Answer: 
<u>Step-by-step explanation:</u>
![\ \ \dfrac{3}{4}-x\bigg(\dfrac{1}{2}-\dfrac{5}{8}\bigg)+\bigg(-\dfrac{3}{8}x\bigg)\\\\\\=\dfrac{3}{4}\bigg(\dfrac{2}{2}\bigg)-x\bigg[\dfrac{1}{2}\bigg(\dfrac{4}{4}\bigg)-\dfrac{5}{8}\bigg]+\bigg(-\dfrac{3}{8}x\bigg)\\\\\\=\dfrac{6}{8}-x\bigg(\dfrac{4}{8}-\dfrac{5}{8}\bigg)-\dfrac{3}{8}x\\\\\\=\dfrac{6}{8}-x\bigg(-\dfrac{1}{8}\bigg)-\dfrac{3}{8}x\\\\\\=\dfrac{6}{8}+\dfrac{1}{8}x-\dfrac{3}{8}x\\\\\\=\dfrac{6}{8}-\dfrac{2}{8}x\\\\\\=\dfrac{3}{4}-\dfrac{1}{4}x\quad \text{(reduced both fractions)}](https://tex.z-dn.net/?f=%5C%20%5C%20%5Cdfrac%7B3%7D%7B4%7D-x%5Cbigg%28%5Cdfrac%7B1%7D%7B2%7D-%5Cdfrac%7B5%7D%7B8%7D%5Cbigg%29%2B%5Cbigg%28-%5Cdfrac%7B3%7D%7B8%7Dx%5Cbigg%29%5C%5C%5C%5C%5C%5C%3D%5Cdfrac%7B3%7D%7B4%7D%5Cbigg%28%5Cdfrac%7B2%7D%7B2%7D%5Cbigg%29-x%5Cbigg%5B%5Cdfrac%7B1%7D%7B2%7D%5Cbigg%28%5Cdfrac%7B4%7D%7B4%7D%5Cbigg%29-%5Cdfrac%7B5%7D%7B8%7D%5Cbigg%5D%2B%5Cbigg%28-%5Cdfrac%7B3%7D%7B8%7Dx%5Cbigg%29%5C%5C%5C%5C%5C%5C%3D%5Cdfrac%7B6%7D%7B8%7D-x%5Cbigg%28%5Cdfrac%7B4%7D%7B8%7D-%5Cdfrac%7B5%7D%7B8%7D%5Cbigg%29-%5Cdfrac%7B3%7D%7B8%7Dx%5C%5C%5C%5C%5C%5C%3D%5Cdfrac%7B6%7D%7B8%7D-x%5Cbigg%28-%5Cdfrac%7B1%7D%7B8%7D%5Cbigg%29-%5Cdfrac%7B3%7D%7B8%7Dx%5C%5C%5C%5C%5C%5C%3D%5Cdfrac%7B6%7D%7B8%7D%2B%5Cdfrac%7B1%7D%7B8%7Dx-%5Cdfrac%7B3%7D%7B8%7Dx%5C%5C%5C%5C%5C%5C%3D%5Cdfrac%7B6%7D%7B8%7D-%5Cdfrac%7B2%7D%7B8%7Dx%5C%5C%5C%5C%5C%5C%3D%5Cdfrac%7B3%7D%7B4%7D-%5Cdfrac%7B1%7D%7B4%7Dx%5Cquad%20%5Ctext%7B%28reduced%20both%20fractions%29%7D)
Answer:
m = 11/7
Step-by-step explanation:
m = y2-y1/x2-x1
m = 7-(-4)/2-(-5)
m = 7+4/2+5
m = 11/7
Answer:
54
Step-by-step explanation:
Multiply:
<em>37x37=72</em>
<em>Subtract:</em>
<em>180-72=108</em>
<em>Divide:</em>
<em>108/2=54</em>
The answer to your question is 54 because you have to bring over 2 to 3 and it would be 1x then bring over 52 and add it to 2 = 54
Answer:
2H2S (g) + 3O2 (g) → 2H2O (l) + 2SO2 (g)
Calculate ΔH° from the given data. Is the reaction exothermic or endothermic?
ΔH°f (H2S) = -20.15 kJ/mol; ΔH°f (O2) = 0 kJ/, mol; ΔH°f (H2O) = -285.8 kJ/mol; ΔH°f (SO2) = -296.4 kJ/mol
