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postnew [5]
3 years ago
7

Find two consecutive integers whose product is 50

Mathematics
1 answer:
GarryVolchara [31]3 years ago
3 0

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

-----------------------------------------------------

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

----------------------------------------------------------

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.

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Jessica has 48 coins, some of them are nickels and some are dimes. How many of each does she have if she has $3.25 total? 69 POI
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<u>Answer:</u>

Nickels = 44.6

Dimes = 3.4

<u>Step-by-step explanation:</u>

We know that Jessica has 48 coins of which some are nickels and some are dimes so we can write it as:

n = amount of nickels

d = amount of dimes

n+d=48 --- (1)

We are to find the number of nickels and dimes if their total worth is $3.25.

Since a dime is worth 10¢ and a nickel is 5¢, so it should be:

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Read 2 more answers
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