F(x)=6x^4-10x^3+40x-50, plug 2 in for x
f(2)=6(2)^4-10(3)^3+40(2)-50
f(2)=12^4-30^3+80-50
f(2)=20735-27,000+80-50
f(2)=-6,235
Answer:
ok thx
Step-by-step explanation:
Answer:
D
Step-by-step explanation:
our basic Pythagorean identity is cos²(x) + sin²(x) = 1
we can derive the 2 other using the listed above.
1. (cos²(x) + sin²(x))/cos²(x) = 1/cos²(x)
1 + tan²(x) = sec²(x)
2.(cos²(x) + sin²(x))/sin²(x) = 1/sin²(x)
cot²(x) + 1 = csc²(x)
A. sin^2 theta -1= cos^2 theta
this is false
cos²(x) + sin²(x) = 1
isolating cos²(x)
cos²(x) = 1-sin²(x), not equal to sin²(x)-1
B. Sec^2 theta-tan^2 theta= -1
1 + tan²(x) = sec²(x)
sec²(x)-tan(x) = 1, not -1
false
C. -cos^2 theta-1= sin^2
cos²(x) + sin²(x) = 1
sin²(x) = 1-cos²(x), our 1 is positive not negative, so false
D. Cot^2 theta - csc^2 theta=-1
cot²(x) + 1 = csc²(x)
isolating 1
1 = csc²(x) - cot²(x)
multiplying both sides by -1
-1 = cot²(x) - csc²(x)
TRUE
The answer is 7, your welcome :).
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