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In America, the Jacks, Queens and Kings are called face cards for the obvious reason that those are the cards with people's faces on them. Aces are not face cards.
Did you mean "n is GREATER than or equal to 2"?
This is not a "proof" and this is not a strong area for me, but....
So we just need to show that n! is divisble by k....
n! = 1 * 2 * 3 * ... * k * .... * n-1 * n Since k is a factor of n!, n! must be divisble by k.
Lightning - Not well named
Better to call him thunder
All bark and no bite.
I believe it would be 72 / (72 + 49) which would be 59.5%.
An angle plus it's complement is equal to 90. Let's call the angle X. It's complement is therefore X also. And the sum of them is 90, so 2X = 90 and X = 45.
A third way (actually just a variation of mathsyperson's first method) is to use substitution. In the first equation, you have y isolated and being equal to 2x-9. Substitute 2x-9 in for y in the second equation.
3x - 4y = 16
3x - 4(2x-9) = 16
3x - 8x + 36 =16
-5x = -20
x = 4
X,Y, and Z are Pythagorean triplets if
. Like Mathsyperson mentioned, one triplet is 3, 4, and 5 since . Once you know that triplet, you can multiply each term of the tripletby the same to come up with a new triplet. For example, multiple each of the 3, 4, 5 triplet by 2 to get 6, 8 and 10. . So 6, 8, and 10 are also a triplet. That answers both questions a and b for you. Now just find one that contains the number 14.There aren't any shortcuts that I know of. Just start with the smallest primes (2, 3, 5, 7, 11) because they're the easiest. Once you get to the point where the prime numbers are greater the square root of the number you're factoring, you can quit. For example, if you're factoring 401 you don't have to try any factors greater than 21.
I don't see how least common multiple applies here but factorization would help.
1156 = 2 * 2 * 17 * 17 = (2*17) * (2* 17) = (2*17)^2 = 34^2
4761 = 3 * 3 * 23 * 23 = (3*23) * (3*23) = (3*23)^2 = 69^2
EDIT: As Jane notes in the next post, I made a mistake. Thanks Jane. It's been edited now and hopefully is correct. And hopefully corrected before you had a chance to read the flawed one which would have confused you further.
The gradient, or slope, is the "rise over the run". Plot the point that you know (-4, -2). Now go up 2 on Y axis (the rise) and to the right 1 (the run). That should take you to (-3, 0). There is only one line which passes though any two points. Draw that line and you have your graph. To put your line in the y = mx + b format, you need to know 2 things: the slope (m) which is 2, and the y-intercept (b). The y intercept is where the line crosses the x axis, i.e. the value of y when x = 0. You can look at the line you drew and see it passes through (0,6) so 0 is your y-intercept. The equation of the line would then be y = 2x + 6. (You could also figure out the y intercept by noting that from your original point of (-4,-2) you need to increase x by 4 to get to 0. For every unit change in x (the rise), y increases by 2 since the slope is 2. So y increase by 8 from -2 to +6).
If you're aksing where does this new line intersect with the line y = -x + 2, here's how to do that. You have 2 equations that are equal to y so set those 2 equal to each other.
y = 2x + 6
y = -x + 2
2x + 6 = -x + 2
3x = -4
x = -4/3
Now plug that value of x into one of your original equations to get y.
y = 2x + 6
y = 2(-4/3) + 6
y = -8/3 + 6
y = 10/3
To double check, plug your x and y values into the other equation.
y = -x + 2
10/3 = -(-4/3) + 2
10/3 = 10/3.
Looks right to me.
What's wrong with using a mnuemonic device to help you remember something?
Sally (sin) = Oscar (opposite) \ Has (Hypotenuse)
Can (cosine) = A (adjacent) \ Hat (Hypotenuse)
Tell (Tangent) = On (Oppposite) \ Always (adjacent)
You're right, Maelwys. I wrote it incorrectly. Should be as above. You have to read down for the first column (Sally Can Tell) and then start reading across (Oscar has a hat on always).
Sally (sin) = Oscar (opposite) \ Hat (Hypotenuse)
Can (cosine) = Has (hypotenuse) \ On (Oppposite)
Tell (Tangent) = A (adjacent) \ Always (adjacent)
"Sally Can Tell Oscar Has A Hat On Always" (except we used a different word instead of "hat". Think about, it's not hard. Or maybe it is).
An even number can be expressed as 2x. The next even number would be 2x + 2. The product would be 4x^2 + 4x = 4x( x+ 1). Either the x or the (x+1) has to be even and therefore divisible by 2. Combine that with the 4, and you have your 8.
That's a classic puzzle. It's usually written something like below:
Three ladies go to a restaurant for a meal. They receive a bill for $30. They each put $10 on the table, which the waiter collects and takes to the till. The cashier informs the waiter that the bill should only have been for $25 and returns $5 to the waiter in $1 coins. On the way back to the table the waiter realizes that he cannot divide the coins equally between the ladies. As they didnt know the total of the revised bill, he decides to put $2 in his own pocket and give each of the ladies $1.
Now that each lady has been given a dollar back, each of the ladies has paid $9. Three times 9 is 27. The waiter has $2 in his pocket. Two plus 27 is $29. The ladies originally handed over $30. Where is the missing dollar?
Jane's answe is right on the money. More importantly, that's the only way it can be done.
Bayogang, you say you can figure it out if you pull from the crate labeled oranges but I don't think it's possible. Let's say you pulled a orange from the orange crate. That means that crate has to be apples and oranges. The crate labeled apples can't be apples because it's mislabeled so therefore it contains oranges. And the crate labeled apples and oranges must have apples.
But you could also haved pulled an apple from the crate labeled oranges. If that happens then that crate could either have just apples or both apples and oranges. Let's first assume it has apples. Then the crate labeled apples and oranges must have oranges. And the crate labeled apples must have oranges.
Now the second case where the orange crate has apples and oranges. Then the apple crate must have oranges. And the crate labeled apples and oranges must have apples.
So if you pull an orange from the crate labeled oranges, you can figure it out (similarly, you could figure it out if you pulled an apple from the crate labeled apple). But if you pull an apple from the orange crate, there are the 2 possibilities shown above. So picking one from the orange crate won't necessarily enable you to figure out what's in the three crates. You have to draw from the crate labeled "apples and oranges".
You roll m dice. What is the probability that any one of them is a six?
By this, do you mean: one and only one of the dice is a six OR at least one of the dice is a six?
If you mean "one and only one", the number of ways to roll one 6 is
. That's out of different possible rolls.If you mean at least one, the answer Maelwys gives is the correct probability.
a)
6H + 7C = 2500
1H + 13C = 2310
b) rewriting the second equation to be H = 2310 - 13C. Now substitute 2310-13c in for H in the 1st equation:
6(2310-13C) + 7C = 2500
13860 - 78C + 7C = 2500
71C = 11360
C = 160
Now substitue 160 in for C in either equation to find H:
6H + 7(160) = 2500
6H + 1120 = 2500
6H = 1380
H = 230
c) Take those value for C and H and substitute that back into one of the equations and see if it's true:
1H + 13C = 2310
230 + 13(160) = 2310
230 +2080 =2310
2310 = 2310
That's it!
The slope is 0.104895. I noticed you rounded your answer to the nearest hundredth. Is that was the instructions said (round to the nearest hundredth instead of to the nearest hundred)? If so, you should round to 0.10.
And I would leave the answer as a positive number. You're going downhill (rise) which is a expressed as a negative number but there is no way to determine whether the horizontal movement (run) is positive or negative. Besides, the question asks for the "slope of the descent" which already implies the direction.