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scZoUnD [109]
3 years ago
14

What is the solution to the system of equations below? 5 x + 2y = -3 3x + 6y = 6

Mathematics
1 answer:
Orlov [11]3 years ago
3 0
Eqn (1)*3: 15x + 6y = -9 -> Eqn (3)
Eqn (3) - Eqn (2): 12x = -15
x = -5/4 = -1 1/4
Eqn (2): -15/4 + 6y = 6
6y = 9 3/4
y = 1 5/8

x = -1 1/4
y= 1 5/8
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Answer:

1,800$

Step-by-step explanation:

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Andrej [43]

The slope is 2.

Equation if needed: y= 2x + 1

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at a noodle shop 45 people equally share 6 trays of noodles. what fraction of the noodles does each person get?
aleksley [76]

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Step-by-step explanation:

5 0
3 years ago
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2. Which of the following equations are perpendicular to 2y = -3x + 1 I. II. III. y=-x-1 - 2x + 3y = -5 2x + 3y = 2 (A) I only (
Novay_Z [31]

The equation in given as ;

2y = -3x + 1

This can be written as ;

y= -3/2 x + 1/2

This means the equation has a gradient of -3/2

Let this slope , be , ---------m1

For perperdicular lines , the product of their slopes = -1 .This means if the other line has a gradient of m2 then : m1 * m2 = -1

So from the answers :

i) y= 2/3 x - 1 the slope is 2/3

m2 = 2/3

m1 * m2 = -1 -------check the if this is true by using the two values of gradient as;

-3/2 * 2/3 = - 1 ------ This is true-----equation i

II.

-2x + 3y = -5

3y = 2x -5

y= 2/3 x -5/3 -----m2 here is 2/3

m1*m2 = -1

-3/2 * 2/3 = -1 -----this is true , so ----equation ii

iii)

2x + 3y = 2

3y = -2x + 2

y= -2/3 x + 2/3 -----m2 = -2/3

m1*m2 = -1

-3/2 * -2/3 = 1 -----this is not true,,,equation iii is not perpendicular to our equation.

so, equation i and ii are perpendicular to our equation .

Answer : B i and ii only

4 0
1 year ago
No link or bot answer the question
kupik [55]

Answer:

1

Step-by-step explanation:

5 0
2 years ago
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