Answer:
A
Step-by-step explanation:
Here, we want to choose the option that best suit the question.
Let’s consider the cases individually;
for (x-6) = 0
we have a root at x = 6 which is a real root
Secondly
x^2 + 8x + 16 = x^2 + 4x + 4x + 16 = 0
That would be;
x(x + 4) + 4(x + 4)
(x + 4)(x + 4) = 0
Thus x = - 4 twice
So the roots of the given equation would be;
6 , -4 , -4
Which is 3 real roots but only 2 real distinct roots
By null factor Law factorised eqn=
(x-3+i)(x-3-i)
x^2-6x+10
<h3>
Answer:</h3>
1/17 or 0.0588 (without replacement)
<h3>
Step-by-step explanation:</h3>
To answer this question we need to know the following about a deck of cards
- A deck of cards contains 4 of each card (4 Aces, 4 Kings, 4 Queens, etc.)
- Also there are 4 suits (Clubs, Hearts, Diamonds, and Spades).
- Additionally, there are 13 cards in each suit (Clubs/Spades are black, Hearts/Diamonds are red)
.
In this case, we are required to determine the probability of choosing two diamonds.
- There are 13 diamonds in the deck.
- Assuming, the cards were chosen without replacement;
P(Both cards are diamonds) = P(first card is diamond) × P(second card is diamond)
P(First card is diamond) = 13/52
If there was no replacement, then after picking the first diamond card, there are 12 diamond cards remaining and a total of 51 cards remaining in the deck.
Therefore;
P(Second card is diamond) = 12/51
Thus;
P(Both cards are diamonds) = 13/52 × 12/51
= 156/2652
= 1/17 or 0.0588
Hence, the probability of choosing two diamonds at random (without replacement) is 1/17 or 0.0588.
Answer:
7 and 14
Step-by-step explanation:
