I have ti find the two solutions to this equation. This problem deals with imaginary numbers i [tex]{3x}^{2} - x + 4 = 0[/tex]
I have ti find the two solutions to this equation. This problem deals with imaginary numbers i [tex]{3x}^{2} - x + 4 = 0[/tex]
x=5, y=4. (5, 4).
Step-by-step explanation:
2x-3y=-2
4x+y=24
-2(2x-3y)=-2(-2)
4x+y=24
-4x+6y=4
4x+y=24
7y=28
y=28/7
y=4
4x+4=24
4x=24-4
4x=20
x=20/4
x=5
1. Yes
2. Yes
Step-by-step explanation:
1. Lets test Lisa's different solution. Lets times 5 to the first equation:
[tex]5\cdot{x}+10\cdot{y}=35[/tex]
and 2 times the second equation:
[tex]4\cdot{x}-10\cdot{y}=10[/tex]
lets add the two equations:
[tex]9\cdot{x}=45[/tex]
[tex]x=5[/tex]
The second method is multiplying the first equation by -2:
[tex]-2\cdot{x}-4\cdot{y}=-14[/tex]
and add the second equation:
[tex]-9\cdot{y}=-9[/tex]
[tex]y=1[tex]</p<pSubstitute into equation 1:</p<p[tex]x+2\cdot{1}=7[/tex]
[tex]x=5[/tex]
The answer to Lisa's question is yes she wull get the same solution if she uses a different method.
2. Yes, The answer would change if the same amout of x and y values are the same and therefore we cannot solve for x and y. If there was infinitely many solutions we would have quadratic equations.
1. g. 2nd. order non-linear differential equation
2. a. 1st. order linear differential equation
3. c. 3rd. order linear differential equation
4. g. 2nd. order non-linear differential equation
Step-by-step explanation:
QUESTION 1
[tex](1+y^2)(\frac{d^2y}{dt^2})+t\frac{dy}{dt} +y=e^t[/tex]
Order: The highest derivative present in this differential equation is the second derivative ([tex]\frac{d^2y}{dt^2}[/tex]). Hence the order of this differential equation is 2.
Linearity: There is the presence of the product of the dependent variable , [tex]y[/tex] and its derivative [tex][(1+y^2)(\frac{d^2y}{dt^2})][/tex].
Hence this differential equation is non-linear.
Classification: Second order non-linear ordinary differential equation.
QUESTION 2
The given differential equation is
[tex]t^2\frac{d^2y}{dt^2}+t\frac{dy}{dt} +2y=\sin t[/tex]
Order: The highest derivative present in this differential equation is [tex]\frac{d^2y}{dt^2}[/tex]
Hence it is a second order differential equation.
Linearity: There is no presence of the product of the dependent variable and/or its derivative. There is no presence of higher powers of the dependent variable or its derivative. There is no transcendental function of the dependent variable.
Classification: First order linear differential equation
QUESTION 3
The given differential equation is
[tex]\frac{d^3y}{dt^3}+t\frac{dy}{dx} +\cos (2t)y=t^3[/tex]
Order: The highest derivative present in this differential equation is [tex]\frac{d^3y}{dt^3}[/tex]
Hence it is a third order differential equation.
Linearity: There is no presence of the product of the dependent variable and/or its derivative. There is no presence of higher powers of the dependent variable or its derivative. There is no transcendental function of the dependent variable.
Classification: Third order linear ordinary differential equation
QUESTION 4
The given differential equation is;
[tex]y"-y+y^2=0[/tex]
Order: The highest derivative present in this differential equation is [tex]y"[/tex]
Hence it is a second order differential equation.
Linearity: There is the presence of higher power of the dependent variable, [tex]y^2[/tex]. Hence the differential equation is non-linear.
Classification: Second order non-linear differential equation
A. Yes
B. No
C. No
D. The graph of the equations is attached
x = 2
Step-by-step explanation:
A. Yes, she will get the same solution
Given the solution is based on the two solutions, the solution will be the same for all algebraic methods used
B. No, the answer in A does not change
A system of two equations in two unknowns with no solution is an inconsistent system of equations and there should be no solutions for all correct algebraic methods used
C. No, the answer in A does not change
If there are infinitely many solutions, there will be infinitely many solutions for all correct algebraic methods used
D. The graph of the equations;
y = 2·x + 4...(1) and y = 3·x + 2.....(2) is attached
Subtracting equation (1) from equation (2), we have;
3·x + 2 - (2·x + 4) = y - y = 0
x - 2 = 0
∴ x = 2
The graphs of the two equations also intersect at x = 2.
