2. 8x -28 = -140
8x -28 + 28 = -140 + 28
8x = -112
8x/8 = -112/8
x = -14
3. -9 + x/3 = -23
-9 + 9 + x/3 = -23 + 9
x/3 = -14
x/3/3 = -14/3
x = -14/3
4. x/-1.5 - 3.5 = -13.5
x/-1.5 - 3.5 + 3.5 = -13.5 + 3.5
x/1.5 = -10
x/-1.5/-1.5= -10/-1.5
x = 20/3
5.-6(x + 3) = -36
-6x - 18 = -36
-6x - 18 + 18 = -36 + 18
-6x = -18
-6x/-6 = -18/-6
x = 3
6. k + 3.7/9.8 = -0.22
k + 3.7/9.8/9.8 = -0.22/9.8
k + 3.7 = -2.156
k + 3.7 - 3.7 = -2.156 - 3.7
k = -5.856
7. 12(x - 6) = -108
12x - 72 = -108
12x - 72 + 72 = -108 + 72
12x = -36
12x/12 = -36/12
x = -3
8. -21.83x - -19.9 = -23.83
-21.83x + 19.9 = -23.83
-21.83x + 19.9 - 19.9 = -23.83 - 19.9
-21.83x = -43.73
-21.83x/-21.83 = -43.73/-21.83
x = 2
9. -10x - 68 + x = 40
-9x - 68 = 40
-9x -68 + 68 = 40 + 68
-9x = 108
-9x/-9 = 108/-9
x = -12
10. -34 - 3x - 2x = 71
-34 - 5x = 71
-34 + 34 - 5x = 71 + 34
-5x = 105
-5x/-5 = 105/-5
x = -21
11. 3x - 77 - 8x = 23
-5x - 77 = 23
-5x - 77 + 77 = 23 + 77
-5x = 100
-5x/-5 = 100/-5
x = -20
12. 3x - 5(2x - 12) = 123
3x - 10x + 60 = 123
-7x + 60 = 123
-7x + 60 - 60 = 123 - 60
-7x = 63
-7x/-7 = 63/-7
x = -9
13. -3x + 6(x + 6) = 15
-3x + 6x + 36 = 15
3x + 36 = 15
3x + 36 - 36 = 15 - 36
3x = -21
3x /3 = -21/3
x = -7
14. 5x + 2(4x - 9) = -174
5x + 8x - 18 = -174
13x - 18 = -174
13x - 18 + 18 = -174 + 18
13x = -156
13x/13 = -156/13
x = -12
15. -3x + 6(5x + 3) = -171
-3x + 30x + 18 = -171
27x + 18 = -171
27x + 18 - 18 = -171 - 18
27x = -189
27x/27 = -189/27
x = -7
Answer:
y = 50x
Step-by-step explanation:
the equation for this is y = 50x because for every 50 units we go up by on the graph we go 1 unit to the right so:
slope-intercept form: y = mx + b where m is the slope and b is the y-intercept. Our y-intercept is 0 because that is where our line intersects the y-axis and because it is 0 we do not need to put it in the equation.
y = 50x
Answer:
2.
the distance from E to D,D toF and F to E are equal.
Answer: b and d
Step-by-step explanation:
Since the roots are x=2 and x=6, we can write the equation as
Substituting in the coordinates of the vertex,
So, the equation is 
On expanding, we get 
The technique of matrix isolation involves condensing the substance to be studied with a large excess of inert gas (usually argon or nitrogen) at low temperature to form a rigid solid (the matrix). The early development of matrix isolation spectroscopy was directed primarily to the study of unstable molecules and free radicals. The ability to stabilise reactive species by trapping them in a rigid cage, thus inhibiting intermolecular interaction, is an important feature of matrix isolation. The low temperatures (typically 4-20K) also prevent the occurrence of any process with an activation energy of more than a few kJ mol-1. Apart from the stabilisation of reactive species, matrix isolation affords a number of advantages over more conventional spectroscopic techniques. The isolation of monomelic solute molecules in an inert environment reduces intermolecular interactions, resulting in a sharpening of the solute absorption compared with other condensed phases. The effect is, of course, particularly dramatic for substances that engage in hydrogen bonding. Although the technique was developed to inhibit intermolecular interactions, it has also proved of great value in studying these interactions in molecular complexes formed in matrices at higher concentrations than those required for true isolation.
