Answer:
58.5
Step-by-step explanation:
your welcome
Use
y-y1=m(x-x1)
y-(-2)=4(x-1) Now why 4??Because parallel lines have the exact same slope
y+2=4x-4
y=4x-4-2
y=4x-6
Hope it helps!

~Just a teen who listens to music
Answer:
Correct answer: F. graph F or x ∈ |-5 ; 5| (including endpoints)
Step-by-step explanation:
Let us first define the absolute value:
| x | = 1. { x with condition x ≥ 0 }
or 2. { - x with condition x < 0 }
This is a linear inequality
1. x ≤ 5 ∧ x ≥ 0 ⇒ 0 ≤ x ≤ 5 or interval x ∈ |0 ; 5| (including endpoints)
2. - x ≤ 5 when we multiply both sides of the equation by -1 we get:
x ≥ -5 ∧ x < 0 ⇒ -5 ≤ x < 0 or interval x ∈ |-5 ; 0) (including -5)
The solution to this linear inequality is the union of these two intervals:
x ∈ |-5 ; 0) ∪ |0 ; 5| ⇒ x ∈ |-5 ; 5| (including endpoints)
x ∈ |-5 ; 5| (including endpoints)
God is with you!!!
Answer:
Step-by-step explanation:
The complete question is
Water flows into a tank according to the rate F(t)= (t+6)/(1+t), and at the same time empties out at the rate E(t)= (ln(t+2))/(t+1), with both F(t) and E(t) measured in gallons per minute. How much water, to the nearest galllon, is in the tank at time t=10 minutes.
Let C(t) be the amount of water in the tank at time t. We now that the rate of change of the tank is given by
![\frac{dC}{dt}=[\tex]rate at which water flows in- rate at which water flows out. Then [tex]\frac{dC}{dt}=\frac{t+6}{t+1}-\frac{\ln(t+2)}{(t+1)}[\tex]so, the desired expression is obtained by integrating with respect to t. This leads us to [tex]C(t) = \int \frac{t+1}{t+1}+ \frac{5}{t+1} - \frac{\ln(t+2)}{(t+1)} dt=t+ 5 \ln (|t+1|)-\int \frac{\ln(t+2)}{(t+1)} dt +C](https://tex.z-dn.net/?f=%5Cfrac%7BdC%7D%7Bdt%7D%3D%5B%5Ctex%5Drate%20at%20which%20water%20flows%20in-%20rate%20at%20which%20water%20flows%20out.%20%3C%2Fp%3E%3Cp%3EThen%20%3C%2Fp%3E%3Cp%3E%5Btex%5D%5Cfrac%7BdC%7D%7Bdt%7D%3D%5Cfrac%7Bt%2B6%7D%7Bt%2B1%7D-%5Cfrac%7B%5Cln%28t%2B2%29%7D%7B%28t%2B1%29%7D%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3Eso%2C%20the%20desired%20expression%20is%20obtained%20by%20integrating%20with%20respect%20to%20t.%20%3C%2Fp%3E%3Cp%3EThis%20leads%20us%20to%20%3C%2Fp%3E%3Cp%3E%5Btex%5DC%28t%29%20%20%3D%20%5Cint%20%5Cfrac%7Bt%2B1%7D%7Bt%2B1%7D%2B%20%5Cfrac%7B5%7D%7Bt%2B1%7D%20-%20%5Cfrac%7B%5Cln%28t%2B2%29%7D%7B%28t%2B1%29%7D%20dt%3C%2Fp%3E%3Cp%3E%3Dt%2B%205%20%5Cln%20%28%7Ct%2B1%7C%29-%5Cint%20%5Cfrac%7B%5Cln%28t%2B2%29%7D%7B%28t%2B1%29%7D%20dt%20%2BC)
Unfortunately, the integral
cannot be expressed using fundamental functions. So, the problem cannot have an specific function (if you are willing to know the complete answer, the integral of this function uses the polylogarithm function with n=2).
Since you want the exact amount of water at time, you need to give C a value, that is, you need to know an initial condition for the problem. This means, you need to know the amount of water in the tank at time 0
Answer:
There seems to be a typo error but still this has solution.
Let the number be = x
A positive real number is 6 less than another. Means the second will be 
The sum of the square of the two numbers is 38.
=> 
=> 
=> 
=> 
Taking out 2 common;
=> 
Solving this quadratic equation, we get;
and 
As positive number is needed, we have
or x = 
x = 0.162277
And other number is 6.162277.