X^2+3x+12
3x^2-5x-2
X^2+4x+9
From the given problem, I gather that there are only two groups of forces. These are:
13 lb, 35 lb, resultant force 30 lb
20 lb, 15 lb, resultant force 25 lb
We use the pythagorean theorem to determine if the three forces are grouped correctly, such that the resultant force is the hypotenuse.
Resultant Force = √(a² + b²)
So, for the first group,
30 ? √(13² + 35²)
30 ≠ 37.33
Thus, the first group does not pull at right angles to each other.
For the second group,
25 ? √(20² + 15²)
25 = 25
Thus, the second group does pull at right angles to each other.
Answer:
See below (I hope this helps!)
Step-by-step explanation:
Because odd numbers are always 1 greater than even numbers, we can call the two odd numbers x + 1 and y + 1 where x and y are even integers. Multiplying the two gives us:
(x + 1) * (y + 1)
= x * y + x * 1 + 1 * y + 1 * 1
= xy + x + y + 1
We know that x * y will be even because x and y are also even and the sum of two even numbers will be even, and we also know that x and y are even and that 1 is odd. Since the sum of even and odd numbers is always odd, the product of any two numbers is always odd.
*NOTE: I put a limitation on x and y in my proof (the limitation was that x and y must be EVEN integers) but you don't have to do that, you could make the odd integers 2x + 1 and 2y + 1 where x and y could be any integer from the set Z like mirai123 did. I simply gave this proof because it was the first thing that came to mind. While mirai123's proof and mine are different, they are still both correct.
Answer: 30 percent of 20 is 6.
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Step-by-step explanation:
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(n/100) * 20 = 6 ?
Solve for "n" .
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Divide each side of the equation by "20" :
{ (n/100) * 20 } / 20 = 6 / 20 ;
to get:
n/100 = 6/20 ;
↔ 6/20 = n/100 ; solve for "n" ;
Look at the denominators:
20 * ? = 100 ?
100/ 20 = ? ;
100/20 = 5 ;
So, 20 * " 5" = 100 ;
Then 6* 5 = "n" (for the numerator, "n"; in : "n /100" ) ;
→ 6 * 5 = 30 ;
→ n = 30 .
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Answer: 30 percent of 20 is 6.
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