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Marina CMI [18]
2 years ago
14

Which systems of equations have no solutions? (Check all that apply. ) x – y = 1, –3x 3y = 3 2y = 3 – 4x, x y = 0 12x 2 = 4y – 1

0, 2x y = –11 –3x – y = 5, 15x = 10 – 5y x y = 1, –4x 2y = 7.
Mathematics
2 answers:
rjkz [21]2 years ago
6 0

Answer:

A, D

Step-by-step explanation:

mash [69]2 years ago
3 0

To solve such problems we need to know about the system of equatons.

<h3 /><h3>System of Equation</h3>

When the lines of the system of the equation are parallel to each other the system of equations has no solution.

Two lines are parallel lines if they have the same slope and different y-intercept.

<h3>Given options</h3>

A.)  x – y = 1, –3x+3y = 3

B.)  2y = 3 – 4x, x+y = 0

C.)  12x+2 = 4y – 10, 2x+y = –11

D.)  –3x – y = 5, 15x = 10 – 5y

E.)   x+y = 1, –4x+2y = 7.

<h3 /><h3>Solutions</h3>

A.) x-y=1, -3x+3y = 3

The two lines when plotted on the graph will give two parallel lines. the graph is been plotted below in image(A).

therefore, the system of equations has no solution.

B.) 2y = 3 – 4x, x+y = 0

The two lines when plotted on the graph will give a point of intersection. the graph is been plotted below in image(B)

therefore, the system of equations has a unique solution.

C.) 12x+2 = 4y – 10, 2x+y = –11

The two lines when plotted on the graph will give a point of intersection. the graph is been plotted below in image(C)

therefore, the system of equations has a unique solution.

D.) –3x – y = 5, 15x = 10 – 5y

The two lines when plotted on the graph will give two parallel lines. the graph is been plotted below in image(D).

therefore, the system of equations has no solution.

E.)  x+y = 1, –4x+2y = 7

The two lines when plotted on the graph will give a point of intersection. the graph is been plotted below in image(E)

therefore, the system of equations has a unique solution.

Learn more about system of equations:

brainly.com/question/12895249

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#SPJ4

6 0
1 year ago
What is the inverse of the logarithmic function of f(x)=log2x
neonofarm [45]

Answer:

y=2^{x}

6 0
3 years ago
“encontrar la integral indefinida y verificar el resultado mediante derivación”
Oliga [24]

I=\displaystyle\int\frac x{(1-x^2)^3}\,\mathrm dx

Haz la sustitución:

y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx

\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{y^3}=\frac1{4y^2}+C=\frac1{4(1-x^2)^2}+C

Para confirmar el resultado:

\dfrac{\mathrm dI}{\mathrm dx}=\dfrac14\left(-\dfrac{2(-2x)}{(1-x^2)^3}\right)=\dfrac x{(1-x^2)^3}

I=\displaystyle\int\frac{x^2}{(1+x^3)^2}\,\mathrm dx

Sustituye:

y=1+x^3\implies\mathrm dy=3x^2\,\mathrm dx

\implies I=\displaystyle\frac13\int\frac{\mathrm dy}{y^2}=-\frac1{3y}+C=-\frac1{3(1+x^3)}+C

(Te dejaré confirmar por ti mismo.)

I=\displaystyle\int\frac x{\sqrt{1-x^2}}\,\mathrm dx

Sustituye:

y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx

\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{\sqrt y}=-\frac12(2\sqrt y)+C=-\sqrt{1-x^2}+C

I=\displaystyle\int\left(1+\frac1t\right)^3\frac{\mathrm dt}{t^2}

Sustituye:

u=1+\dfrac1t\implies\mathrm du=-\dfrac{\mathrm dt}{t^2}

\implies I=-\displaystyle\int u^3\,\mathrm du=-\frac{u^4}4+C=-\frac{\left(1+\frac1t\right)^4}4+C

Podemos hacer que esto se vea un poco mejor:

\left(1+\dfrac1t\right)^4=\left(\dfrac{t+1}t\right)^4=\dfrac{(t+1)^4}{t^4}

\implies I=-\dfrac{(t+1)^4}{4t^4}+C

4 0
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VladimirAG [237]

Answer:

Anahdbd8enduemahehnwkh2vehnsnbvcjuygvvnic

Hinde ko po talaga alam

Step-by-step explanation:

Bcbvcujrirjrhjd26yehfnx

Sorry po talaga

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