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Oct 1, 2020

#1
+428
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Ok, I'll help!

a) 35s+20p=150

b) Because she buys 3 sweaters, her total for sweaters is 35*3=105. [My] equation now is 105+20p=150, where 20p is the total money Holly spends on the sweaters. Subtract 150 from both sides, so 20p=45, and p=2.25. But she can't buy 2.25 pants, so this is the total whole number of pants she can buy without going over her money limit. Therefore it is 2, because the quotient when 45 is divided by 20 is 2.

c) Again, [my] equation is when p=0 35s=150. Holly can buy 1 sweater safely without going over the limit, 2 sweaters, 3, and 4. But when[ I ]reach 5, she can't, as she will spend 15*5=\$175 on her sweaters, which is OVER her limit. So the greatest number of sweaters she can buy when she buys no pants is 4.

Oct 1, 2020
#2
0

1: Your equation would be ; 35s + 20p= 150.

2: If each sweater is \$35, and she wants to buy 3, you will multiply 35 by 3 to get 105, then you'd subtract 105 from 150 to get 45. A pair of pants are \$20, and she only has \$45 left. Holly would be able to buy 2 pairs of pants.

3: Her overall amount of money is \$150. To see the number of sweaters Holly can buy, you could do one of two things:

- Divide 150 by 35

Or

- Multiply 35 by one, then two, then three, and so on.

The easier of the two is the division. You'd get 4 with multiple numbers after the decimal point, but the most sweaters that Holly could buy with \$150 is 4.

Oct 1, 2020