A machine in a factory produces 25 gismos a day. After a $174 repair job, the machine improves its gizmo output by 16%.If the sale of a gizmo makes a profit of 75cents, the number of days it will take to recover the machine's repair is??? PLEAS HELP ME!!!

Using a system of equation, the number of days it will take to recuvwr the repair cost is 58 days.

The increase in amount of gizmo produced after repair :

Prize per gizmo = 75 cent = $0.75

Repair cost = $174

The number of days it will take to recover the repair fee can be calculated thus :

(Increase on gizmo per day × cost per gizmo)× number of days = repair fee

(4 × 0.75)d= 174

3d = 174

d = 174 / 3

d = 58

Therefore, it will take 58 days to recover the repair fee.

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Around 10 days ( 9.28 days: 9 days 6 hours 43 minutes 12 seconds).

Step-by-step explanation:

Given that 25 gizmos are produced each day, and 75 ¢ are made off of each one, a 174$ repair job will take away money that will need to be paid off.

We can create the following equation by using the amount of gizmos earned per day(x or time) multiplied the money made off of each one(in cents ¢), is equal to the repair job since it will be paid off once it is canceled out.

25x(75¢) = 174$.

Since the money made by each gizmo is in cents and not dollars, it is best to convert this into dollars since the repair job is in this form. Because 100¢ are in each dollar, 75¢ = 75/100 dollars = 3/4 dollars.

3/4 * 25 x = (3/4) * (25x) = (25*3) / 4 = 75/4x

(75/4)x = 174$

(75/4x) * (4/75) = (174$) * (4/75)

x = (174 * 4) / 75

x = 696 / 75

x = 9.28

We are multiplying by the reciprocal (numerator and denominator are switched) since it will make your coefficient 1 which is the variable itself. anything times it's reciprocal is equal to one.

75/4x = 174$

×4 ×4

(cancel out the denominator)

75x = 696$

÷ 75 ÷ 75

(cancel out 75 to get the variable by itself)

x = 696/75

x = 232/25

x = 928/100

x = 9.28

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,