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.
Step-by-step explanation:
(1/12)^-2b x 12^-2b+2=12
12^-(-2b) x 12^-2b+2=12
12^2b+(-2b+12)=12
2b-2b+12=12
=1
Answer:
the answer is 2.
Step-by-step explanation:
i dont know how to explain sorry
Answer:
$300
Step-by-step explanation:
has: $11,200 plus $400
owes: $4500+ $800+ $6000
$11,600- $11, 300= $300
Answers and explanations:
87. The domain of added functions includes the restrictions of both. So the range of the added function in this question is [-4, 3]
88. When finding the domain of a divided function we do the same as adding, but with an extra rule: g can't equal zero. So for this question the domain is (-4, 3)
89. To graph f + g you add the y-values for each x-value. I added a picture to help explain this one!
