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x - y = 1 so x = y+1 15(y+1) - 4y = -3 15y + 15 - 4y = -3 11y = 12 y = 12/11 x = 12/11 + 1 = 23/11 Ordered pair, then, is (23/11, 12/11)
An equation and a graph are connected because with any equation if y is 11 and x is 12 then you go on the graph on the x axis over 12 and up 11. So if you plug in the x and then whatever y is that is the cororderents on your graph.For example the equation is y=5x+1 then you plug in 5 to x so now the equation is now y=5(5)+1 and y will equal 26 so your cororderents are (5,26)
x = -6 y = -11
You'll need another equation to solve this one. The equation you give has a graph that is a line. Every point on the line is a solution to the equation so there are infinitely many solutions.
(0,7)
There are an infinite number of ordered pairs that satisfy the equation.
yes ORDERED PAIR
x - y = 1 so x = y+1 15(y+1) - 4y = -3 15y + 15 - 4y = -3 11y = 12 y = 12/11 x = 12/11 + 1 = 23/11 Ordered pair, then, is (23/11, 12/11)
An equation and a graph are connected because with any equation if y is 11 and x is 12 then you go on the graph on the x axis over 12 and up 11. So if you plug in the x and then whatever y is that is the cororderents on your graph.For example the equation is y=5x+1 then you plug in 5 to x so now the equation is now y=5(5)+1 and y will equal 26 so your cororderents are (5,26)
x = -6 y = -11
You'll need another equation to solve this one. The equation you give has a graph that is a line. Every point on the line is a solution to the equation so there are infinitely many solutions.
-11
Yes, for example if you have y=x but you shifted the equation up 3 units hence: y=x+3. than you will receive a different y from every instance (point) of x. Reference: collegemathhelper.com/2015/11/horizontal-graph-transformations-for.html
(0,7)
-b +/- sqrt(b^2 - 4ac)/2a a = 1 b = -7 c = -18 7 +/- sqrt( 49 + 72 )/2 7 +/- 11/2 roots are; 9 and -2 ( my TI-84 confirms this answer )
y = x - 11 is one possible equation.
There is no unique pair of numbers that satisfies these requirements. Suppose a and b is such a pair, and sqrt(105) = x then you want a < x < b But a < (a+x)/2 < x < (b+x)/2 < b So that (a+x)/2 and (b+x)/2 are a closer pair. and you can then find a closer pair still - ad infinitum. The question can be answered (sort of) if it asked about "integers" rather than "numbers". 100 < 105 < 121 Taking square roots, this equation implies that 10 < sqrt(105) < 11 so the answer could be 10 and 11. But (and this is the reason for the "sort of") the above equation also implies that -11 < sqrt(105) < -10 giving -11 and -10 as a pair of consecutive integers. So, an unambiguous answer is possible only if the question specifies positive integers.