Given:
A Hudson Valley farmer has 26 employees.
Salary of each employee = $410 per week.
Remaining amount after one week = $126
To find:
The initial amount of money he begins with.
Solution:
We have,
Amount he pays to one employee in one week = $410
Amount he pays to 26 employee in one week = 26 × $410
= $10660
Let x be the initial amount of money he begins with.
Remaining amount = Initial amount - Amount he pays to 26 employee in one week
![126=x-10660](https://tex.z-dn.net/?f=126%3Dx-10660)
![126+10660=x](https://tex.z-dn.net/?f=126%2B10660%3Dx)
![10786=x](https://tex.z-dn.net/?f=10786%3Dx)
Therefore, the initial amount of money he begins with is $10786.
Answer:
Y=x(5)+120
Step-by-step explanation:
Y= the money you have to pay for the party
X(5)= you multiply x by 5
+120= you add the fee
a. Given that y = f(x) and f(0) = -2, by the fundamental theorem of calculus we have
![\displaystyle \frac{dy}{dx} = 3x^2 + 4x + k \implies y = f(0) + \int_0^x (3t^2+4t+k) \, dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%203x%5E2%20%2B%204x%20%2B%20k%20%5Cimplies%20y%20%3D%20f%280%29%20%2B%20%5Cint_0%5Ex%20%283t%5E2%2B4t%2Bk%29%20%5C%2C%20dt)
Evaluate the integral to solve for y :
![\displaystyle y = -2 + \int_0^x (3t^2+4t+k) \, dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%20%3D%20-2%20%2B%20%5Cint_0%5Ex%20%283t%5E2%2B4t%2Bk%29%20%5C%2C%20dt)
![\displaystyle y = -2 + (t^3+2t^2+kt)\bigg|_0^x](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%20%3D%20-2%20%2B%20%28t%5E3%2B2t%5E2%2Bkt%29%5Cbigg%7C_0%5Ex)
![\displaystyle y = x^3+2x^2+kx - 2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%20%3D%20x%5E3%2B2x%5E2%2Bkx%20-%202)
Use the other known value, f(2) = 18, to solve for k :
![18 = 2^3 + 2\times2^2+2k - 2 \implies \boxed{k = 2}](https://tex.z-dn.net/?f=18%20%3D%202%5E3%20%2B%202%5Ctimes2%5E2%2B2k%20-%202%20%5Cimplies%20%5Cboxed%7Bk%20%3D%202%7D)
Then the curve C has equation
![\boxed{y = x^3 + 2x^2 + 2x - 2}](https://tex.z-dn.net/?f=%5Cboxed%7By%20%3D%20x%5E3%20%2B%202x%5E2%20%2B%202x%20-%202%7D)
b. Any tangent to the curve C at a point (a, f(a)) has slope equal to the derivative of y at that point:
![\dfrac{dy}{dx}\bigg|_{x=a} = 3a^2 + 4a + 2](https://tex.z-dn.net/?f=%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cbigg%7C_%7Bx%3Da%7D%20%3D%203a%5E2%20%2B%204a%20%2B%202)
The slope of the given tangent line
is 1. Solve for a :
![3a^2 + 4a + 2 = 1 \implies 3a^2 + 4a + 1 = (3a+1)(a+1)=0 \implies a = -\dfrac13 \text{ or }a = -1](https://tex.z-dn.net/?f=3a%5E2%20%2B%204a%20%2B%202%20%3D%201%20%5Cimplies%203a%5E2%20%2B%204a%20%2B%201%20%3D%20%283a%2B1%29%28a%2B1%29%3D0%20%5Cimplies%20a%20%3D%20-%5Cdfrac13%20%5Ctext%7B%20or%20%7Da%20%3D%20-1)
so we know there exists a tangent to C with slope 1. When x = -1/3, we have y = f(-1/3) = -67/27; when x = -1, we have y = f(-1) = -3. This means the tangent line must meet C at either (-1/3, -67/27) or (-1, -3).
Decide which of these points is correct:
![x - 2 = x^3 + 2x^2 + 2x - 2 \implies x^3 + 2x^2 + x = x(x+1)^2=0 \implies x=0 \text{ or } x = -1](https://tex.z-dn.net/?f=x%20-%202%20%3D%20x%5E3%20%2B%202x%5E2%20%2B%202x%20-%202%20%5Cimplies%20x%5E3%20%2B%202x%5E2%20%2B%20x%20%3D%20x%28x%2B1%29%5E2%3D0%20%5Cimplies%20x%3D0%20%5Ctext%7B%20or%20%7D%20x%20%3D%20-1)
So, the point of contact between the tangent line and C is (-1, -3).
Answer:
no solution
Step-by-step explanation:
ya gotta use your eyes and brain huney. it aint that hard. kk? but ill still explain for you.
since x+y=6 and x+y=4. then 4=6. but that is false so no solution.
YOUR WELCOME!
Answer:
LCM (7, 18 and 21) = 126
Step-by-step explanation:
Step 1: Address input parameters & values
Intergers: 7 18 21
LCM (7, 18, 21) = ?
Step 2: Arrange the group of numbers in the horizontal form with space or comma separated format
7, 18 and 21
Step 3: Choose the divisor which divides each or most of the integers of in the group (7, 18 and 21), divide each integers separately and write down the quotient in the next line right under the respective integers. Bring down the integer to the next line if the integer is not divisible by the divisor. Repeat the same process until all the integers are brought to 1.
Step 4: Multiply the divisors to find the LCM 7, 18 and 21
7 × 3 × 6 = 21
LCM(7, 18, 21) = 126
The least common multiple for three numbers 7, 18, and 21 is 126