Sub x = 2-y^2 to Q, we get:
Q = 3(2-y^2)*y^2
let y^2 = k
Q = 3(2-k)k = 3(2k-k^2)
2k-k^2 has a max when k = 1
Then y^2 = 1 -> y = 1 or -1
Answer:
option 2.
m1 = 75 , m2 = 129 , m3 = 100
Step-by-step explanation:
with the rule that the internal angles of a triangle add up to 180 ° we can calculate the missing angles
x + 46 + 29 = 180
x = 180 - 46 -29
x = 105
a flat angle has 180 °
m1 + 105 = 180
m1 = 180 - 105
m1 = 75
46 + 54 + y = 180
y = 180 - 46 -54
y = 80
80 = z + 29
z = 80 - 29
z = 51
as they are two crossed lines the angle is reflected from the opposite side
with that principle and knowing that the angle of a turn is 360 °, if we subtract the 2 known angles and divide it by 2 we will obtain the missing angle (m2)
m2 * 2 = 360 - 51 * 2
m2 = 258/2
m2 = 129
m2 = 29 + m3
129 = 29 + m3
m3 = 129 - 29
m3 = 100
Answer:
(c) III
Step-by-step explanation:
If you simplify the equations and the left side is identical to the right side, then there are an infinite number of solutions: the equation is true for all values of x.
Another way to simplify the equation is to subtract the right side from both sides. If that simplifies to 0 = 0, then there are an infinite number of solutions.
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<h3>I. </h3>
2x -6 -6x = 2 -4x . . . . eliminate parentheses
-4x -6 = -4x +2 . . . . no solutions (no value of x makes this true)
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<h3>II.</h3>
x +2 = 15x +10 +2x . . . . eliminate parentheses
x +2 = 17x +10 . . . . one solution (x=-1/2)
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<h3>III.</h3>
4 +6x = 6x +4 . . . . eliminate parentheses
6x +4 = 6x +4 . . . . infinite solutions
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<h3>IV.</h3>
6x +24 = 2x -4 . . . . eliminate parentheses; one solution (x=-7)
Answer:
M. (4x-1) (x+4)
K. (3x-1)(2x-1)
L.(2x-7)(x+5)
N.3x^2-4x+39
O.(3x-1)(3x+4)
Sorry if it's wrong but that's what I got