Answer:
B. Angles ABD and BAD
Step-by-step explanation:
Markings are different
Lets say we have
P(x)/q(x)
vertical assymtotes are in the form x=something, not y=0
y=0 are horizontal assemtotes
so verticall assymtotes
reduce the fraction
set the denomenator equal to zero
those values that make the deomenator zero are the vertical assymtotes
the horizontal assymtote
when the degree of P(x)<q(x), then HA=0
when the degree of P(x)=q(x), then divide the leading coefient of P(x) by the leading coeficnet of q(x)
example, f(x)=(2x^2-3x+3)/(9x^2-93x+993), then HA is 2/9
ok so for vertical assymtote example
f(x)=x/(x^2+5x+6)
the VA's are at x=-3 and x=-2
horizontal assymtote
make degree same
f(x)=(3x^2-4)/(8x^2+9x),
the HA is 3/8
hope I helped, read the whole thing then ask eusiton
Answer:
-16 +19w
Step-by-step explanation:
-8(2-3w) -5w
Distribute the -8
-8*2 -8*(-3w) -5w
-16 +24w -5w
-16 +19w
There is no solution to this problem
Answer:
<h2>B = 18°</h2>
Step-by-step explanation:
To find angle B we use tan
tan ∅ = opposite / adjacent
From the question
AC is the opposite
BC is the side adjacent to angle B
So we have
tan B = AC / BC
tan B = 6/9
tan B = 1/3
B = tan-¹ 1/3
B = 18.43°
B = 18° to the nearest hundredth
Hope this helps you