Answer:
All rhombuses are similar.
Answer:
After 2 minutes the temperature of the hot chocolate will be 149.46 degrees Fahrenheit.
Step-by-step explanation:
We are going to use the Newton's law of cooling to solve this exercise. The Newton's law of cooling states that the amount of heat lost by a body is proportional to the difference of temperature between the body and the enviroment. We are going to use the following function :
T(t)=T_{0}+(T_{i}-T_{0}).e^{-kt}T(t)=T
0
+(T
i
−T
0
).e
−kt
Where ''T(t)'' is the temperature of the body that depends of the variable ''t'' which is time.
Where T_{0}T
0
is the temperature of the surroundings
In this case T_{0}T
0
is the temperature of the freezer
Where T_{i}T
i
is the initial temperature of the body which is cooling. In this case, T_{i}T
i
is the temperature of the hot chocolate
And where ''k'' is a constant. In this case, k=0.12k=0.12 is a data of the exercise
If we replace all the values in the equation and replacing t=2minutest=2minutes
⇒
T(2minutes)=0+(190-0).e^{-(0.12).(2)}T(2minutes)=0+(190−0).e
−(0.12).(2)
T(2minutes)=(190).(e^{-0.24})=149.46T(2minutes)=(190).(e
−0.24
)=149.46
We find that the temperature of the hot chocolate after 2 minutes is 149.46 degrees Fahrenheit
The answer is 221. And it’s solved with PEMDAS
Answer:
Its 81
Step-by-step explanation:
Answer:
82.4 cm
Step-by-step explanation:
The computation of the height is given below:
As we know that each right triangle contains the top angle of 20 degrees
So, the right angle & the bottom corner angle equivalent to
= 180 - 90 - 20
= 70 degrees
Now the height is
tangent = opposite leg ÷ adjacent leg
tan (70°) = height ÷ 30 cm
height = 30 cm × tan (70°)
= 82.4 cm
