You are asking for the condition needed for the
<span>two segments to be perpendicular to each other. </span>
<span>If the slope of one segment is m, then the slope of a perpendicular segment would be -1/m </span>
<span>this means that m2 = -1/m and so m2 x m = -1 </span>
<span>If you look carefully at your choices, the 3rd answer involves the two slope and has the -1. It's </span>
<span>the correct answer.</span>
The answer is True. In geometry, perpendicular lines are defined as two lines that meet or intersect each other at right angles (90°).
Answer:
The answer is 3.7
Step-by-step explanation:
Use cross multiplication to set up the equation then simplify,
17y = 63
y = 63/17
y = 3.705
Round to the nearest tenth.
Solution is y = 3.07
Good luck!
Answer:
-2, 8/3
Step-by-step explanation:
You can consider the area to be that of a trapezoid with parallel bases f(a) and f(4), and width (4-a). The area of that trapezoid is ...
A = (1/2)(f(a) +f(4))(4 -a)
= (1/2)((3a -1) +(3·4 -1))(4 -a)
= (1/2)(3a +10)(4 -a)
We want this area to be 12, so we can substitute that value for A and solve for "a".
12 = (1/2)(3a +10)(4 -a)
24 = (3a +10)(4 -a) = -3a² +2a +40
3a² -2a -16 = 0 . . . . . . subtract the right side
(3a -8)(a +2) = 0 . . . . . factor
Values of "a" that make these factors zero are ...
a = 8/3, a = -2
The values of "a" that make the area under the curve equal to 12 are -2 and 8/3.
_____
<em>Alternate solution</em>
The attachment shows a solution using the numerical integration function of a graphing calculator. The area under the curve of function f(x) on the interval [a, 4] is the integral of f(x) on that interval. Perhaps confusingly, we have called that area f(a). As we have seen above, the area is a quadratic function of "a". I find it convenient to use a calculator's functions to solve problems like this where possible.