n, n+1 - two consecutive integers
n(n + 1) = 50 <em>use distributive property</em>
n² + n = 50 <em>subtract 50 from both sides</em>
n² + n - 50 = 0
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ax² + bx + c =0
if b² - 4ac > 0 then we have two solutions:
[-b - √(b² - 4ac)]/2a and [-b - √(b² + 4ac)]/2a
if b² - 4ac = 0 then we have one solution -b/2a
if b² - 4ac < 0 then no real solution
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n² + n - 50 = 0
a = 1, b = 1, c = -50
b² - 4ac = 1² - 4(1)(-50) = 1 + 200 = 201 > 0 → two solutions
√(b² - 4ac) = √(201) - it's the irrational number
Answer: There are no two consecutive integers whose product is 50.
9514 1404 393
Answer:
y = 3x
Step-by-step explanation:
A proportional relation has the equation ...
y = kx
To write the desired equation, you need to know the value of k. That can be found from the given information:
for x = 2, y = 6
6 = k·2
3 = k . . . . divide by 2
Now, we know the equation can be written ...
y = 3x
21,935,483.87 rounded to the nearest million is 22,000,000.00
Sqrt[x] = -x
x= x ^ 2
x - x^2 = 0
-x(x-1)=0
x(x-1)=0
x = 1 or x = 0
sqrt 1 isn not equal to 0
x= 0
