Answer:
(6,20) because both lines pass through this point
Step-by-step explanation:
To solve this you can use the substitution method.
Since both of them are equal to y, substitute one of the equations for y, that way you have x+14 = 3x+2. From here you continue to simplify.
Eliminate x from either side of the equation. For example, subtract x from the side that says "x+14". (Make sure you are also subtracting it from the other side of the equal sign as well.
Once you do this you should now have 14 = 2x+2
In order continue, you now have to get x by itself. So now subtract 2 from both sides of the equation. After doing this, you should have 12 = 2x
You then simplify x by dividing both sides by 2. This will get you x = 6
Now that you have the x-value, substitute that into either of the two equations (it is recommended you substitute it into both equations to make sure you have the correct x-value). For example: If I substitute x into Line C's equation, I will now have y = 6 + 14.
6 + 14 is 20, therefore you're y-value is y = 20
Explanation:
Factoring to linear factors generally involves finding the roots of the polynomial.
The two rules that are taught in Algebra courses for finding real roots of polynomials are ...
- Descartes' rule of signs: the number of positive real roots is equal to the number of coefficient sign changes when the polynomial is written in standard form.
- Rational root theorem: possible rational roots will have a numerator magnitude that is a divisor of the constant, and a denominator magnitude that is a divisor of the leading coefficient when the coefficients of the polynomial are rational. (Trial and error will narrow the selection.)
In general, it is a difficult problem to find irrational real factors, and even more difficult to find complex factors. The methods for finding complex factors are not generally taught in beginning Algebra courses, but may be taught in some numerical analysis courses.
Formulas exist for finding the roots of quadratic, cubic, and quartic polynomials. Above 2nd degree, they tend to be difficult to use, and may produce results that are less than easy to use. (The real roots of a cubic may be expressed in terms of cube roots of a complex number, for example.)
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Personally, I find a graphing calculator to be exceptionally useful for finding real roots. A suitable calculator can find irrational roots to calculator precision, and can use that capability to find a pair of complex roots if there is only one such pair.
There are web apps that will find all roots of virtually any polynomial of interest.
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<em>Additional comment</em>
Some algebra courses teach iterative methods for finding real zeros. These can include secant methods, bisection, and Newton's method iteration. There are anomalous cases that make use of these methods somewhat difficult, but they generally can work well if an approximate root value can be found.
Answer:
A) 99.5%
B) 42.1%
Step-by-step explanation:
A) Since the probability of 3 sixes is 0.5%, the probability of less than 3 sixes (that is, zero, one, or two sixes) is 100% - 0.5%, or 99.5%
B ) Since the probability of no sixes is 57.9%, the probability of at least 1 six is 100% - 57.9%, or 42.1%
X = 45 is the answer
180-x = 3(90-x)
=> 180-x = 270 - 3x
=> -x+3x = 270-180
=> 2x = 90
=> x = 90/2
=> x = 45
Please mark brainliest.
Have good day :)
Answer:
8
Step-by-step explanation:
