Answer:
a = 3 & b = -10
Step-by-step explanation:
(X - 5)^2 + a = X^2 +bX + 28
Solving the left hand side
(X - 5)^2 + a
= (X - 5)(X - 5) + a
= X^2 - 5X - 5X + 25 + a
= X^2 - 10X + 25 + a
Equating the the right hand side
X^2 - 10X + 25 + a = X^2 +bX + 28
: -10X = bX
X = -10 and
25 + a = 28
a = 28 - 25
a = 3
Answer:
x ⠀⠀200⠀⠀ ⠀500
y⠀⠀⠀ 11⠀⠀⠀⠀⠀a
As x directly varies with y
200 × a = 500 × 11
a = 5/2 × 11 = 27.5
A way of undoing any operation, is using the inverse of the operation. So if you want to ‘undo’ addition you subtract, and if you want to ‘undo’ multiplication you divide.
Hope this helps!!
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
Answer: 46.8
Step-by-step explanation:
