I am pretty sure the answer is 10. Someone correct me if I'm wrong
Answer:
5.1
Step-by-step explanation:
Compounded Annually:
A=P(1+r)^t
A=P(1+r)
t
A=27200\hspace{35px}P=20000\hspace{35px}r=0.062
A=27200P=20000r=0.062
Given values
27200=
27200=
\,\,20000(1+0.062)^{t}
20000(1+0.062)
t
Plug in values
27200=
27200=
\,\,20000(1.062)^{t}
20000(1.062)
t
Add
\frac{27200}{20000}=
20000
27200
=
\,\,\frac{20000(1.062)^{t}}{20000}
20000
20000(1.062)
t
Divide by 20000
1.36=
1.36=
\,\,1.062^t
1.062
t
\log\left(1.36\right)=
log(1.36)=
\,\,\log\left(1.062^t\right)
log(1.062
t
)
Take the log of both sides
\log\left(1.36\right)=
log(1.36)=
\,\,t\log\left(1.062\right)
tlog(1.062)
Bring exponent to the front
\frac{\log\left(1.36\right)}{\log\left(1.062\right)}=
log(1.062)
log(1.36)
=
\,\,\frac{t\log\left(1.062\right)}{\log\left(1.062\right)}
log(1.062)
tlog(1.062)
Divide both sides by log(1.062)
5.1116317=
5.1116317=
\,\,t
t
Use calculator
t\approx
t≈
\,\,5.1
5.1
Answer:
115m
Step-by-step explanation:
Note that one of the face of a pyramid is a right angled triangle.
Since the pyramid was about 147m tall, the height of triangle will be 147m. If one of the faces is built with one of its faces at 52° to the incline, we can find the original length of one of its sides using SOH CAH TOA
tan(theta) = opposite/adjacent
Given theta = 52°, opposite = 147m (the side directly opposite the angle)
Substituting we have;
Tan52° = 147/adjacent
Adjacent = 147/tan52°
Adjacent = 114.8m
= 115m
Therefore 115m is the original length of one of its side
Answer:
Carl's checked bags way 15.5 kilograms each.
Step-by-step explanation:
35-4=31 -> 31÷2=15.5
