To win a contest the number of beans in a jar has to be guessed within 20 of the actual number. If the number of beans in the jar is 645 which equation can be used to find the minimum and maximum number of beans that will win the contest what is the maximum guess that could win.

- Equation of maximum or minimum: |n - 645| = ±20

- The maximum guess that could win is 665

Step-by-step explanation:

Given

Number of Beans = 645

Guess has to be within 20 more or less of 645

Required

- Equation to find minimum and maximum number of beans

- Maximum guess that could win;

Let the number of guess be represented by n

This question requires the use of absolute value

For maximum guess;

n has to be greater than 645;

So, the equation goes thus: |n - 645| = 20

And For minimum guess;

n has to be lesser than 645;

So, the equation goes thus: |n - 645| = -20

To calculate maximum or minimum, the above equations can be merged to give

|n - 645| = ±20

Hence, equation of maximum or minimum: |n - 645| = ±20

Calculating the maximum guess that could win;

Since, the question specifically request for the maximum guess;

We make use of the following formula;

Maximum Guess: |n - 645| = 20

Solve for n

|n - 645| = 20

n - 645 = 20

Add 645 to both sides

n - 645 + 645 = 20 + 645

n = 20 + 645

n = 665

Hence, the maximum guess that could win is 665

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !

the correct answer is c

good luck love