Pure tin was mixed with a4%in tin alloy to produce an ally that was 16%tin .how much pure in and how much 4% alloy were used to produce 32 kg of 16% alloy ?

4 kg of pure tin

Step-by-step explanation:

Let x kg be the amount of pure tin and y kg be the amount of 4% tin alloy mixed.

1. The weight of the mixture is 32 kg, then

2. There are 4% of tin in y kg of 4% tin alloy, so there are kg of tin in this alloy. There are 16% of tin in 32 kg of the mixture, then there are kg of tin in alloy. Thus,

3. Solve the system of two equations:

From the first equation

Substitute it into the second equation:

The amount of 4% alloy were used to produce 32 kg of 16% alloy is 4

Explanation:

let x kg be the amount of pure tin and y kg be the amount of 4% tin alloy mixed.

The weight of the mixture is 32 kg

then  ----- -(1)

there are 4% of tin in y kg of 4% tin alloy, so there are 0.04y kg of tin in this alloy.

There are 16% of tin in 32 kg of the mixture, then there are 0.16x0.32 = 5.12 kg of tin in alloy;

------------------- - (2);

solving (1) and (2)

we get,

substituting in eq-(2)

we get;

multiplying with 100 on both sides we get;

y=28

X=4

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this

3100

step-by-step explanation: