EXPENSES: Camille has no more than $50.00 to spend each week for lunch and bus fare. Lunch costs $3.00 each day, and bus fare is $1.75 each ride. Write an inequality for this situation. Can Camille buy lunch 7 times and ride the bus 14 times in one week

Inequality:

She can afford to buy the lunch and bus rides

Step-by-step explanation:

Given

daily

daily

Solving (a) The inequality

Let bus ride be represented with B and Lunch be represented with L

If 1 bus ride costs $1.75, then

B bus ride would cost $1.75 * B

If 1 lunch costs $3.00, then

L lunch would cost $3.00 * L

Total Spending is then the sum of the above expression i.e.

$3.00 * L + $1.75 * B

The question says Camille can not spend more than $50.00.

This is represented by less than or equal to.

So, the expression is:

Solving (b): Can she afford 7 lunch and 14 bus ride?

To do this, we simply substitute 7 for L and 14 for B

The above inequality is true.

Hence, she can afford to buy the lunch and bus rides

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

33

step-by-step explanation: g=6