Find two consecutive numbers who squares differ by 31 I got 2 and 31 but I got it wrong someone help me please
consecutive number are one apart
so let the first be x
then the 2nd is x+1
their squares differ by 1 ---> (x+1)^ - x^2 = 31
x^2 + 2x + 1 - x^2 = 31
2x = 30
x = 15
so the first is 15, the next one is 16
check:
16^2 - 15^2
= 256 - 225
= 31
all is good!
We know in January it was 11 degrees Celsius
In December it is -1 degree Celsius
Take that and put -1 on the left.
Now put a < sign next to the one.
After that put the 11 next to the < sign.Now you have your inequality of -1<11.
Write an inequality for the statement:The average temperature in January was 11° C which is warmer than Decembers average temperature of -1° C.
To find two consecutive numbers whose squares differ by 31, let's set up an equation based on the problem:
Let's assume that the first number is "x." The next consecutive number will be "x + 1."
According to the problem, the squares of these two numbers differ by 31. Therefore, we can write the equation as:
(x + 1)^2 - x^2 = 31
Expanding the equation, we get:
(x^2 + 2x + 1) - x^2 = 31
Simplifying further:
2x + 1 = 31
Now, let's solve for x:
2x = 31 - 1
2x = 30
Dividing both sides of the equation by 2:
x = 15
So, the first number is 15. Now, we can find the next consecutive number by adding 1:
x + 1 = 15 + 1 = 16
Therefore, the consecutive numbers whose squares differ by 31 are 15 and 16.