Distribute and simply 7-2(x-5)+3x
1 answer:
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Consider the functions m and n:
i)
for x≠1.
the domain of m is all the real numbers except 1.
ii) n(x)=x-3 is a polynomial function, because x-3 is a linear polynomial.
The domain of any polynomial function is all real numbers. To determine the Range of this function, let c be a value in the range:
x-3=c, then x=3+c.
so if we want the function to produce c, we just set x=3+c, check:
n(3+c)=(3+c)-3=c.
This means that the Range of n is all real numbers.
What this means, is that the input (x) of the function n(x) can be any real number, and the output (n(x)) can also be any number.
so let n(x)=a, consider m(a):
m(a)=(a+5)/(a-1), clearly a≠1, this means n(x)=x-3≠1, that is x≠-4
This means that the domain of the composed function, m(n(x)) = say f(x),
is all Real numbers except 4.
Check: letting x be 4, would mean n(4)=4-3=1
which would mean m(n(4))=m(1)=(1+5)/(1-1)=6/0, which makes no sense.
Answer: Any function with domain R-{4}, has the same domain of (mon)(x)
i)
the domain of m is all the real numbers except 1.
ii) n(x)=x-3 is a polynomial function, because x-3 is a linear polynomial.
The domain of any polynomial function is all real numbers. To determine the Range of this function, let c be a value in the range:
x-3=c, then x=3+c.
so if we want the function to produce c, we just set x=3+c, check:
n(3+c)=(3+c)-3=c.
This means that the Range of n is all real numbers.
What this means, is that the input (x) of the function n(x) can be any real number, and the output (n(x)) can also be any number.
so let n(x)=a, consider m(a):
m(a)=(a+5)/(a-1), clearly a≠1, this means n(x)=x-3≠1, that is x≠-4
This means that the domain of the composed function, m(n(x)) = say f(x),
is all Real numbers except 4.
Check: letting x be 4, would mean n(4)=4-3=1
which would mean m(n(4))=m(1)=(1+5)/(1-1)=6/0, which makes no sense.
Answer: Any function with domain R-{4}, has the same domain of (mon)(x)
Answer:
The difference of 27x³ and 8y³ → [3x - 2y][9x² + 6xy + 4y²]
The difference of 27x³ and 64y³ → [3x - 4y][9x² + 12xy + 16y²]
The sum of 27x³ and 64y³ → [3x + 4y][9x² - 12xy + 16y²]
The sum of 27x³ and 8y³ → [3x + 2y][9x² - 6xy + 4y²]
Step-by-step explanation:
For the first two, we use the Difference of Cubes [(a³ - b³)(a² + 2ab + b²)] first by taking the cube root of the given expression to get our first factor[s]:
Then, use the acronym of SOAP {whether the next operations symbols will be negative or positive [SAME (your cube-rooted factor has an IDENTICAL OPERATION SYMBOL as your given expression), OPPOSITE (the first sign in your second factor in the second set of parentheses will be the opposite of what the sign in your given expression, which will be a plus sign), ALWAYS POSITIVE (the last sign in your second factor in the second set of parentheses will ALWAYS stay positive NO MATTER WHAT)]} to get the second factor:
Given: 27x³ - 64y³
[3x - 4y][9x² + 12xy + 16y²]
↑ ↑ ↑
same as opposite ALWAYS POSITIVE
given of given
Given: 27x³ - 8y³
[3x - 2y][9x² + 6xy + 4y²]
↑ ↑ ↘
same as opposite ALWAYS POSITIVE
given of given
Now, for the last two, we use the Sum of Cubes [(a³ + b³)(a² - 2ab + b²)] first by taking the cube root of the given expression to get our first factor[s]:
Then, use the acronym of SOAP {whether the next operations symbols will be negative or positive [SAME (your cube-rooted factor has an IDENTICAL OPERATION SYMBOL as your given expression), OPPOSITE (the first sign in your second factor in the second set of parentheses will be the opposite of what the sign is in your given expression, which will be a minus sign), ALWAYS POSITIVE (the last sign in your second factor in the second set of parentheses will ALWAYS stay positive NO MATTER WHAT)]} to get the second factor:
Given: 27x³ + 64y³
[3x + 4y][9x² - 12xy + 16y²]
↑ ↑ ↘
same as opposite ALWAYS POSITIVE
given of given
Given: 27x³ + 8y³
[3x + 2y][9x² - 6xy + 4y²]
↑ ↑ ↘
same as opposite ALWAYS POSITIVE
given of given
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* As you can see, when using the <em>Difference\Sum of Cubes</em>, SOAP can vary depending on how an expression is given to you. They resemble each other though.