good evening ,
Answer:
p = (1÷8)×(1÷7)×(1÷6)
Step-by-step explanation
the probability you will draw 135 is:
(1÷8)×(1÷7)×(1÷6) = 0,002976190476
:)
<span>I note that this problem starts out with "Which is a factor of ... " This implies that you were given several answer choices. If that's the case, it's unfortunate that you haven't shared them.
I thought I'd try finding roots of this function using synthetic division. See below:
f(x) = 6x^4 – 21x^3 – 4x^2 + 24x – 35
Please use " ^ " to denote exponentiation. Thanks.
Possible zeros of this poly are factors of 35: plus or minus 1, plus or minus 5, plus or minus 7. Use synthetic division; determine whether or not there is a non-zero remainder in each case. If none of these work, form rational divisors from 35 and 6 and try them: 5/6, 7/6, 1/6, etc.
Provided that you have copied down the function
</span>f(x) = 6x^4 – 21x^3 – 4x^2 + 24x – 35 properly, this approach will eventually turn up 1 or 2 zeros of this poly. Obviously it'd be much easier if you'd check out the possible answers given you with this problem.
By graphing this function, I found that the graph crosses the x-axis at 7/2. There is another root.
Using synth. div. to check whether or not 7/2 is a root:
___________________________
7/2 / 6 -21 -4 24 -35
21 0 -14 35
----------- ------------------------------
6 0 -4 10 0
Because the remainder is zero, 7/2 (or 3.5) is a root of the polynomial. Thus, (x-3.5), or (x-7/2), is a factor.
Hello!
Since D is the midpoint and the two equations are and both sides of point D the equations equal each other
9x - 7 = 3x + 17
Now you solve it algebraically
Add 7 to both sides
9x = 3x + 24
Subtract 3x from both sides
6x = 24
Divide both sides by 6
x = 4
Now we put this into both equations and add them
9(4) - 7 = 29
3(4) + 17 = 29
29 + 29 = 58
The answer is 58 units
Hope this helps!
Answer:
32,292,000
Step-by-step explanation:
In your question, it asks how many license plate combinations we could make WITHOUT repeats.
We need some prior knowledge to answer this question.
We know that:
- There are 26 letters in the alphabet
- We can make 10 digits (0 - 9)
With the information we know above, we can solve the question.
Since we CAN'T have repeats, we would be excluding a letter or number for each license plate.
We're going to need to multiply each "section" in order to find how many combinations of license plates we can make.
We decrease by one letter and one number in each section since we can't have repeats.
Now, we can solve.
Work:
When you're done multiplying, you should get 32,292,000.
This means that there could be 32,292,000 different combinations of license plates.
<h3>I hope this helps you out.</h3><h3>Good luck on your academics.</h3><h3>Have a fantastic day!</h3>
