution" by considering the domains of the left- and right-side expressions. The 3/2 power is equivalent to the square root of the cube. That will generally only be defined for non-negative values (x ≥ -5). The cube on the right will only be positive for x > 1, so the practical domain of this equation is x ≥ 1.
