Helppppppppppppppppppppppppppppppppp

This problem addresses some common algebraic errors. For the equalities stated below assume that x and y stand for real numbers.

weqwewe [10]

Answer:

The answers are:

a) $(x+y)^2 = x^2 + y^2$ is FALSE.

b) $(x+y)^2 = x^2 + 2xy + y^2$ is TRUE.

c) $\frac{x}{x+y}= \frac{1}{y}$ is FALSE.

d) $x-(x+y) = y$ is FALSE.

e) $\sqrt{x^2} = x$ is FALSE.

f) $\sqrt{x^2} = |x|$ is TRUE.

g) $\sqrt{x^2+4} = x+2$ is FALSE.

h) $\frac{1}{x+y} = \frac{1}{x} + \frac{1}{y}$ is FALSE.

Step-by-step explanation:

We can show that the FALSE statement are, in fact, false finding an example where the equality fails.

a) Take x=2 and y=3 and notice that (2+3)² = 25, while 2²+3²=4+9=13.

b) (x+y)² = (x+y)(x+y)=x²+xy+xy+y² = x² +2xy +y².

c) Take x=2 and y=3 and notice that 2/(2+3) = 2/5 which is different from 1/3.

d) Take x=2 and y=3 and notice that 2-(2+3) = 2-5=-3 which is different from 3. Recall that x-(x+y) = x-x-y=-y.

e) Take x=-2, and notice that [tax]\sqrt{(-2)^2} = \sqrt{4} = 2[/tex] which is different from -2.

f) The previous example illustrate why this is true.

g) Take x=1, and notice that $sqrt{1^2+4} = \sqrt{5}$ which is different from 3.

h) Take x=2 and y=3 and notice that 1/(2+3)=1/5 and 1/3+1/2 = 5/6.

3 0

3 years ago

you receive an allowance of a penny the first day of the month and the amount you receive doubles every day for the rest of the

nignag [31]

Answer:

56

Step-by-step explanation:

8 0

3 years ago

For what value of c is the function defined below continuous on (-\infty,\infty)?

kozerog [31]

$f(x)= \left \{ {{x^2-c^2,x \ \textless \ 4} \atop {cx+20},x \geq 4} \right
$

It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
A function, f, is continuous at x = 4 if
 $\lim_{x \rightarrow 4} \ f(x) = f(4)$

In notation we write respectively
 $\lim_{x \rightarrow 4-} f(x) \ \ \ \text{ and } \ \ \ \lim_{x \rightarrow 4+} f(x)$

Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just 4c + 20.

On the other hand, for x < 4, f(x) = x^2 - c^2. Hence
$\lim_{x \rightarrow 4-} f(x) = \lim_{x \rightarrow 4-} (x^2 - c^2) = 16 - c^2$

Thus these two limits, the one from above and below are equal if and only if
4c + 20 = 16 - c²
Or in other words, the limit as x --> 4 of f(x) exists if and only if
4c + 20 = 16 - c²

$c^2+4c+4=0
\\(c+2)^2=0
\\c=-2$

That is to say, if c = -2, f(x) is continuous at x = 4.

Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers $(-\infty, +\infty)$