For this case we must find the solution of the following quadratic equation:
Where:
Then, the solution is given by:
Substituting the values:
By definition we have:
Thus, we have two complex roots.
Answer:
The equation has no real roots.
7/7, solve for k, explain too please
1 answer:
You might be interested in
7We can see that line NM = JK so NM also equals 10.
We know NL = 23 and NM = 10 therefore, ML = NL -NM ML = 23 -13
ML = 10
The last thing to solve is the length of KL
KL = ML / sine 34
KL = 10 / 0.55919
KL = 17.9
Total fence length required = JK + KL + NL =
13 + 17.9 + 23
= 53.9 yards
Cost of Fencing = $3.50 * 53.9 =
<span> <span> <span> $188.65 </span> </span> </span>
We know NL = 23 and NM = 10 therefore, ML = NL -NM ML = 23 -13
ML = 10
The last thing to solve is the length of KL
KL = ML / sine 34
KL = 10 / 0.55919
KL = 17.9
Total fence length required = JK + KL + NL =
13 + 17.9 + 23
= 53.9 yards
Cost of Fencing = $3.50 * 53.9 =
<span> <span> <span> $188.65 </span> </span> </span>
Answer:
There are 35 different sets of 3 books Joe could choose
Step-by-step explanation:
* Lets explain how to solve the problem
- Combination is a collection of the objects where the order doesn't
matter
- The formula for the number of possible combinations of r objects from
a set of n objects is nCr = n!/r!(n-r)!
- n! = n(n - 1)(n - 2)................. × 1
Lets solve the problem
- Joe has one book each for algebra, geometry, history, psychology,
Spanish, English and Physics in his locker
∴ He has <em>seven</em> books in the locker
- He wants to chose <em>three</em> of them
∵ The order is not important when he chose the books
∴ We will use the combination <em>nCr</em> to find how many different sets
of three books he can choose
- The total number of books is 7
∴ n = 7
∵ He chooses 3 of them
∴ r = 3
∵ 7C3 = 7!/3!(7 - 3)! = 7!/3!(4!)
∴
∴ 7C3 = 35
* There are 35 different sets of 3 books Joe could choose