Answer:
Step-by-step explanation:
![D =\left[\begin{array}{cc}8&-4\\-14&12\end{array}\right] -\left[\begin{array}{cc}-1&-15\\0&7\end{array}\right] \\\\\\\\=\left[\begin{array}{ccc}8+1&-4+15\\-14-0&12-7\end{array}\right]\\\\\\=\left[\begin{array}{ccc}9&11\\-14&5\end{array}\right]](https://tex.z-dn.net/?f=D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D8%26-4%5C%5C-14%2612%5Cend%7Barray%7D%5Cright%5D%20-%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-1%26-15%5C%5C0%267%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5C%5C%5C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D8%2B1%26-4%2B15%5C%5C-14-0%2612-7%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D9%2611%5C%5C-14%265%5Cend%7Barray%7D%5Cright%5D)
Your equation is: 5h + 4 = 10h + 8
First subtract 5h from both sides:(So that variable terms stays in one side only)
=> 5h + 4 -5h = 10h + 8 -5h
=> 4 = 5h + 8
Now subtract 8 from both sides:(So constant term stays in the opposite side of variable terms)
=> 4 - 8 = 5h + 8 - 8
=> -4 = 5h
Now divide both sides by 5:(division by coefficient of variable)
=> -4/5 = 5h/5
=> -4/5 = h
=> h = -4/5 = -0.8
Step-by-step explanation:
A. cos4x-sin4x=2-cos2x
Answer:
A
Step-by-step explanation:
Let A represent amount of Type A coffee pounds used.
Let B represent amount of Type B coffee pounds used
A + B = 156
B = 156 - A
A = 156 - B
5.80A + 4.65B = 826.60
580 (156 - B) + 4.65B - 826.60 = 0
904.8 - 5.80B + 4.65B - 826.60 = 0
904.8 - 1.15B - 826.60 = 0
78.2 - 1.15B = 0
78.2/1.15 = 1.15B/1.15
68 = B
B = 68 pounds of Type B coffee
There's many more steps you can take to check and etc but am too lazy to put down sorry.
