Answer:
one point
Step-by-step explanation:
A system of two linear equations will have one point in the solution set if the slopes of the lines are different.
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When the equations are written in the same form, the ratio of x-coefficient to y-coefficient is related to the slope. It will be different if there is one solution.
- ratio for first equation: 1/1 = 1
- ratio for second equation: 1/-1 = -1
These lines have <em>different slopes</em>, so there is one solution to the system of equations.
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<em>Additional comment</em>
When the equations are in slope-intercept form with the y-coefficient equal to 1, the x-coefficient is the slope.
y = mx +b . . . . . slope = m
When the equations are in standard form (as in this problem), the ratio of x- to y-coefficient is the opposite of the slope.
ax +by = c . . . . . slope = -a/b
As long as the equations are in the same form, the slopes can be compared by comparing the ratios of coefficients.
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If the slopes are the same, the lines may be either parallel (empty solution set) or coincident (infinite solution set). When the equations are in the same form with reduced coefficients, the lines will be coincident if they are the same equation.
x - y = - 11 .............( 1 )
y + 7 = - 2x ...............( 2 )
from equation ( 1 )
x - y = - 11
x = -11 + y ...........( 3)
putting x in equation ( 2 )
y + 7 = - 2 x
y + 7 = -2 ( -11 + y ) y + 7 = 22 - 2 y y + 2 y =22 - 7
3 y = 15 y = 15 / 3
putting value of y in equation 3
x = -11+ ( 13 /5 ) x = -33 /3 + 15 / 3 ( l.c.m)
x = -17 / 5
check
x = -11 + y
- 17 / 5 = -11 + 13 / 5
-17 /5 = -17 / 5
We see that they are made up of identical squares, meaning that each side of each square is the same length. There are 12 square sides that make up the perimeter of A, so we can do 72/12 which is 6. We see there are ten total square lengths on the perimeter of b, so we can do 6 times 10 which is 60 cm. That’s your perimeter.
We can solve this problem by seeing at which part both of the parts of the graphs of the function are discontinued. Both of the parts of the graph of the function are discontinued at -2, so we will have to find a function that has a value that is undefined for x = -2. We can do this using the denominator of the fraction that's in each of the functions. The function where x = -2 will cause it to be undefined is the third one, so the answer to this question is C.,
f (x) = 
+5.
