Answer:
Step-by-step explanation:
f * g = (x^2 + 3x - 4) (x+4)
open bracket
x((x^2 + 3x - 4) + 4 (x^2 + 3x - 4)
x³ +3x²-4x+x²+12x-16
x³+3x²+x²-4x+12x-16
x³+4x²+8x-16 (domain is all real numbers.
f/g = (x^2 + 3x - 4)/(x+4)
factorising (x^2 + 3x - 4)
x²+4x-x_4
x(x+4) -1 (x+4)
(x+4)(x-1)
f/g = (x^2 + 3x - 4)/(x+4) =(x+4)(x-1)/(x+4) = (x-1)
Before factorisation, this was a rational function so the domain is all real numbers excluding any value that would make the denominator equal zero.
Hence I got x - 1, and x cannot equal -4
So the domain is just all real numbers without -4
Let Ch and C denote the events of a student receiving an A in <u>ch</u>emistry or <u>c</u>alculus, respectively. We're given that
P(Ch) = 88/520
P(C) = 76/520
P(Ch and C) = 31/520
and we want to find P(Ch or C).
Using the inclusion/exclusion principle, we have
P(Ch or C) = P(Ch) + P(C) - P(Ch and C)
P(Ch or C) = 88/520 + 76/520 - 31/520
P(Ch or C) = 133/520
Answer:
B=48 white flowers
Step-by-step explanation:
For 3 pink flowers=2 white flowers
1 pink flower=2/3 white flowers
Now
72 pink flower=(2/3)×72 white flowers
72 pink flowers=2×24 white flowers
72 pink flowers=48 white flowers
Answer:
y=2x-11
Step-by-step explanation:
Answer:
<h2>3</h2>
Step-by-step explanation:
IF point E lies on the line segment DF, this means that all the points DEF are collinear and DE+EF = DF.
Given parameter
DE = 6
DF = 9
Required
EF
Substituting the given parameter into the expression above to get the required will be;
DE+EF = DF.
EF = DF-DE
EF = 9-6
EF = 3
Hence the length of EF is equivalent to 3
