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
![x+y=100](https://tex.z-dn.net/?f=x%2By%3D100)
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
![0.3x+0.6y=0.54(x+y)=54](https://tex.z-dn.net/?f=0.3x%2B0.6y%3D0.54%28x%2By%29%3D54)
Now,
![x+y=100\implies y=100-x](https://tex.z-dn.net/?f=x%2By%3D100%5Cimplies%20y%3D100-x)
![\implies0.3x+0.6(100-x)=54](https://tex.z-dn.net/?f=%5Cimplies0.3x%2B0.6%28100-x%29%3D54)
![\implies60-0.3x=54](https://tex.z-dn.net/?f=%5Cimplies60-0.3x%3D54)
![\implies0.3x=6](https://tex.z-dn.net/?f=%5Cimplies0.3x%3D6)
![\implies x=20](https://tex.z-dn.net/?f=%5Cimplies%20x%3D20)
![\implies y=100-20=80](https://tex.z-dn.net/?f=%5Cimplies%20y%3D100-20%3D80)
The correct answer is 70 because 420 divided by 6 is 70.
Step-by-step explanation:
ABC is an isoceles triangle (both legs are equally long). and AB is its baseline.
OC is now a median for ABC splitting the angle at C and AB in half.
so, we have 2 right-angled triangles : OAC and OBC.
the half-angles at C are 42/2 = 21°.
the angles at A and B are 90°.
and the half-angles at O are 180 - 90 - 21 = 69°.
remember that the sum of all angles in a triangle must always be 180°.
AB are the heights of both of these triangles.
the single height is sin(69)×7 = 6.535062985... cm
and so,
AB = 2× height = 13.07012597... cm.