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insens350 [35]
3 years ago
5

PLEASEE HELL find the solution to the system of equations

Mathematics
2 answers:
seropon [69]3 years ago
7 0

Answer:

(-3, -4)

Step-by-step explanation:

y= -4

so you plus y in for the y=2x+2 equation where you get x=-3

Sergio [31]3 years ago
4 0

\huge\text{Hey there!}

\large\text{If y = -4 then SUBSTITUTE it into the EQUATION}

\large\text{-4 =  2x + 2}

\large\text{TURN  the EQUATION}

\large\text{2x + 2 = -4}

\large\text{SUBTRACT by 2 to BOTH SIDES}

\large\text{2x + 2 - 2 = -4 - 2}

\large\text{CANCEL out: 2 - 2 because that gives you 0}

\large\text{Keep: -4 - 2 because it helps us solve for x}

\large\text{2x = -6}

\large\text{DIVIDE by 2 to BOTH SIDES }

\mathsf{\dfrac{2x}{2}=\dfrac{-6}{2}}

\large\text{CANCEL out: }\mathsf{\dfrac{2}{2}} \large\text{because that gives you 1}

\large\text{KEEP: }\mathsf{\dfrac{-6}{2}}  \large\text{because it gives us the value of x }

\boxed{\boxed{\large\text{Answer: \bf x = -3}}}\huge\checkmark\\\\\boxed{(-3,-4)}

\text{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

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Can you solve the system of equation using substitution y=x+8 <br> X+y=2
Vsevolod [243]
X + y = 2
Substitute x + 8 into y
x + x + 8 = 2
2x + 8 =2
(Subtract 8 from both sides)
2x = -6
(Divide both sides by 2)
x = -3

Subsitute x = -3 into either equation to find y
x + y = 2
-3 + y = 2
(Add 3 to both sides)
y = 2 + 3
y = 5

Or y can be solved for using:
y = x + 8
y = -3 + 8
y = 5
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4 years ago
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ololo11 [35]

Answer:

y = \frac{6-3x}{2}

Step-by-step explanation:

Given

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y = \frac{6-3x}{2}

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The <em>correct answer</em> is:

A) A tangent is never a secant.

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Determine the equations of the vertical and horizontal asymptotes, if any, for y=x^3/(x-2)^4
djverab [1.8K]

Answer:

Option a)

Step-by-step explanation:

To get the vertical asymptotes of the function f(x) you must find the limit when x tends k of f(x). If this limit tends to infinity then x = k is a vertical asymptote of the function.

\lim_{x\to\\2}\frac{x^3}{(x-2)^4} \\\\\\lim_{x\to\\2}\frac{2^3}{(2-2)^4}\\\\\lim_{x\to\\2}\frac{2^3}{(0)^4} = \infty

Then. x = 2 it's a vertical asintota.

To obtain the horizontal asymptote of the function take the following limit:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4}

if \lim_{x \to \infty}\frac{x^3}{(x-2)^4} = b then y = b is horizontal asymptote

Then:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4} \\\\\\lim_{x \to \infty}\frac{1}{(\infty)} = 0

Therefore y = 0 is a horizontal asymptote of f(x).

Then the correct answer is the option a) x = 2, y = 0

3 0
3 years ago
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