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Veseljchak [2.6K]
3 years ago
14

Malloy solved the equation −5x − 16 = 8; his work is shown below. Identify the error and where it was made. −5x − 16 = 8 Step 1:

−5x − 16 + 16 = 8 + 16 Step 2: −5x = 24 Step 3: negative five times x over five = twenty-four over five Step 4: x = twenty-four over five Step 1: He should have subtracted 16 from each side. Step 1: He should have divided both sides by −5. Step 3: He should have multiplied both sides by −5. Step 3: He should have divided both sides by −5.
Mathematics
2 answers:
s2008m [1.1K]3 years ago
5 0
The correct answer is : Step 3: He should have divided both sides by -5 because in order to isolate x from -5x, you must divide its coefficient, which is -5. Dividing 5 from both sides leaves -1x or -x.
Artemon [7]3 years ago
4 0

Answer: I believe the answer your looking for

is D, Step 3: He should have divided both sides by −5.

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What is the problem of this solving?!
nika2105 [10]
     This question can be solved primarily by L'Hospital Rule and the Product Rule.

y= \lim_{x \to 0}  \frac{x^2cos(x)-sin^2(x)}{x^4}
 
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y= \lim_{x \to 0} \frac{[2xcos(x)-x^2sin(x)]-2sin(x)cos(x)}{4x^3}
 
     II) Product Rule and L'Hospital Rule:

y= \lim_{x \to 0} \frac{[-2xsin(x)+2cos(x)]-[2xsin(x)+x^2cos(x)]-[2cos^2(x)-2sin^2(x)]}{12x^2} \\ y= \lim_{x \to 0} \frac{2cos(x)-4xsin(x)-x^2cos(x)-2cos^2(x)+2sin^2(x)}{12x^2}
 
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]y= \alpha + \beta \\ \\ \alpha =\lim_{x \to 0} \frac{-2sin(x)-[4sin(x)+4xcos(x)]-[2xcos(x)-x^2sin(x)]}{24x} \\ \beta = \lim_{x \to 0} \frac{4sin(x)cos(x)+4sin(x)cos(x)}{24x} \\  \\ y = \lim_{x \to 0} \frac{-6sin(x)-4xcos(x)-2xcos(x)+x^2sin(x)+8sin(x)cos(x)}{24x}
 
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3 years ago
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