An amount of R351 is made up with 114 coins (R2 coins and R5 coins only). How many R2 coins are there? How many R5 coin are there?

41

Step-by-step explanation:

For this, you have to make two equations;

One that calculates the amount of coins

and one that calculates the amount of R

I'm going to set

R2 = x

R5 = y

To make this a bit easier to look at

So to do the number of coins, you want to do this:

x + y = 114

This is saying that the total amount of R2 coins plus the total amount of R5 coins will add up to 114

Then, you want to calculate the worth of them.

2(x) + 5(y) = 351

This is saying that for ever R2 coin there is, it will increase the total by 2, and every R5, it will increase the total by 5. When you add these together, it will come out to 351.

Then you have to solve the system of equations:

x + y = 114

2x + 5y = 351

-2 * (x + y = 114)

2x + 5y = 351

-2x + -2y = -228

2x + 5y = 351

3y = 123

y = 41

Since y is the amount of R5 coins, we know we have 41 R5 coins.

So the answer is

41

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