Answer:
<em>The answer is 13</em>
Step-by-step explanation:
Two positive and consecutive old numbers are x and x - 2.
=> x(x - 2) = 143
=> x^2 - 2x - 143 = 0
=> x^2 + 11x - 13x - 143 = 0
=> x(x + 11) - 13(x + 11) = 0
=> (x + 11)(x - 13) = 0
=> x = -11 (invalid)
or x = 13 (valid), the remaining number is 13 - 2 = 11
=> The two numbers are 11 and 13, and the greater number is 13.
Hope this helps!
:)
Answer:
it's a sooooooo it's the first one
Answer: the equations are
k = d + 33
k - 2 = 4(d-2)
Step-by-step explanation:
Let the present age of Kevin be represented by k
Let the present age of Daniel be represented by d
Kevin is 33 years older than Daniel. This means that the expression for their current age is
k = d + 33
Two years ago, Kevin was 4 times as old as Daniel. This means that 2 will be subtracted from their present ages to depict two years ago. Therefore, Two years ago for Daniel will be d-2
Two years ago for Kevin will be k - 2
Remember that Kevin was 4 times as old as Daniel two years ago, it becomes
k - 2 = 4(d-2)
This is a common factor problem.
Pencils come in a pack of 12
Erasers come in a pack of 10
First, break the number into their prime factors(the idea is that we will break the number down into its smallest multiples, which are prime numbers):
10 = 2 * 5
12 = 2 * 2 *3
So now we take the unique multiples of each number, and when we multiply them together, we will get the smallest number that both 10 and 12 can be divided into(this is what the problem is asking for)
We have (2*2*3) that comes from 12, and the only unique number that comes from the 10 is (5)
So now, we multiply:
2*2*3*5=60
However, this isn't exactly out answer. Now we have to divide our answer by the number of each this in the pack to know how many packs to buy.
60/12=5 packs of pencils
60/10= 6 packs of erasers
I hope this helps. Let me know if you have any questions!!
