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-73
Alanis Price
Wiki User
∙ 7y agoThe two numbers are:
-7/2 - (√369)/2 ≈ -13.105
-7/2 + (√369)/2 ≈ 6.105
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What are two numbers that have a sum of -18 and a product of 80?
The numbers are: -10 and -8
What two numbers sum is 80 and has a difference of 10?
45 - 35 = 10 45 + 35 = 80
Which pair of numbers whose sum is 80 has the largest product?
41 & 39 = 1599. (40 x 40 = 1600 but I assume you meant different numbers)
What are the product of the prime numbers between 61-80?
141
What is the sum of the prime numbers between 80 and 90?
172
Two numbers have a product of 80 their sum is 24?
Two numbers have a product of 80 their sum is 24?
What are two numbers that have a sum of -18 and a product of 80?
The numbers are: -10 and -8
What is The sum of two numbers is 18 and their product is 80?
8 + 10 = 188 x 10 = 80Therefore, the two numbers are 8 and ten
What two negative integers have a sum of -18 and a product of 80?
-10 + -8 = -18 -10 x -8 = 80 The two numbers are therefore -10 and -8.
Two integers have a difference of -18 and a sum of 2 What is the product of the two integers?
The product of the two integers is -80.
What two numbers sum is 80 and has a difference of 10?
45 - 35 = 10 45 + 35 = 80
The quotient of two numbers is 20. Their sum is 84. What are the 2 numbers?
The numbers are: 80 and 4
What is the product of 80?
A product is the result of multiplying two or more numbers. A single number cannot have a product.
What three numbers equal 80?
There's only one number that's equal to 80. It's 80.There's a good chance that there could be three numbers whose sum is 80, or threedifferent numbers whose joint product is 80, but the question doesn't ask for those.
Two negative integers have a sum of -18 and a product of 80?
The answer is -10 and -8. -8 + -10= -18 -8x-10 =80.
Which pair of numbers whose sum is 80 has the largest product?
41 & 39 = 1599. (40 x 40 = 1600 but I assume you meant different numbers)
What is 74 plus 6?
The sum of 74 plus 6 is a total of 80. Sum is the result you get when adding two numbers together.
