Answer:
f'(x) = 2[3tan²(x)sec²(x) - 10csc⁴(x)cot(x)]
Step-by-step explanation:
f' of tan(x) = sec²(x)
f' of csc(x) = -csc(x)cot(x)
General Power Rule: uⁿ = xuⁿ⁻¹ · u'
Step 1: Write equation
2tan³(x) + 5csc⁴(x)
Step 2: Rewrite
2(tan(x))³ + 5(csc(x))⁴
Step 3: Find derivative
d/dx 2(tan(x))³ + 5(csc(x))⁴
- General Power Rule: 2 · 3(tan(x))² · sec²(x) + 5 · 4(csc(x))³ · -csc(x)cot(x)
- Multiply: 6(tan(x))²sec²(x) - 20(csc(x))³csc(x)cot(x)
- Simplify: 6tan²(x)sec²(x) - 20csc⁴(x)cot(x)
- Factor: 2[3tan²(x)sec²(x) - 10csc⁴(x)cot(x)]
Hey there!
x^2-18x+79=0
To complete the square, you need to take half of 18, and square it. You have to have complete squares for C for this to work.
-18/2= -9
-9^2= 81
(x^2-18x+81)-81+79=0
The factor is 9, and the -18 is negative.
(x-9)^2 -81+79=0
Add the numbers.
-81+79= -2
(x-9)^2=-2
Now, unsquare both sides.
√(x-9)^2= √-2
Negative square roots do not exist, so it is an imaginary number. Put tha tin front of the 2.
(x-9)=± i√2
x= -9±i√2 <=== the answer
I hope this helps!
~kaikers
Answer:
0.008 meters
Step-by-step explanation:
The answers is d. The right way to notate the yellow region indicated is D.
Answer:
The area will be nine times bigger than the original area.
Step-by-step explanation:
Brainliest appreciated.