Let's prove each one:
A) y=x-3
x=0→y=0-3→y=-3=-3 ok
x=2→y=2-3→y=-1 different to 5 No

B) y=x+3
x=0→y=0+3→y=3 different to -3 No

C) y=4x-3
x=0→y=4(0)-3=0-3→y=-3=-3 ok
x=2→y=4(2)-3=8-3→y=5=5 ok
x=3→y=4(3)-3=12-3→y=9=9 ok
x=4→y=4(4)-3=16-3→y=13=13 ok

D) y=4x+3
x=0→y=4(0)+3=0+3→y=3 different to -3 No

Answer: Option C) y=4x-3

8 0

3 years ago

2+cotA=1 homework help

Alex Ar [27]

$2+\cot A=1\implies \cot A=-1\implies \tan A=-1$

This happens whenever $A=\dfrac{3\pi}4$ or $A=\dfrac{7\pi}4$ . More generally, $\tan A=-1$ whenever you start with one of these angles and add any multiple of $\pi$ , so the general solution would be $A=\dfrac{3\pi}4+n\pi$ , where $n$ is any integer. (Notice that when $n=1$ , you end up with $\dfrac{7\pi}4$ .)

3 0

3 years ago