Im pretty sure that the answe would be northern plains
Assuming that <span>4^2-6(2^x)-16=0 is correct, we can rearrange it as:
</span><span>-6(2^x) + 4^2 - 16=0
Are you sure it's not </span>6(2^x) + 4^2 - 16=0 ?
If 6(2^x) + 4^2 - 16=0 is correct, then
6(2^x) + 4^2 - 16=16 - 16 = 0, that is, 6(2^x) = 0. Then x = 0 (answer)
Yes because it shows that y is proprtional equal to 99 times x. so whenever you need to work out y, you get x and times it by 99.
Step-by-step explanation:
Remember when expanding radicals,

When expanding radicals into two radicals, we don't let our radicand have two negative answers.


We don't do this



<span><span><span><span><span><span><span><span><span><span><span><span><span>this is the answer(</span></span><span><span><span><span>4</span></span><span><span /></span><span><span><span><span>x</span></span></span><span><span><span>2</span></span></span></span></span></span><span><span>)</span></span></span></span><span><span><span>2
</span></span></span></span></span></span></span></span></span></span></span></span><span>4x22</span>