If the sales price of a toy car is $64, including a 20% discount, what was the regular price of the toy card before the discount ?

Hi there! So the toy car was on sale for 20%, and the sale price is $64. To find the original price of the toy car, we can write and solve a proportion. Set it up like this:

64/x = 80/100

This is because 64 is part of the whole, which is the original price, and although you get a 20% discount, you still have to pay 80% of the original price, which is why 80 is over the 100, in stead of 20. Let's cross multiply the values. 64 * 100 is 6,400 and 80 * x is 80x. That leaves 6,400 = 80x. Now, divide each side by 80 to isolate the x. 80x/80 cancels out. 6,400/80 is 80. We found the number, but let's check it. 80 * 80% (0.8) is 64. There. x = 80. The regular price of the toy car was $80.

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

ac ≅ ab (given)< cax ≅ < bax   (given)ax ≅ ax (common side)

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

ac ≅ ab (given)< cay ≅ < bay (since < cax ≅ < bax is given and xy is the extension of line ax)ay ≅ ay (common side)

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)