Answer:
b (-3,3)
Step-by-step explanation:
it's where the two lines intersect
hope this helps
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 :)
ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work
EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same
For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y
For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4
Therefore, for the two expressions to be conjugates, we must satisfy the two conditions.
Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the
x²y = -4 ... (I)
Condition 2: Real parts are the same
x² + y = -3 ... (II)
We have a system of equations since both conditions must be satisfied
x²y = -4 ... (I)
x² + y = -3 ... (II)
We can rearrange equation (II) so that we have
y = -3 - x² ... (II)
Substituting into equation (I)
x²y = -4 ... (I)
x²(-3 - x²) = -4
-3x² - x⁴ = -4
x⁴ + 3x² - 4 = 0
(x² + 4)(x² - 1) = 0
(x² + 4)(x-1)(x+1) = 0
Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.
Solve for y:
y = -3 - x² ... (II)
y = -3 - (±1)²
y = -3 - 1
y = -4
So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:
-3 + ix²y
= -3 + i(±1)²(-4)
= -3 - 4i
x² + y + 4i
= (±1)² - 4 + 4i
= 1 - 4 + 4i
= -3 + 4i
They result in conjugates
The prime number will be 69 and 89
