F(x)=x^3-9x
and
g(x)=x^2-2x-3
so you just need to divide f(x) by g(x)
Therefore:
f(x)/g(x) = (x^3-9x) / (x^2-2x-3)
and of course you need to factor these two function to see if some factor would cancel another
x^3-9x = x(x^2-9)=x(x-3)(x+3)
and
x^2-2x-3 = (x-3)(x+1)
F(x)=5x
normal domain: all real numbers
practical domain: <span>all positive integers
</span>becasue we can substituent with any positive integer in the place of x
We know that these two angles are equal to each other (There is the "congruent" sign) so we can set them equal to each other and solve for x
3x - 17 = 25 - 3x
(3x + 3x) - 17 = 25 + (-3x + 3x)
6x + (- 17 + 17) = 25 + 17
6x/6 = 42/6
x = 7
Hope this helped!
Answer:
KL = 27
JK = 16
MK = 30
NL = 23
m∠JKL = 132°
m∠KLJ = 22°
m∠KMJ = 54°
m∠KJL = 26°
Step-by-step explanation:
The given parameters of the quadrilateral JKLM are;
JM = 27, ML = 16, JL = 46, NK = 15, KLM = 48, JKM = 78, MJL = 22
Taking the sides as parallel, we have that quadrilateral JKLM is a parallelogram
Therefore;
KL = JM = 27
JK = ML = 16
m∠KLJ = m∠MJL = 22°
MN = NK = 15
MK = MN + NK = 15 + 15 = 30
NL = JL/2 = 46/2 = 23
m∠KJM = m∠KLM = 48°
m∠KJL = m∠KLM - m∠MJL = 48° - 22° = 26°
m∠KML = m∠JKM = 78°
m∠MKL = 180° - m∠KML - m∠KLM = 180° - 78° - 48° = 54°
m∠MKL = 54°
m∠JKL = m∠JKM + m∠MKL = 78° + 54° = 132°
m∠KMJ = m∠MKL = 54°
