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dsp73
3 years ago
12

Find the solution to the system of equations. y= 2x + 3 y= -3x + 3

Mathematics
1 answer:
Alexxandr [17]3 years ago
3 0

Answer:

(0, 3 )

Step-by-step explanation:

Given the 2 equations

y = 2x + 3 → (1)

y = - 3x + 3 → (2)

Since both equations express y in terms of x we can equate them, that is

2x + 3 = - 3x + 3 ( add 3x to both sides )

5x + 3 = 3 ( subtract 3 from both sides )

5x = 0 ⇒ x = 0

Substitute x = 0 into (1)

y = 0 + 3 = 3

Solution is (0, 3 )

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Find the conjugate of 2 - 5i and then calculate the product of the given complex number and its conjugate. (1 point)
sergejj [24]

Answer:

29

Step-by-step explanation:

conjugate of a+ib=a-ib

conjugate of 2-5i=2+5i

(2+5i)(2-5i)=2²-(5i)²=4-25i²=4-25(-1)=4+25=29

7 0
2 years ago
Read 2 more answers
3x - 20 = 4 PLEASE EXPLAIN HOW YOU GOT THE ANSWER IN STEPS!
sammy [17]

Answer:

x = 8

Step-by-step explanation:

3x - 20 = 4

add 20 to both sides

3x - 20 + 20 = 4 + 20

3x = 24

divide both sides by 3

\frac{3x}{3} = \frac{24}{3}

x = 8

7 0
2 years ago
What is the answer when you evaluate n + m -2(m-m) when m = 3 and n = 2?
Effectus [21]

Answer:

5

Step-by-step explanation:

n + m -2(m-m)

Combine like terms

n+m+2(0)

n+m

Let m=3 and n=2

2+3

5

7 0
2 years ago
Evaluate |-11| + |-7|.<br><br> -18<br> -4<br> 4<br> 18
Marta_Voda [28]

Answer:  D) 18

======================================================

Explanation:

|-11| simplifies to 11

|-7| simplifies to 7

Whatever is inside the absolute value, you just remove the negative and that's the result of that absolute value expression. It represents the distance on a number line. So for instance -11 is 11 units from 0, which is why |-11| = 11.

Overall,

|-11| + |-7| = 11 + 7 = 18

4 0
2 years ago
The general solution of 2 y ln(x)y' = (y^2 + 4)/x is
Sav [38]

Replace y' with \dfrac{\mathrm dy}{\mathrm dx} to see that this ODE is separable:

2y\ln x\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{y^2+4}x\implies\dfrac{2y}{y^2+4}\,\mathrm dy=\dfrac{\mathrm dx}{x\ln x}

Integrate both sides; on the left, set u=y^2+4 so that \mathrm du=2y\,\mathrm dy; on the right, set v=\ln x so that \mathrm dv=\dfrac{\mathrm dx}x. Then

\displaystyle\int\frac{2y}{y^2+4}\,\mathrm dy=\int\dfrac{\mathrm dx}{x\ln x}\iff\int\frac{\mathrm du}u=\int\dfrac{\mathrm dv}v

\implies\ln|u|=\ln|v|+C

\implies\ln(y^2+4)=\ln|\ln x|+C

\implies y^2+4=e^{\ln|\ln x|+C}

\implies y^2=C|\ln x|-4

\implies y=\pm\sqrt{C|\ln x|-4}

4 0
2 years ago
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