Eight times the sum of the digits of a certain two-digit number exceeds the number by 19. the tens digit is two less than the units digit. find the number.

number is 13

Step-by-step explanation:

Let

the digit at units place is b

the digit at tens place is a

The number is: ab

Now According to condition:  The tens digit is two less than the units digit

a = b -2   eq1

From the condition: Eight times the sum of the digits of a certain two-digit number exceeds the number by 19.

8(a+b)=10a+b+19

By simplifying

Putting a= b-2

8a+8b=10(b-2)+b+19

8(b-2)+8b=10b-20+b+19

8b-16+8b=10b-20+b+19

By adding like terms we get:

5b=15

Dividing both sides by 5

b = 3

Now putting value of b in eq1

a = b - 2

a = 3  -2

a =1

hence the number is : 13

i hope it will help you!

u=3

t=u-2=1

step-by-step explanation:

if you want to solve percent error follow these steps.

1. subtract the accepted value from the experimental value.

2. take the absolute value of step 1.

3. divide that answer by the accepted value.

4. multiply that answer by 100 and add the % symbol to express the answer as a percentage.

if you want to solve percent of change just do this.

1. work out the difference (increase) between the two numbers you are comparing.

2. divide the increase by the original number and multiply the answer by 100. if your answer is a negative number then this is a percentage decrease.

hope this : )

probably .5

i can on others if necessary.