Such that |x-a|<\varepsilon and $x-b|<\varepsilon. can $a$ be countable? can $a$ be uncountable?
1 answer:
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Answer:
<u>y-intercept</u><u> </u><u>is</u><u> </u><u>(</u><u>0</u><u>,</u><u> </u><u>-</u><u>2</u><u>1</u><u>)</u>
Step-by-step explanation:
For y-intercept, x = 0:
If we say that side "a" is the shortest (adjacent), side "b" is the second shortest (opposite), and side "c" is the longest (hypotenuse): then angle A would be 30°, angle B would be 60°, and angle C would be 90°
tan B = b/a
or
a tan B = b
they have 4 as the smallest value in all four options, so we know that a=4
4 tan 60° = b
tan 60° = √3
so, 4 tan 60° = 4√3
b=4√3
that leaves two options left, so now we find "c":
c^2 = (a^2) + (b^2)
OR
so a=4, b=4√3, and c=8
Making your answer B
tan B = b/a
or
a tan B = b
they have 4 as the smallest value in all four options, so we know that a=4
4 tan 60° = b
tan 60° = √3
so, 4 tan 60° = 4√3
b=4√3
that leaves two options left, so now we find "c":
c^2 = (a^2) + (b^2)
OR
so a=4, b=4√3, and c=8
Making your answer B