Answer:
B) -3.464
Step-by-step explanation:
∫₀³ g'(x) cos²(2g(x) + 1) dx
Using u substitution:
u = 2g(x) + 1
du = 2g'(x) dx
½ du = g'(x) dx
When x = 0, u = 11.
When x = 3, u = -3.
½ ∫₁₁⁻³ cos²(u) du
You can use a calculator to solve this, or you can evaluate algebraically.
Use power reduction formula:
½ ∫ (½ + ½ cos(2u)) du
¼ ∫ du + ¼ ∫ cos(2u) du
¼ ∫ du + ⅛ ∫ 2 cos(2u) du
¼ u + ⅛ sin(2u) + C
Evaluating from u = 11 to u = -3:
[¼ (-3) + ⅛ sin(-6) + C] − [¼ (11) + ⅛ sin(22) + C]
-⁷/₂ + ⅛ sin(-6) − ⅛ sin(22)
−3.464
The 6 in the hundreds place tells you the quotient is at least 100. The non-zero digits in the 1s and 10s places tell you the quotient is more than 100.
In order to answer your question AND dissapoint you... :P
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It is impossible to write 0.02 in tenths.
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Let's review.
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Let's say we have decimal 0.123.
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In decimal format...
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1 is the tenths place.
2 is the hundredths place.
3 is the thousandths place.
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And so on.
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So... unfortunately, you cannot write 0.02 in tenths.
The answer is -16 - 10i.
Using the distributive property on the first part, we have:
-2i*7--2i*4i + (3+i)(-2+2i)
-14i+8i² +(3+i)(-2+2i)
Using FOIL on the last part,
-14i+8i²+(3*-2+3*2i+i*-2+i*2i)
-14i+8i²-6+6i-2i+2i²
-10i+8i²-6+2i²
Since we know that i = -1,
-10i+8(-1)-6+2(-1)
-10i-8-6-2
-16-10i