y a positive one will give us a negative number. Similarly, –1/n3 will also be negative. The last choice is n3 – 1\. We are told that 0 < n < 1\. Because n is a positive value less than one, we know that 0 < n3 < n2 < n < 1\. In other words, n3 will be a small positive value, but it will still be less than one. Thus, because n3 < 1, if we subtract 1 from both sides, we see that n3 – 1 < 0\. Therefore, n3 \- 1 is a negative value. All negative numbers are less than positive numbers. Thus, we can eliminate n2 and 1/n2, which are both positive. We are left with –1/n, –1/n3, and n3 – 1. Let us compare –1/n and –1/n3. First, let us assume that –1/n < –1/n3. –1/n < –1/n3 Multiply both sides by n3. We don't need to switch the signs because n3 is positive. –n2 < –1 Multiply both sides by –1\. n2 \> 1. We know that n2 will only be bigger than 1 when n > 1 or if n < –1\. But we know that 0 < n < 1, so –1/n is not less than –1/n3. Therefore, –1/n3 must be smaller. Finally, let's compare –1/n3 and n3 – 1\. Let us assume that –1/n3 < n3 – 1. –1/n3 < n3 – 1 Multiply both sides by n3. –1 < n6 – n3 Add one to both sides. n6 – n3 \+ 1 > 0\. If this inequality is true, then it will be true that –1/n3 is the smallest number. Here it will be helpful to try some values for n. Let's pick n = 1/2 and see what happens. It will help to use our calculator. (1/2)6 – (1/2)3 \+ 1 = 0.891 > 0. Therefore, we suspect that because n6 – n3 \+ 1 > 0, –1/n3 is indeed the smallest number. We can verify this by trying more values of n. The answer is –1/n3.2a:T48c,We calculate the slopes of the lines using the slope formula. The slope of line ![](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/221919/gif.latex "l") is ![](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/221920/gif.latex "m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{-3 - (-6)}{4-8} = \frac{3}{-4} = - \frac{3}{4}") The slope of line !

PSAT Math Diagnostic Test 8

Solve for :

Solve for :

$\frac{24x-8}{3}=12$