[tex]1. Lisa is working with the system of equations x+2y=7 and 2x−5y=5. She multiplies the first equatio[/tex]
1, During a 1-hr television program, there were 22 commercials, Some commercials were 15 sec and some were 30 sec long. Find the number of 15- sec commercials and the number of 30 sec commercial if the total playing time for commercial was 9.5 minutes.
a) show how to formulate your system of equations for your problem,
- state the variables: x number of 15 sec commercials, y number of 30 sec commercials
- translate the word statement into algebraic language
b) state the two equations:
- translate the word statement into algebraic equations:
* there were 22 commercials => x + y = 22
* the total playing time for commercial was 9.5 minutes => 15x + 30y = 9.5*60
Equation (1) x + y = 22
Equation (2) 15x + 30y = 570
c) state which method you will use to solve either addition or substitution
adition: multiply the first equation times - 15 and add the two equations.
d) solve your system showing and explaining each step of the process
- muliply eq (1) times - 15
-15x - 15y = - 330
- add that to the eq (2):
-15x + 15x - 15y + 30y = -330 + 570
- add like terms: 15y = 240
- divide both members by 15: 15y/15 = 240 / 15 => y = 16
- replace y = 16 in eq I(1) => x + 16 = 22 => x = 22 - 16 = 6
Solution: x = 6, y = 16
e) Be sure to check your solution!
x + y = 22
6 + 16 = 22
22 = 22 --> check
15x + 30y = 9.5*60
15(6) + 30(16) = 570
90 + 480 = 570
570 = 570 ---> check
f) Once a solution is found explain what this solution means in the context of the problem, write the answer to the word problem in words
The solution means that during a 1-hr television program, there were 6 commercials of 15 sec and 16 commercials of 30 sec.
2, How many quarts of water should be mixed with a 30% vinegar solution to obtain 12 qt of a 25% vinegar solution. (Hint: water is 0% vinegar).
a) variables:
- state the variables
x: number of quarters of water
y: number of quarters of vinegar
- translate the words into mathematical language
12 qt 25% vinegar solution
=> total number of quarts = 12 = x + y
balance in vinegar:
vinegar from water + vinegar from 30% vinegar solution = vinegar in the 25% vinegar solution
0 + 0.3y = 0.25 (x + y)
c) Equations
Eq (1) x + y = 12
Eq (2): 0.3y = 0.25x + 0.25y
=> 0.25x - 0.05y = 0
d) solve
- multiply eq (1) times 0.05 and add to eq (2)
0.05x + 0.05y = 0.6
0.25x - 0.05y = 0
0.30x = 0.6
x = 0.6 / 0.3 = 2
x + y = 12 => y = 12 - y = 12 - 2 = 10
Solution: x = 2, y = 10
e) check:
x + y = 12
2 qt + 10qt = 12 qt
12 qt = 12 qt ---> check
vinegar:
10*0.3 = 12*0.25
3 = 3 -> check
f) Explanation of the solution
The solution means that you have to mix 2 qts of water with 10 qts of 30% vinegar solution to obtain 12 qt of 25% vinegar solution.
Beyonce will get the same result.
Step-by-step explanation:
The given system is
2x-3y=-2
and 4
x+y=24
This system contains two lines that are not parallel to each other therefore the solution x=5, y=4 is unique.
It doesn't matter which method he used or which variable he eliminated first.
The solution will still be the same.
If the system of two equations with has no solution, the Beyonce will not get a solution in the first place, but the fact that the system has no solution remains the same even if Beyonce uses a different method.
If there are infinite solutions, Beyonce might get different solutions using a different method, but the fact that the system has infinite solution remains the same.
All methods are supposed to be correct in most if not all situations and it should not matter if you use a different method given the options
Part B. see the procedure
Part C. see the procedure
Step-by-step explanation:
we have
[tex]f(x)=x^{2}+2x+1[/tex] -----> equation A
[tex]g(x)=3-x-x^{2}[/tex] -----> equation B
Part B. Solve the system algebraically
equate the equation A and the equation B
[tex]x^{2}+2x+1=3-x-x^{2}[/tex]
[tex]2x^{2}+3x-2=0[/tex]
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]2x^{2}+3x-2=0[/tex]
so
[tex]a=2\\b=3\\c=-2[/tex]
substitute in the formula
[tex]x=\frac{-3(+/-)\sqrt{3^{2}-4a(2)(-2)}} {2(2)}[/tex]
[tex]x=\frac{-3(+/-)\sqrt{25}} {4}[/tex]
[tex]x=\frac{-3(+/-)5} {4}[/tex]
[tex]x1=\frac{-3(+)5} {4}=0.5[/tex]
[tex]x2=\frac{-3(-)5} {4}=-2[/tex]
Find the values of y
For x=0.5
[tex]f(0.5)=0.5^{2}+2(0.5)+1=2.25[/tex]
For x=-2
[tex]f(-2)=(-2)^{2}+2(-2)+1=1[/tex]
the solutions are the points
(0.5,2.25) and (-2,1)
Part C. Solve the system by graph
using a graphing tool
we know that
The solution of the non linear system is the intersection point both graphs
The intersection points are (0.5,2.25) and (-2,1)
therefore
The solutions are the points (0.5,2.25) and (-2,1)
see the attached figure
[tex]Problem: a non-linear system consists of two functions: f(x)=x²+2x+1 and g(x)=3-x-x². solve this s[/tex]
1.) y intercept-16
X intercept-24
(0,16)
(24,0)
(9,10)
(3,14)
The first number is the number of teams of 4 while the second is the number of teams of 6
2.)
Y intercept-75
X intercept-30
Ordered 30 wristbands
5(18)+2y=150
2y=60
———
2
Y=30
answer:
45
Step-by-step explanation: