HELP ASAP

The product of two consecutive natural numbers is greater than their sum by 155. Find these numbers

numbers are 13 and 14

Step-by-step explanation:

Let the numbers be y and y+1

The products of the numbers is written as:

y(y + 1)

The sum of the numbers is written as:

y + (y + 1) = 2y + 1

From the question,

The products of the two number is great than the sum by 155. This can be written as:

y(y + 1) = 2y + 1 + 155

y^2 + y = 2y + 156

y^2 + y — 2y —156 = 0

y^2 — y —156 = 0

Multiply the first term (i.e y^2) and the last term (—156) together. This gives —156y^2.

Now find two factors of —156y^2 such that when you add the two factors together, it will result to the 2nd term (ie —y) in the equation. These numbers are 12y and — 13y

Now substitute these numbers (i.e 12y and — 13y) in place of —y in the equation above

y^2 — y —156 = 0

y^2 + 12y — 13y —156 = 0

y(y + 12) — 13(y + 12) = 0

(y —13)(y + 12) = 0

y —13 =0 or y +12 =0

y = 13 or y = —12

Since the numbers are positive numbers, y =13

The first number = y = 13

The 2nd number = y + 1 = 13 +1 = 14

Therefore, the consecutive numbers are 13 and 14

hello from mrbilldoesmath!

answer:   x^2 + y^2 = 1024

discussion:

the equation of a circle with center (h,k) and radius "r" is

(x-h)^2 + (y-k)^2 = r^2.

in our case

(x-0)^2 + (y-0) ^2 = 32^2 = 1024

you,

mrb

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.