Let 
be the amount of the 30% alloy and 
the amount of the 60% alloy the metalworker will use. However much is used, the final alloy will have a mass of
kilograms. For each kg of the 30% alloy used, 0.3 kg is copper; similary, each kg of the 60% alloy contributes 0.6 kg, so that
Now,
Answer:
5.9125in^2
Step-by-step explanation:
Step one:
Given data
Dimension of square= 5.25 in
Area of sqaure= 5.25^2= 27.5625 in^2
We are told that the sides of the circle touch the sides of the square
hence the diameter of the circle is 5.25in
radius= 5.25/2= 2.625 in
Area of circle= πr^2
Area of circle = 3.142*2.625^2
Area of circle=3.142*6.890625
Area of circle= 21.65 in^2
Step two:
The area that is not inside the circle
=Area of square-Area of circle
=27.5625- 21.65
=5.9125in^2
9x+ 5y=35
2x + 5y=0 |* -1
9x +5y= 35
-2x -5y= 0
-----------------
7x / = 35
x=35:7
x=5
2x+5y=0
2*5+5y=0
10+5y=0
5y=-10
y= -10:5
y=-2
I hope this helps you
(1-cosx)(1+cosx)=1+cosx-cosx-cos²x=1-cos²x=sin²x
sin²x
____
cos²x
tg²x
