5+7 what strategy did you use doubles near doubles make 10 or my way think 5 and 7 are two apart

Mathematics

1 answer:

rusak2 [61]3 years ago

8 0

Answer:

double plus two , two apart facts make ten extended strategy is used.

Step-by-step explanation:

Sum of 5 + 7 = 12 less two = 12; OR to make 10, I need to take 3 from the 5, add it to the 7 to make 10 and that leaves 2 from the 5. I add the 2 to the 10 I now have and that = 12.

You might be interested in

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 $(-\infty, +\infty)$