How do u add/subtract/multiply/divide fractions with whole numbers?

plz

To Add fractions with whole numbers You have to 
1. separate the whole number from the fraction part of each number 
2.Add the whole numbers and then add the fractions again ,We will need to get common denominators so that we can add the fractions .The second fraction already has 10 as the denominator ,so we are ready to add .
 

OK. let's say you have 4 2/5+ 3 4/7. I am going to teach you how to add fractions with whole numbers.

step 1: take the whole numbers, and multiply them by the denominators, the numbers on the bottom of fractions. like this: on 4 2/5. multiply 4, (the whole number), by the denominator. 5 would be on the bottom. 4x5=20. then add the numerator to it. Now we have 22. take 22 as the numerator. put it over your original denominator, which was 5. you have 22/5. Do the same thing to 3 4/7.

step 2: Now you have 22/5 + 25/7. Find a number both of the denominators both multiply into. 5 and seven both go into 35. change 35 to both of the fraction's denominators.

step 3: Now you have 22/35 and 25/35. We know that to make 5 turn into 35, you multiplied it by seven. you also multiplied 7 by 5 to get 35. Now do the same thing to their numerators, the numbers on the top of the fractions. you changed 22/5 to 22/35. 5 times 7 equals 35. Now whatever you do to the denominator, you have to do to the numerator. so, you need to multiply 22 by 7 as well. Now you have 154/35. Do the same thing to 25/35 except multiply 25 by 5.

step 4: Now you have 154/35 + 125/35. If the denominators are the same, keep them the same on your final answer. Just add the numerators at this point. It would be 279/35.

step 5: Now to simplify! All you have to do is divide 279 by 35. 35 goes into 279 seven times with 34 left over. 7 will be the whole number. 34 will be part of the final fraction.
Take 35, our denominator from just a little while ago, keep it their, as the denominator. Since their is 34 left over, that is the numerator. The final answer is 7 34/35.

Atranslation 6 units to the right and 4 down

< 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)