Answer:
T = 3.967 C
Step-by-step explanation:
Density = mass / volume
Use the mass = 1kg and volume as the equation given V, we will come up with the following equation
D = 1 / 999.87−0.06426T+0.0085043T^2−0.0000679T^3
= (999.87−0.06426T+0.0085043T^2−0.0000679T^3)^-1
Find the first derivative of D with respect to temperature T
dD/dT = 
Let dD/dT = 0 to find the critical value we will get
= 0
Using formula of quadratic, we get the roots:
T = 79.53 and T = 3.967
Since the temperature is only between 0 and 30, pick T = 3.967
Find 2nd derivative to check whether the equation will have maximum value:
Substituting the value with T=3.967,
d2D/dT2 = -1.54 x 10^(-8) a negative value. Hence It is a maximum value
Substitute T =3.967 into equation V, we get V = 0.001 i.e. the volume when the the density is the highest is at 0.001 m3 with density of
D = 1/0.001 = 1000 kg/m3
Therefore T = 3.967 C
Answer:
4-25
Step-by-step explanation:
So what you would do is you would divide the top number (numerator) by the bottom number (denominator) Like you would normally. So take 3/10 for example you would punch into a calculator 3 divided by 10 or do the work which would be
0.3
and that is how you convert a fraction into a decimal with a denominator of 10
I hope that this helped you. (:
Answer:
$3283.2
Step-by-step explanation:
Given data
Principal= $2700
Rate= 4%
Time= 5 years
Required
the final Amount A
The compound interest formula is
A=P(1+r)^t
Substitute
A=2700(1+0.04)^5
A=2700(1.04)^5
A=2700*1.216
A=$3283.2
Hence the balance in the account after 5 years is $3283.2
Answer:
Second choice:
Fifth choice:
Step-by-step explanation:
Let's look at choice 1.
I'm going to subtract 1 on both sides for the first equation giving me 
. I will replace the 
in the second equation with this substitution from equation 1.
Expand using the distributive property and the identity 
:
So this not the desired result.
Let's look at choice 2.
Solve the first equation for by dividing both sides by 2:
.
Let's plug this into equation 2:
This is the desired result.
Choice 3:
Solve the first equation for by adding 3 on both sides:
.
Plug into second equation:
Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:
Not the desired result.
Choice 4:
I'm going to solve the bottom equation for since I don't want to deal with square roots.
Add 3 on both sides:
Divide both sides by 2:
Plug into equation 1:
This is not the desired result because the 
variable will be squared now instead of the 
variable.
Choice 5:
Solve the first equation for by subtracting 1 on both sides:
.
Plug into equation 2:
Distribute and use the binomial square identity used earlier:
.
This is the desired result.
