Upper Tolerance
Remark
The 11/16 is the only thing that will be affected. The three won't go up or down when we add 1/64 so we should just work with the 11/16. We need only add 11/16 and 1/64 together to see what the upper range is. Later on we can add 3 into the mix.
Solution
<u>Upper Limit</u>
Now change the 11/16 into 64. Multiply numerator and denominator or 11/16 by 4
Which results in
With a final result for the fractions of 45/64
So the upper tolerance = 3 45/64
<u>Lower Tolerance</u>
Just follow the same steps as you did for the upper tolerance except you subtract 1/64 like this.
Your answer should be 3 and 43/64
(a) Yes all six trig functions exist for this point in quadrant III. The only time you'll run into problems is when either x = 0 or y = 0, due to division by zero errors. For instance, if x = 0, then tan(t) = sin(t)/cos(t) will have cos(t) = 0, as x = cos(t). you cannot have zero in the denominator. Since neither coordinate is zero, we don't have such problems.
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(b) The following functions are positive in quadrant III:
tangent, cotangent
The following functions are negative in quadrant III
cosine, sine, secant, cosecant
A short explanation is that x = cos(t) and y = sin(t). The x and y coordinates are negative in quadrant III, so both sine and cosine are negative. Their reciprocal functions secant and cosecant are negative here as well. Combining sine and cosine to get tan = sin/cos, we see that the negatives cancel which is why tangent is positive here. Cotangent is also positive for similar reasons.
Answer:
2nd option
Step-by-step explanation:
- 5 = 
← in the form 
Answer:
The value of a = - 1 , And b = 
Step-by-step explanation:
Given as :
Function f(x) = 
And f(f(1)) = 0 And f(2) = - 3
Now For , x = 2 , y = - 3
I.e f(2) = 
or, - 3 =
I.e 2 + a = - 6 - 3b
Or, a + 3b = - 8 ....... 1
Again f(f(1)) = 0
So, 
= 0
Or, 1 + a = 0
∴ a = - 1
So , put htis value of a i n eq 1 , we get value of b
So , - 1 + 3b = - 8
Or, 3b = - 7
∴ b =
Hence The value of a = - 1 , And b = Answer
