Xue Ying had 143 more sweets than Joanne. After Joanne gave away 22 sweets, Xue Ying had 6 times as many sweets as Joanne. How many sweets did Joanne have in the end?

In the end Joanne had 55 sweets.

Step-by-step explanation:

Let the number of sweets that Xue Ying had be y and let the number of sweets Joanne had be x.

It is given that Xue Ying had 143 more sweets than Joanne. We can write this information as an equation.

               y = x + 143.

It is also given that Joanne gave away 22 of her sweets and after that Xue Ying had 6 times as many more sweets. We can write this as an equation.

             y = 6(x - 22) = 6x - 132.

Therefore we can equate the two equations and solve for x.

                 x + 143  = 6x  - 132

                5x  = 275

        ∴       x =   sweets.

Therefore in the end Joanne had 55 sweets.                        

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.

7/2

(decimal: 3.5)

welcome ; )

step-by-step explanation: