RTP: [a tan(u) + b]² + [b tan(u) - a]² = (a² + b²) sec²(u)
Proving LHS = RHS:
LHS = [a tan(u) + b]² + [b tan(u) - a]²
= a² tan²(u) + 2ab tan(u) + b² + b² tan²(u) - 2ab tan(u) + a²
= (a² + b²) tan²(u) + (a² + b²)
= (a² + b²)[tan²(u) + 1]
= (a² + b²) sec²(u), using the identity: tan²(x) + 1 = sec²(x)
= RHS
Answer:
David might be 40
Step-by-step explanation:
Brianna = 20 + 20 = 40 = David
James = 10 x 2 = Brianna
David = 40 - 20 = Brianna
It’s the absolute value of 0 and 5 because it’s the same absolute value the absolute vale is 5 for both 5 and -5 because absolute value of a number is the distance from 0 you can’t have a negative value
Answer:
Step-by-step explanation:
Given that;
Compressed Soil 2.85 2.66 3 2.82 2.76 2.81 2.78 3.08 2.94 2.86 3.08 2.82 2.78 2.98 3.00 2.78 2.96 2.90 3.18 3.16
Intermediate Soil 3.17 3.37 3.1 3.40 3.38 3.14 3.18 3.26 2.96 3.02 3.54 3.36 3.18 3.12 3.86 2.92 3.46 3.44 3.62 4.26
Compressed soil intermediate soil
Mean 2.907 3.286
SD 0.1414 0.2377
Here we have,
The pooled Standard deviation
So standard error for difference in population mean is
by inputting the values we get 0.0623
Degree of freedom for t is
df = 19 + 20 - 2 = 37,
so t-critical value is 2.715.
So required confidence interval for
will be
So required confidence interval is (0.2103 , 0.5483).
Answer:
A: 82.5
Step-by-step explanation:
Average score of the 4 regular tests = (78 + 72 + 74 + 88)/4 = 78
We are told;
- the regular tests account for 40% of the final grade.
- final exam counts for 20%
- project counts for 10%
- homework counts for 30%
Thus;
weighted mean grade = (40% × 78) + (20% × 79) + (10% × 88) + (30% × 89) = 82.5
