One positive number is 2 less than another. The sum of their squares is 34. What is the smaller number?

Mathematics

1 answer:

Viktor [21]2 years ago

3 0

Answer:
The smaller number is 3. The number 3 squared is 9, plus the other number 5 squared, which is 25. when you add them, you get 34

Step-by-step explanation:
x = first number
y = second number
a. x+2 = y
b. x^2 + y^2 = 34
Replace y from equation a into equation b:
x^2 + (x+2)^2 = 34
Expand it
x^2 + x^2 + 4x + 4 = 34
2*x^2 + 4x = 30
2x^2 + 4x - 30=0
solving x:
x = -5 and x =3 , it says it is a positive number, then it is x is 3
and the second number, y = x + 2
y = 5
Lets test it!
3^2 + 5^ 2 = 9 + 25 = 34

You might be interested in

Lim t^4 - 6 / 2t^2 - 3t + 7

Harman [31]

I think you meant to say

$\displaystyle \lim_{t\to2}\frac{t^4-6}{2t^2-3t+7}$

(as opposed to x approaching 2)

Since both the numerator and denominator are continuous at t = 2, the limit of the ratio is equal to a ratio of limits. In other words, the limit operator distributes over the quotient:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)}$

Because these expressions are continuous at t = 2, we can compute the limits by evaluating the limands directly at 2:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)} = \frac{2^4-6}{2\cdot2^2-3\cdot2+7} = \boxed{\frac{10}9}$