karina is showing her steps to solve the expression -26.7 divided by 1/3 in which step did she make an error
1 answer:
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Answer:
A. System C simplifies to 2x - 3y = 4 and 4x - y = 54 by dividing the second equation by three (3).
B. Systems A and C have the same solutions.
C. Systems A and B have different solutions.
Step-by-step explanation:
Given the following system of equations;
<em><u>System A</u></em>
2x - 3y = 4 .....equation 1
4x - y = 18 .......equation 2
We would solve the above equations using the elimination method;
Multiplying eqn 1 by 2;
2*(2x) - 2*(3y) = 4*2
4x - 6y = 8 ...... equation 3
Subtracting eqn 2 from eqn 3;
(4x - 4x) + (-6y - (-y)) = 8 - 18
0 + (-6y + y) = -10
-5y = -10
5y = 10
y = 2
To find the value of x;
2x - 3y = 4
2x - 3(2) = 4
2x - 6 = 4
2x = 4 + 6
2x = 10
x = 5
<em><u>Solutions (x, y) = (5, 2)</u></em>
<u><em>System B</em></u>
3x - 4y = 5 ..... equation 1
y = 5x + 3 ..... equation 2
We would solve the equations using substitution method;
Substituting eqn 2 into eqn 1, we have;
3x - 4(5x + 3) = 5
3x - 20x - 12 = 5
-17x - 12 = 5
-17x = 12 + 5
-17x = 17
x = -1
To find the value of y;
y = 5x + 3
y = 5(-1) + 3
y = -5 + 3
y = -2
<u><em>Solutions (x, y) = (-1, -2)</em></u>
<u><em>System C</em></u>
2x - 3y = 4 ..... equation 1
12x - 3y = 54 ...... equation 2
We would solve the above equations using the elimination method;
Subtracting eqn 2 from eqn 1;
(2x - 12x) + (-3y -(-3y)) = 4 - 54
-10x + (-3y + 3y) = -50
-10x + 0 = -50
-10x = -50
10x = 50
x = 5
To find the value of y;
2x - 3y = 4
2(5) - 3y = 4
10 - 3y = 4
3y = 10 - 4
3y = 6
y = 2
<u><em>Solutions (x, y) = (5, 2)</em></u>
Answer:
The point C is 12.68 km away from the point A on a bearing of S23.23°W.
Step-by-step explanation:
Given that AB is 50 km and BC is 40 km as shown in the figure.
From the figure, the length of x-component of AC = |AB sin 80° - BC cos 20°|
=|50 sin 80° - 40 cos 20°|=11.65 km
The length of y-component of AC = |AB cos 80° - BC sin 20°|
=|50 cos 80° - 40 sin 20°|= 5 km
tan= 5/11.65
=23.23°
AC= km
Hence, the point C is 12.68 km away from the point A on a bearing of S23.23°W.
