Answer: Yes, if there is no number in front of a variable(or letter, as you said), it is actually 1.
a = 1a
x^2 = 1x^2
2m + n = 0 means 2m + 1n = 0
Step-by-step explanation:
For (-3, 2): 2 > 5(-3) + 2; 2 > -15 + 2; 2 > -13 (i.e. not correct)
for (-1, 3): 2 > 5(-1) + 2; 2 > -5 + 2; 2 > -3 (i.e. not correct)
for (0, 2): 2 = 5(0) + 2; 2 = 0 + 2; 2 = 2 (i.e. not correct)
for (1, 2): 2 < 5(1) + 2; 2 < 5 + 2; 2 < 7 (i.e. correct): 2 > 1/2(1) + 1; 2 > 1/2 + 1; 2 > 3/2 (i.e. correct)
for (2, -1): -1 < 5(2) + 2; -1 < 10 + 2; -1 < 12 (i.e. correct): -1 < 1/2(2) + 1; -1 < 1 + 1; -1 < 2 (i.e. not correct)
for (2, 2): 2 < 5(2) + 2; 2 < 10 + 2; 2 < 12 (i.e. correct): 2 = 1/2(2) + 1; 2 = 1 + 1; 2 = 2 (i.e. not correct)
Therefore, both ineaqualities are tru for only the ordered pair (1, 2)
If you would like to know what is the value of x - 10 when x = 35, you can calculate this using the following step:
x = 35
x - 10 = 35 - 10 = 25
The correct result would be A. 25.
Check the picture below.
so the area of the hexagon is really just the area of two isosceles trapezoids.
![\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{parallel~sides}{bases}\\[-0.5em] \hrulefill\\ a=2\\ b=4\\ h=2 \end{cases}\implies \begin{array}{llll} A=\cfrac{2(2+4)}{2}\implies A=6 \\\\\\ \stackrel{\textit{twice that much}}{2A = 12} \end{array}](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20trapezoid%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7Bh%28a%2Bb%29%7D%7B2%7D~~%20%5Cbegin%7Bcases%7D%20h%3Dheight%5C%5C%20a%2Cb%3D%5Cstackrel%7Bparallel~sides%7D%7Bbases%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D2%5C%5C%20b%3D4%5C%5C%20h%3D2%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%20A%3D%5Ccfrac%7B2%282%2B4%29%7D%7B2%7D%5Cimplies%20A%3D6%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Btwice%20that%20much%7D%7D%7B2A%20%3D%2012%7D%20%5Cend%7Barray%7D)
