We can use a system of equations in order to solve for both of the numbers. Let's start off by assigning variables to each number. The bigger number can be 'x', and the smaller number can be 'y'.
We can make two equations from the given:
x = 18 + y
("One number is 18 more than another number")
x + y = 36
("The sum of the numbers is 36")
If you look at the first equation, the variable 'x' already has a value (18 + y). We can input its value into the second equation in order to solve for y:
x + y = 36
(18 + y) + y = 36
18 + 2y = 36
2y = 18
y = 9
Input the value of 'y' into the first equation:
x = 18 + y
x = 18 + 9
x = 27
<u>One number is 27 and the other number number is 9.</u>
<u></u>
Let me know if you'd like me to explain anything I did here.
- breezyツ
Answer:
Step-by-step explanation:
<u><em>(1).</em></u> (x - <em>h</em>)² + (y - <em>k</em>)² = <em>r</em>²
where <em>(h, k)</em> are coordinates of a center of the circle with radius <em>r</em>
<u><em>(2).</em></u> ( a ± b )² = a² ± 2ab + b²
~~~~~~~~~~~~~~
( 1 , - 2 ) ; r = 3
( x - 1 )² + ( y + 2 )² = 3²
x² - 2x + 1 + y² + 4y + 4 = 9
x² + y² - 2x + 4y - 4 = 0
a = - 2
b = 4
c = - 4
The formula for Slope is m =
where m is the slope and the x's and y's are your given coordinates. So, plug the given information into the formula and solve for y.
m =
Plug in the given values
=
Cross multiply
10 - 6 = -4 - 4y Subtract
4 = -4 - 4y Add 4 to both sides
8 = -4y Divide both sides by -4
-2 = y Swich the sides to make it easier to read
y = -2
Check your answer by plugging -2 back into the equation with the other values.
[<span>tex] \frac{1}{4} [/tex] =
Simplify the double negative
</span><span>
=
Simplify the numerator and denominator
</span><span>
=
</span>
Since both sides equal each other,
y = -2.
A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:
n!/((n-r)!r!)
In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.
5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5
5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10
5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10
5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5
Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.
Now we add together the combinations
1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.
The answer is 32.
-------------------------------
Note: There is also an equation for permutations which is:
n!/(n-r)!
Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.
We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.
I hope this helps! If you have any questions, let me know :)
