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

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204

step-by-step explanation:

i found the koficent of the linear function and i foun c acordind to the strict formula y=kx+c and then i took 2 as an x and replacet it on the formula i found and then foun the y