An angle is 10° more than the measure of its complement. Find the number of degrees in each angle. The angles measure degrees.
1 answer:
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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
Ax²+bx+c=0
The formula to solve for x is:
So for 3x²+13x+10=0, plug in the data
x=-13±√13²-4(3)(10) / 2(3)
x=-13±√169-120 / 6
x=-13±√49 /6
x=-13<span>±7 / 6
x=-13+7/6
x=-6/6
x=-1
x=-13-7/ 6
x=-20/6 or 3.3repeating
3x</span>²+14x+15=0
x=-14±√14²-4(3)(15) / 2(3)
x=-14±√196-180 /6
x=-14<span>±</span>√16 / 6
x=-14+4/6
x=-10/6 or 1.6repeating
x=-14-4/6
x=-18/6
x=-3
The formula to solve for x is:
So for 3x²+13x+10=0, plug in the data
x=-13±√13²-4(3)(10) / 2(3)
x=-13±√169-120 / 6
x=-13±√49 /6
x=-13<span>±7 / 6
x=-13+7/6
x=-6/6
x=-1
x=-13-7/ 6
x=-20/6 or 3.3repeating
3x</span>²+14x+15=0
x=-14±√14²-4(3)(15) / 2(3)
x=-14±√196-180 /6
x=-14<span>±</span>√16 / 6
x=-14+4/6
x=-10/6 or 1.6repeating
x=-14-4/6
x=-18/6
x=-3
Given:
Nick has x toy race cars. This is one-third of what Bob has.
So, the number of Nick toys = one-third of what Bob has.
Let the Bob has y toys
So, we can write the following expression:
Solve for y, multiply both sides by (3):
So, the answer will be:
Bob has (3x) toy race cars.