Answer:
£191.36
Step-by-step explanation:
Sum the parts of the ratio, 1 + 4 = 5 parts
Divide 320 by 5 to find the value of one part of the ratio.
320 ÷ 5 = 64 ← value of 1 part of the ratio , thus
4 parts = 4 × 64 = 256
Thus he bought 64 first class and 256 second class.
Cost = (64 × £0.67) + (256 × £0.58)
= £42.88 + 148.48
= £191.36
Answer: (b) A
<u>Step-by-step explanation:</u>
A: the center of the circle
B: the diameter of the circle
C: the radius of the circle
D: <em>not sure, could be the perimeter</em>
<em />
Circles are named by their center, so this is named "Circle A".
Answer:
A) y=-6x+1
Step-by-step explanation:
m=(y2-y1)/(x2-x1)
m=(-17-(-5))/(3-1)
m=(-17+5)/2
m=-12/2
m=-6
y-y1=m(x-x1)
y-(-5)=-6(x-1)
y+5=-6(x-1)
y=-6x+6-5
y=-6x+1
Equate whats inside (arguments) 
with base period of sine function 
and solve for x to get period,
So the period of the graph of the given function is precisely 2.
Hope this helps :)
Answer:
Option C is correct.
Step-by-step explanation:
-x+3y=2
4x-2y=22
In matrix form is represented as:
![\left[\begin{array}{cc}-1&3\\4&-2\end{array}\right] \left[\begin{array}{c}x&y\end{array}\right] =\left[\begin{array}{c}2&22\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-1%263%5C%5C4%26-2%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%2622%5Cend%7Barray%7D%5Cright%5D)
AX=B
|A| = (-1)(-2)-(3)(4)
|A| = 2-12
|A| = -10
Adj A = ![\left[\begin{array}{cc}-2&-3\\-4&-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%26-3%5C%5C-4%26-1%5Cend%7Barray%7D%5Cright%5D)
A^-1 = -1/10
A^-1 = 1/10![\left[\begin{array}{cc}2&3\\4&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%263%5C%5C4%261%5Cend%7Barray%7D%5Cright%5D)
X= A^-1 B
X = 1/10![\left[\begin{array}{c}2&22\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%2622%5Cend%7Barray%7D%5Cright%5D)
X=1/10![X=1/10\left[\begin{array}{c}2*2+3*22\\4*2+1*22\end{array}\right]\\X=1/10\left[\begin{array}{c}4+66\\8+22\end{array}\right]\\X=1/10\left[\begin{array}{c}70\\30\end{array}\right]\\X=\left[\begin{array}{c}70/10\\30/10\end{array}\right]\\X=\left[\begin{array}{c}7\\3\end{array}\right]](https://tex.z-dn.net/?f=X%3D1%2F10%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%2A2%2B3%2A22%5C%5C4%2A2%2B1%2A22%5Cend%7Barray%7D%5Cright%5D%5C%5CX%3D1%2F10%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%2B66%5C%5C8%2B22%5Cend%7Barray%7D%5Cright%5D%5C%5CX%3D1%2F10%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D70%5C%5C30%5Cend%7Barray%7D%5Cright%5D%5C%5CX%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D70%2F10%5C%5C30%2F10%5Cend%7Barray%7D%5Cright%5D%5C%5CX%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D7%5C%5C3%5Cend%7Barray%7D%5Cright%5D)
So, x = 7 and y =3
Hence Option C is correct.
