1 x 0 = 0?

Mathematics

2 answers:

Montano1993 [528]3 years ago

6 0

Any number multiplied by 0 equals 0. So 1 times 0 equals 0. Also if you multiply 1 times 5( for example) will equal 5.

Alexxandr [17]3 years ago

3 0

Both

1 times a=a
and
b times 0=0

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Please help im so lost Ty btw

klasskru [66]

Answer:

1) gradient (00) (-2 4) = y2-y1 / 2-1 = 4/-2 = -2 m = -2/1 means = m = -2 (negative slope) 2) gradient y2-y1 / x2-x1 = 3-0 / 2-0 = 3/2 = (1 1/2)/1 m = 1 1/2 (positive slope) we use the formula y-values divided by the change in the x-values. The equation of the gradient each goes like this 1) y = -2x as y is at origin nothing else to add The equation of the gradient each goes like this 2) y = 1 1/2x The equation of the point formula 1) we take the y -y1 = m (x +x 1) = y-0 = -2x (x +0) (as m = -2) y = -2(x +0) and The equation of the point formula 2) y - 0 = m ( x +x1) y - 0 = 1 1/2( x +0) = y = 1 1/2( x +0)

5 0

2 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 $(-\infty, +\infty)$