El valor de x en la ecuación 1+4x-3=6x+8
2 answers:
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In short, if the denominator is a decimal or a fraction and you have an integer in the numerator, the numerator gets blown up.
so if you divide by 1/2, the numerator blows up to twice as large, if you divide by 1/5, to five times as large, if you divide by 1/3... well, let's see,
why is that? well, you're dividing, and so is saying "how many times 1/3 goes into 2"?
well, 1/3 + 1/3 + 1/3 = 3/3 or 1, so in 1 whole, the fraction goes there 3 times, so in 2 wholes, it'll be 6 times.
so if you divide by 1/2, the numerator blows up to twice as large, if you divide by 1/5, to five times as large, if you divide by 1/3... well, let's see,
why is that? well, you're dividing, and so is saying "how many times 1/3 goes into 2"?
well, 1/3 + 1/3 + 1/3 = 3/3 or 1, so in 1 whole, the fraction goes there 3 times, so in 2 wholes, it'll be 6 times.
End behavior of a polynomial function is based on the <u>degree of the function</u> and the <u>sign of the leading coefficient</u>.
<u>Sign of the Leading Coefficient</u> determines behavior of right side:
- Positive: right side goes to positive infinity
- Negative: right side goes to negative infinity
<u>Degree of the function</u> determines the behavior of the left side:
- Odd degree: left side is opposite direction of right side
- Even degree: left side is same direction as right side
If you have an expression in the denominator, then you must divide the denominator into the numerator. The result will have a degree and a leading coefficient. Use the rules stated above to determine the end behavior.
For example:
y =
We can factor to get: y =
y = x + 3
Leading Coefficient of y = x + 3 is positive so right side goes to positive infinity.
Degree of y = x + 3 is odd so left side is opposite direction of right side, which means left side goes to negative infinity.
The denominator may not divide evenly into the numerator thus leaving a remainder, but that is ok. We can still use the rules stated above.
Answer:
:Part A) The graph in the attached figure (see the explanation)
Part B) The ordered pair is not included in the solution area for the system (see the explanation)
Step-by-step explanation:
Part A) we have
----> inequality A
The solution of the inequality A is the shaded area below the dashed line
The slope of the dashed line is positive
The y-intercept of the dashed line is (0,-8)
The x-intercept of the dashed line is (2,0)
----> inequality B
The solution of the inequality B is the shaded area above the solid line
The slope of the solid line is negative
The y-intercept of the solid line is (0,5)
The x-intercept of the solid line is (2,0)
The solution of the system of inequalities is the shaded area between the dashed line and the solid line
see the attached figure
Part B) is the point (5, -8) Included in the solution area for the system?
we know that
If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities
Substitute the value of x=5, y=-8 in each inequality and then analyze the results
<em>Inequality A</em>
----> is true
so
the ordered pair satisfy inequality A
<em>Inequality B</em>
---> is not true
so
the ordered pair not satisfy the inequality B
therefore
The ordered pair is not included in the solution area for the system
