Answer:
vertical asymptote at x = 7
horizontal asymptote at y = 6
Step-by-step explanation:
The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.
solve:
x
−
7
=
0
⇒
x
=
7 is the asymptote
Horizontal asymptotes occur as
lim
,f(x)→
c(a constant)
x
→
±
∞
divide terms on numerator/denominator by x
f
(
x
)
=
5/
x
+6=
5
/x
+
6
x/
x
−
7
/x 1
−
7/
x
as
x
±
∞
,
f
(
x
)
→
0
+6
1
−
0
⇒
y
=
6 is the asymptote
graph{((5)/(x-7))+6 [-20, 20, -10, 10]}
7>d
hope this help!!!
ask me if you got a question. hahahah
Answer:
1/2
Step-by-step explanation:
Given:
(2 cubed) (2 superscript negative 4)
= (2³)(2^-4)
= (2³) (1 / 2⁴)
= (2³ * 1) / 2⁴
= 2³ / 2⁴
Both numerator and denominator has the same base. Thus, pick one of the bases
Also, in indices, division sign can be translated to subtraction
Therefore,
2³ / 2⁴
= 2^3-4
= 2^-1
= 1/2¹
= 1/2
(2³)(2^-4) = 1/2
4 to 5, or 4.5. I hoped I helped!!
-Dawn
Use the trig identity
2*sin(A)*cos(A) = sin(2*A)
to get
sin(A)*cos(A) = (1/2)*sin(2*A)
So to find the max of sin(A)*cos(A), we can find the max of (1/2)*sin(2*A)
It turns out that sin(x) maxes out at 1 where x can be any expression you want. In this case, x = 2*A.
So (1/2)*sin(2*A) maxes out at (1/2)*1 = 1/2 = 0.5
The greatest value of sin(A)*cos(A) is 1/2 = 0.5
