The sum of three numbers is 131. the second of three numbers is seven more than twice the first. the third number is 12 less than the first. what equation can be used to solve the problem? x + x + x = 131 x + (x + 7) + (x – 12) = 131 x + (2x + 7) + (x – 12) = 131 x + (2x + 7) – 12 = 131

For value of x

x + (2x + 7) + (x - 12 ) = 131

For y

y = 2x + 7

For z

z = x -12

Step-by-step explanation:

Let three numbers be

x , y and z

Now according to first condition

x + y +z = 131 (i)

According to other condition

The second number i.e. y is 7 more then twice of first

which will be

y = 2x + 7             (ii)

According to last condition

3rd number i.e. z is 12 less then first which is

z = x - 12                (iii)

As we have got The value of y and z from equation (ii) and (iii) respectively we can find the value of x by substituting these values in equation (i)

Now substituting values of x and y in equation (i) gives us

x + (2x + 7) + (x - 12 ) = 131

From this equation we can find the value of x and which could be substituted to the equation (ii) to find the value of y and for value of z we can substitute it to equation (iii)

in problems involving proportions, we can use cross products to test whether two ratios are equal and form a proportion. to find the cross products of a proportion, we multiply the outer terms, called the extremes, and the middle terms, called the means.

hope it

126.66 m²

step-by-step explanation:

the pertinent area formula is a = πr², where r is the radius.

here, a = π(6.35 m)² = 40.32π, or approximately 126.66 m²