Answer:
(f-g)(x)=-x^(2)+2x+8
the solutions are:
<em><u>x=4 or x=-2</u></em>
Step-by-step explanation:
(f-g)(x)=2x+1-(x^(2)-7)
(f-g)(x)=-x^(2)+2x+1+7
(f-g)(x)=-x^(2)+2x+8
does this help or should I solve for the zeros/solutions of this quadratic equation?
then:
-x^(2)+2x+8=0
-(x^(2)-2x-8)=0
x^(2)-2x-8=0
(x-4)(x+2)=0
<em><u>x=4 or x=-2</u></em>
The function that would best represent the given condition above is,
f(x) = 24 / (3x - 2)
To determine the value of f(-2), substitute -2 to all the x's in the function such that,
f(-2) = 24 / ((3)(-2) - 2) = -3
The value of f(-2) is -3.
Does not exist, since the graph of cosine is continuous between the intervals [-1,1]. You do not know where it will be when it “reaches” infinity
Answer/Step-by-step explanation:
27.
✔️Sin 23 = opp/hyp
Sin 23 = t/34
34*sin 23 = t
t = 13.3
✔️Cos 23 = adj/hyp
Cos 23 = s/34
s = 34*cos 23
s = 31.3
28.
✔️Sin 36 = opp/hyp
Sin 36 = s/5
s = 5*sin 36
s = 2.9
✔️Cos 36 = adj/hyp
Cos 36 = r/5
r = 5*cos 36
r = 4.0
29.
✔️Sin 70 = opp/hyp
Sin 70 = w/10
w = 10*sin 70
w = 9.4
✔️Cos 70 = adj/hyp
Cos 70 = v/10
v = 10*cos 70
v = 3.4
