Answer:
A. square root of a^2 + b^2 for both answers
Step-by-step explanation:
The first problem, we are given
a^2 + b^2 = c^2
What we do is solve for c.
sqrt(a^2 + b^2) = c
c = sqrt(a^2 + b^2)
For problem 2,
WE can apply the Pythagorean theorem because we have a right triangle.
The equations is
a^2 + b^2 = c^2 like the first problem
Solving gets us
sqrt(a^2 + b^2) = c
c = sqrt(a^2 + b^2)
Step-by-step explanation:
cot x + 2 tan x + tan³ x
Write in terms of sine and cosine:
(cos x / sin x) + 2 (sin x / cos x) + (sin³ x / cos³ x)
Find the common denominator:
(cos⁴ x / (sin x cos³ x) + 2 (sin² x cos² x / (sin x cos³ x)) + (sin⁴ x / (sin x cos³ x))
Add:
(cos⁴ x + 2 sin² x cos² x + sin⁴ x) / (sin x cos³ x)
Factor:
(sin² x + cos² x)² / (sin x cos³ x)
Pythagorean identity:
1 / (sin x cos³ x)
Multiply top and bottom by cos x:
cos x / (sin x cos⁴ x)
Simplify:
cot x sec⁴ x
The answer is C because if any of those are not parallel will not make the angle similarity true.
Select all the correct answers:
1) Yes
2) No
x=8→h(8)=2(8)^2+5(8)+2=2(64)+40+2=128+40+2→h(8)=170
x=8→f(8)=3^8+2=6,561+2→f(8)=5,563>170=h(8)
3) Yes
4) No
5) Yes
rg=[g(3)-g(2)]/(3-2)=[g(3)-g(2)]/1→rg=g(3)-g(2)
g(3)=20(3)+4=60+4→g(3)=64
g(2)=20(2)+4=40+4→g(2)=44
rg=64-44→rg=20
rf=f(3)-f(2)
f(3)=3^3+2=27+2→f(3)=29
f(2)=3^2+2=9+2→f(2)=11
rf=29-11→rf=18
rh=h(3)-h(2)
h(3)=2(3)^2+5(3)+2=2(9)+15+2=18+15+2→h(3)=35
h(2)=2(2)^2+5(2)+2=2(4)+10+2=8+10+2→h(2)=20
rh=35-20→rh=15
rg=20>18=rf
rg=20>15=rh
6) No
x=4→g(4)=20(4)+4=80+4→g(4)=84
x=4→h(4)=2(4)^2+5(4)+2=2(16)+20+2=32+20+2→h(4)=54
x=4→f(4)=3^4+2=81+2→f(4)=83>54=h(4)
f(4)=83<84=g(4)
5x6=30
6x6=36
You would need 1 more 6
I can't explain this very well, sorry.
