Answer:
(You only need to give one solution)
Step-by-step explanation:
We have the following equation
First, we need to foil out the parenthesis
Now we can combine the like terms
Now, we need to factor this equation.
To factor this, we need to find a set of numbers that add together to get -3 and multiply to give us -4.
The pair of numbers that would do this would be 1 and -4.
This means that our factored form would be
As the first binomial is a difference of squares, it can be factored futher into
Now, we can get our solutions.
The first binomial will produce two complex (Not real) solutions.
So our solutions to this equation are
Week 1-2 so it would be week one
A.) A= (1/2)absinC
b.) sinA/a, sinB/b, sinC/c
sin100/42, sin25/b, sinC/c
use these two fractions.
sin100/42= sin25/b
cross multiply next
b×sin100= 42×sin25
then divide on both sides by sin100
b=18
c.)measurementofA+measurementofB+measurementofC=180
100+25+C=180
125+C=180
-125 -125
The measurement of angle C is 55 degrees.
d.)A=(1/2)42×18×sin55
= 21×18×sin55
A =309.6 units squared
Answer:
Step-by-step explanation: can we get an answer?
A. True. We see this by taking the highest order term in each factor:
B. True. Again we look at the leading term's degree and coefficient. f(x) behaves like -3x⁶ when x gets large. The degree is even, so as x goes to either ± ∞, x⁶ will make it positive, but multiplying by -3 will make it negative. So on both sides f(x) approaches -∞.
C. False. f(x) = 0 only for x=0, x = 5, and x = -2.
D. False. Part of this we know from the end behavior discussed in part B. On any closed interval, every polynomial is bounded, so that for any x in [-2, 5], f(x) cannot attain every positive real number.
E. True. x = 0 is a root, so f(0) = 0 and the graph of f(x) passes through (0, 0).
F. False. (0, 2) corresponds to x = 0 and f(x) = 2. But f(0) = 0 ≠ 2.
