Alternate exterior angles are the pair of angles that lie on the outer side of the two parallel lines but on either side of the transversal line. Illustration: ... Notice how the pairs of alternating exterior angles lie on opposite sides of the transversal but outside the two parallel lines.
I'll solve it algebraically
x+y=4
x-3y=12
multiply first equatoin by -1 and add to second
-x-y=-4
<u>x-3y=12 +</u>
0x-4y=8
-4y=8
divide both sides by -4
y=-2
sub back
x+y=4
x+-2=4
add 2 to both sides
x=6
x=6
y=-2
(6,-2)
Answer:
(-10,-7)
Step-by-step explanation:
first reflect over x=-3
(-7,-10)
then reflect over y=x
(-10,-7)
The function, as presented here, is ambiguous in terms of what's being deivded by what. For the sake of example, I will assume that you meant
3x+5a
<span> f(x)= ------------
</span> x^2-a^2
You are saying that the derivative of this function is 0 when x=12. Let's differentiate f(x) with respect to x and then let x = 12:
(x^2-a^2)(3) -(3x+5a)(2x)
f '(x) = ------------------------------------- = 0 when x = 12
[x^2-a^2]^2
(144-a^2)(3) - (36+5a)(24)
------------------------------------ = 0
[ ]^2
Simplifying,
(144-a^2) - 8(36+5a) = 0
144 - a^2 - 288 - 40a = 0
This can be rewritten as a quadratic in standard form:
-a^2 - 40a - 144 = 0, or a^2 + 40a + 144 = 0.
Solve for a by completing the square:
a^2 + 40a + 20^2 - 20^2 + 144 = 0
(a+20)^2 = 400 - 144 = 156
Then a+20 = sqrt[6(26)] = sqrt[6(2)(13)] = 4(3)(13)= 2sqrt(39)
Finally, a = -20 plus or minus 2sqrt(39)
You must check both answers by subst. into the original equation. Only if the result(s) is(are) true is your solution (value of a) correct.
