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3 years ago

Solve each system using elimination.

Mathematics

2 answers:

Tema [17]3 years ago

5 0

Https://www.mathpapa.com/algebra-calculator.html
Sorry i cant answer im in a hurry! BUT ITS A GREAT ALGEBRA CALCULATOR!

Artyom0805 [142]3 years ago

5 0

A) 3x + 3y = 27 ----- (1)
x - 3y = -11-------(2)
Multiplying (2) by 3 we have 3x - 9y = -33 ----(3)
Subtracting (3) from (1) we have
-9y - 3y = -33 - 27
So y = 5. Substituting value of y into (1) we have 3x + 3(5) = 27. So x = 27 -
15/ 3 = 4.
B) Also 2x + 4y= 22 ----(1)
2x - 2y= -8 -----(2)
Subtracting (2) from (1) we have
-2y - 4y = -8 - 22 = -30
-6y = - 30. Hence y = 5. Substituting into (1) we have 2x + 4(5) = 22. X = 22 - 20/2 = 1
C) 5x - y = 0 -------(1)
3x + y = 24 ------(2)
Adding (2) to (1) we have
3x + 5x = 24. X = 24/8 = 3. So y = 5(3) - y = 0. y = 15

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The sector of a circle with a 12 inches radius has a central angle measure of 60°.

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3 years ago

Vertical and Adjacent Angles (3 Activities) 100 POINTS+BRAINLIEST

Yakvenalex [24]

Answer:

Problem B: x = 12; m<EFG = 48

Problem C: m<G = 60; m<J = 120

Step-by-step explanation:

Problem B.

Angles EFG and IFH are vertical angles, so they are congruent.

m<EFG = m<IFH

4x = 48

x = 12

m<EFG = m<IFH = 48

Problem C.

One angle is marked a right angle, so its measure is 90 deg.

The next angle counterclockwise is marked 30 deg.

Add these two measures together, and you get 120 deg.

<J is vertical with the angle whose measure is 120 deg, so m<J = 120 deg.

Angles G and J from a linear pair, so they are supplementary, and the sum of their measures is 180 deg.

m<G = 180 - 120 = 60

7 0

2 years ago

Read 2 more answers

Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +

Veronika [31]

Answer:

Verified

$y(x) = \frac{3Ln(x) + 3}{x}$

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}$

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

$x^2y' +xy = 3$

- A general solution to the above ODE is also given as:

$y = \frac{3Ln(x) + C }{x}$

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

$y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x) }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}$

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

$-\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3$

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3$

- Therefore, the complete solution to the given ODE can be expressed as:

$y ( x ) = \frac{3Ln(x) + 3 }{x}$

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)$