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3 years ago

X/5 + 7 = -3 Solve for x

Mathematics

2 answers:

LenKa [72]3 years ago

8 0

Answer:

x= -50

Step-by-step explanation:

To solve for x, we need to isolate x

x/5 + 7 = -3

Subtract 7 from both sides to "reverse" the addition

x/5=-10

Multiply both sides by 5 to "reverse" the division being done to x

5*x/5= -10*5

x= -50

charle [14.2K]3 years ago

6 0

Answer:

x = -50

Step-by-step explanation:

First, you minus 7 from both sides toget x/5 = -10, then you multiply 5 on both sides to get x = -50.

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Simplify the complex rational expression

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$2(x+2)/x+1$

Step-by-step explanation:

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LenKa [72]

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Because this may not be the question you want, I'll omit the first step which should be done with latex. The key to all of these except C is to invert the second fraction (turn the second fraction upside down) and multiply.

$\dfrac{6}{5} * \dfrac{-2}{3}$

=(6*-2) / (5*3) = - 12/15 = -4/5

$\dfrac{7}{4} *\dfrac{5}{3}$

= (7*5) / (4*3) = 35 / 12 = 2 11/12

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3 years ago

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3 years ago

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Can you show me how to solve this problem: a. ) 16=4 to the 3rd power x -7

Ghella [55]

We assume you mean
$\ \ \display{16=4^{3x-7}}$
This is solved by writing 16 as a power of 4 and equating exponents.
$\ 4^{2}=4^{3x-7}$
2 = 3x - 7 . . . . . equate exponents
9 = 3x . . . . . add 7
3 = x . . . . . divide by 3

The solution is x = 3.

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3 years ago

Please help quick will give brainliest!

Karo-lina-s [1.5K]

Answer:

The statement that correctly uses limits to determine the end behavior of f(x) is;

$\lim\limits_{x \to \pm \infty} \dfrac{7 \cdot x^2+ x + 1}{x^4 + 1}= \lim\limits_{x \to \pm \infty} \dfrac{7 }{x^2 }$ so the end behavior of the function is that as x → ±∞, f(x) → 0

Step-by-step explanation:

The given function is presented here as follows;

$f(x) = \dfrac{7 \cdot x^2+ x + 1}{x^4 + 1}$

The limit of the function is presented as follows;

$\lim\limits_{x \to \pm \infty} \dfrac{7 \cdot x^2+ x + 1}{x^4 + 1}$

Dividing the terms by x², we have;

$\lim\limits_{x \to \pm \infty} \dfrac{\dfrac{7 \cdot x^2}{x^2} + \dfrac{x}{x^2} + \dfrac{1}{x^2} }{\dfrac{x^4}{x^2} +\dfrac{1}{x^2} }= \lim\limits_{x \to \pm \infty} \dfrac{7 + \dfrac{1}{x} + \dfrac{1}{x^2} }{x^2 +\dfrac{1}{x^2} }$

As 'x' tends to ±∞, we have;

$\lim\limits_{x \to \pm \infty} \dfrac{7 + \dfrac{1}{x} + \dfrac{1}{x^2} }{x^2 +\dfrac{1}{x^2} } = \lim\limits_{x \to \pm \infty} \dfrac{7 + 0 + 0 }{x^2 +0 } = \lim\limits_{x \to \pm \infty} \dfrac{7 }{x^2 }$

However, we have that the end behavior of 7/x² as 'x' tends to ±∞ is 7/x² tends to 0;

Therefore, we have;

$f(x) \rightarrow 0 \ as \lim\limits_{x \to \pm \infty} \dfrac{7 }{x^2 }$

The statement that correctly uses limits to determine the end behavior of f(x) is therefor given as follows;

$\lim\limits_{x \to \pm \infty} \dfrac{7 \cdot x^2+ x + 1}{x^4 + 1}= \lim\limits_{x \to \pm \infty} \dfrac{7 }{x^2 }$ so the end behavior of the function is that as x → ±∞, f(x) → 0.

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3 years ago