Should have ate zero if I’m not wrong
872 times 3=2616, just try to use a Calculator if you can
well, if that function f(x) were to be continuos on all subfunctions, that means that whatever value 7x + k has when x = 2, meets or matches the value that kx² - 6 has when x = 2 as well, so then 7x + k = kx² - 6 when f(2)
![f(x)= \begin{cases} 7x+k,&x\leqslant 2\\ kx^2-6&x > 2 \end{cases}\qquad \qquad f(2)= \begin{cases} 7(2)+k,&x\leqslant 2\\ k(2)^2-6&x > 2 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ 7(2)+k~~ = ~~k(2)^2-6\implies 14+k~~ = ~~4k-6 \\\\\\ 14~~ = ~~3k-6\implies 20~~ = ~~3k\implies \cfrac{20}{3}=k](https://tex.z-dn.net/?f=f%28x%29%3D%20%5Cbegin%7Bcases%7D%207x%2Bk%2C%26x%5Cleqslant%202%5C%5C%20kx%5E2-6%26x%20%3E%202%20%5Cend%7Bcases%7D%5Cqquad%20%5Cqquad%20f%282%29%3D%20%5Cbegin%7Bcases%7D%207%282%29%2Bk%2C%26x%5Cleqslant%202%5C%5C%20k%282%29%5E2-6%26x%20%3E%202%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%207%282%29%2Bk~~%20%3D%20~~k%282%29%5E2-6%5Cimplies%2014%2Bk~~%20%3D%20~~4k-6%20%5C%5C%5C%5C%5C%5C%2014~~%20%3D%20~~3k-6%5Cimplies%2020~~%20%3D%20~~3k%5Cimplies%20%5Ccfrac%7B20%7D%7B3%7D%3Dk)
Infinitely many ways!
Suppose you have the fraction 2/d.
<span>Pick </span>any<span> pair of integers a and b where b ≠ 0.</span>
Then 2b-ad is and integer, as is bd so that (2b - ad)/bd is a fraction.
Consider the fractions a/b and (2b - ad)/bd
<span>Their sum is </span>
a/b + (2b-ad)/bd = ad/bd + (2b-ad)/bd = 2b/bd = 2/d - as required.
<span>Since a and b were chosen arbitrarily, there are infinitely many possible answers to the question.</span>
The answer is A.
hope this helps!
