The textbook is correct.
I cannot tell you what you did wrong because you have not shown me what you actually graphed.
Can you share the address of the desmos graph with us please.
And, if it is not self-evident, explain how you got your answers.
I think thats wrong as well hmm I'm stumpted
could you possibly explain how you get that and what your steps are?
The distance from E to line AC is 9.
b + c = 44.
1. The volume is 108*sqrt(3)
2. The length of the diagonal is 4*sqrt(7).
The vertex is (5,-2).
You tried 4 and were told it isn't right. Tried it where? Who says it's not right? I'm not accepting that, but, just for argument, let's assume your source is correct. We know the above four work, so the answer can't be fewer, it has to be be more. Try 5, and then if necessary 6, 7, 8, etc. until you come to the number that is considered right. Give us that clue. I cannot think of any different ways to make the required value, limited to using an integer base and an integer exponent.
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When I did the long division, I got: 2x2 + 8x + 15 ...
1 - The word B, A, N, A, N, A =6! / 3!.2! =60 permutations. Since each permutations begins with one of the 6 letters, or 60 / 6 = 10 permutations for each letter. Therefore, you will have 60 - 10 =50 permutations without the letter "B" at the beginning of the permutations.
First: given f(x) = ax + b, find f-1(x):
1) let y = f(x) ---> y = ax + b
2) interchange x and y ---> x = ay + b
3) solve for y: ---> x - b = ay ---> y = (x - b) / a
4) exchange f-1(x) for y ---> f-1(x) = (x - b)/a
Since g(x) = f-1(x) - 3 ---> g(x) = (x - b)/a - 3
Since g(x) = 5x - 4 ---> 5x - 4 = (x - b)/a - 3
Rewriting: 5x - 4 = x/a - b/a - 3
Setting the x-terms equal to each other: 5x = x/a
---> 5ax = x
---> 5a = 1
---> a = 1/5
Setting the numbers equal to each other: -4 = -b/a - 3
---> -1 = -b/a
---> -1a = -b
---> a = b
---> b = 1/5
It is very true what you say. However, you should know that I have most these permutations and combinations programmed into my computer and it spits them out in microseconds and milliseconds. I, therefore, don't think about them at all as long as I enter the conditions that are required correctly.
and i got #2, its 20 ways, since if there were no repeating digits, it would be 5!=120 ways, but there are three 2's, so 3! ways, so we just do 5!/3! to get 20 ways!!!:)
We could go ahead and count these directly, but instead we could count in general and then correct for overcounting. That is, if we had 5 distinct digits, there would be 5!=120 orderings. However, we must divide by 3! once for the repetition of the digit 2, and divide by 2! for the repetition of the digit 9 (note that if the repeated digits were different then we could rearrange them in that many ways). So, our answer is 5!/3! x 2!=10.
I understand now! This is the easier way, guest:)
thanks, guest!
You have two arithmetic series: 2 + 4 + 6 + ... + 4004 and 1 + 3 + 5 + ... + 4003.
Both of them can be handled in the same way.
t1 is the first term
tn is the nth term (or the last term)
n is the number of terms
d is the common difference between successive terms
S is the sum of the terms.
For the series in the numerator: 2 + 4 + 6 + ... + 4004
t1 = 2
tn = 4004
n is unknown
d = 2
S is unknown
To find n: use this formula: tn = t1 + (n - 1)·d
4004 = 2 + (n - 1)·2
4002 = 2n - 2
4004 = 2n
n = 2002 [there are 2002 terms in this series]
To find S: use this formula: S = n·(t1 + tn) / 2
S = 2002·(2 + 4004) / 2
S = 2002·(4006) / 2
S = 4,010,006
Now, you need to do the same steps for the denominator and find the answer when the numerator is divided by the denominator.
1 - There are 10 arrangements that can be made as follows:
{2, 2, 2, 9, 9}, {2, 2, 9, 2, 9}, {2, 2, 9, 9, 2}, {2, 9, 2, 2, 9}, {2, 9, 2, 9, 2}, {2, 9, 9, 2, 2}, {9, 2, 2, 2, 9}, {9, 2, 2, 9, 2}, {9, 2, 9, 2, 2}, {9, 9, 2, 2, 2} = 10 arrangements
oh, thanks geno, I didn't understand what guest meant!
It is. 153 pi is. No explanation for why it is wrong was given, yet great explanations for how it is correct are given.
Yes, that is the correct answer.
My explaination is
we know that Triangle ABC is a right triangle because we know the Pythagorean Theorem's converse is true.
The hypotenuse of a right triangle is a diameter of the triangle's circumcircle.
Therefore, the area of the circumcircle shoud be 13^2*pi = 169pi
To find the incircle's area, we see $[ABC] = (10)(24)/2 = 120$
and $s = (10+24+26)/2 = 30$, so $r = [ABC]/s = 120/30 = 4$.
When $[ABC] = rs$, where $r$ is the inradius and $s$ is the semiperimeter of the triangle.
Therefore, the area of the incircle is $4^2\pi = 16\pi$.Finally, the area of the circumcircle is $169\pi - 16\pi = \boxed{153\pi}$
So whoever discredited the guest who got the correct answer with a great explanation, is wrong, and
Melody is absolutly correct although that guest may not be the same guest who asked the problem. :p
So yes, the answer should be 153pi.
2x = 8(y + 1) ---> 2x = (23)(y + 1) ---> 2x = 2(3y + 3) ---> x = 3y + 3
9y = 3(x - 9) ---> (32)y = 3(x - 9) ---> 32y = 3(x - 9) ---> 2y = x - 9 ---> x = 2y + 9
Combining these two equations: 3y + 3 = 2y + 9 ---> y = 6
If y = 6 then x = 3y + 3 ---> x = 21
Please refrain from giving us a list of homework to complete.
you can however select one or two that you really need help with and I'll be glad to help you.
No... I might not have been clear.
what I meant to say was that it could be that the three rolls have a total sum of 6, and your answers include sums of more than 6.
First coin: R/B Second coin: B/G Third coin: G/R
We can list all the possible outcomes for first coin - second coin - third coin:
R - B - G *
R - B - R
R - G - G
B - B - G
B - B - R
B - G - G
B - G - R *
There are 2 successes in 8 possibilities: probability = 2/8 = 1/4
a·b4 = 384 and a2·b5 = 4608 ---> a2·b5 / a·b4 = 4608 / 384 ---> a·b = 12
I tried various whole number values for a and b; none worked.
Square roots won't work either.
I tried cube roots, giving one value the cube root of 2 and the other the cube root of 4.
Finally: a = 3·21/3 and b = 2·41/3