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2 years ago

7

Sheila bought burgers and fries for her children and some friends. the burgers cost $2.05 each and the fries are $.85 each. she

bought a total of 14 items, fir a total cost if $19.10. how many of each did she buy?

1 answer:

jeyben [28]2 years ago

4 0

Let's have B stand for burgers, and F stand for fries.
2.05B+0.85F=$19.10 - This equation is set equal to total price.
B+F=14 - This equation is set equal to total number of items.
Subtract the F in the bottom equation.
B=14-F
Put this in place of B in the first equation.
2.05(14-F)+0.85F=19.10
Make sure you distribute the 2.05 properly, over both the 14 and the F.
28.7 - 2.05F+0.85F=19.10
Collect like terms for the F values.
28.7 - 1.2F=19.10
Subtract 28.7 from both sides.
-1.2F=-9.6
Divide by -1.2 on both sides.
F=8
Put this value in place of F in either equation.
B+8=14
B=6
So F=8 and B=6.

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2 years ago

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I need this done plz

KIM [24]

Answer:

13. $-28 = p$

12. $5 = r$

11. $-12 = t$

10. $-8 = q$

9. $-77 = s$

8. $9 = w$

7. $12 = y$

6. $3 = a$

5. $9 = n$

4. $3 = x$

3. $12 = y$

2. $5 = c$

1. $5 = m$

DON'T GET HOOKED

Step-by-step explanation:

13. $1 + p \div 7 = -3 \\ \\ p\div 7 = -4 \\ \\ -28 = p$

12. $5[r + 4] = 45 \\ \\ r + 4 = 9 \\ \\ 5 = r$

11. $t \div 4 - 3 = -6 \\ \\ t \div 4 = -3 \\ \\ -12 = t$

10. $-4 + 11q = -92 \\ \\ 11q = -88 \\ \\ -8 = q$

9. $s \div 7 + 8 = -3 \\ \\ s \div 7 = -11 \\ \\ -77 = s$

8. $w \div 3 + 3 = 6 \\ \\ w \div 3 = 3 \\ \\ 9 = w$

7. $y \div 3 - 4 = 0 \\ \\ y \div 3 = 4 \\ \\ 12 = y$

6. $3[a + 5] = 24 \\ \\ a + 5 = 8 \\ \\ 3 = a$

5. $14 = 2n - 4 \\ \\ 18 = 2n \\ \\ 9 = n$

4. $3x + 6 = 15 \\ \\ 3x = 9 \\ \\ 3 = x$

3. $6y - 11 = 61 \\ \\ 6y = 72 \\ \\ 12 = y$

2. $9c - 4 = 41 \\ \\ 9c = 45 \\ \\ 5 = c$

1. $2m + 2 = 12 \\ \\ 2m = 10 \\ \\ 5 = m$

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I am delighted to assist you anytime!

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