1)why are these 2 sets of numbers "evenly spaced"?
set#1: x+2,x+4,x+6 set#2: 16,17,18,19,20,21
Because the difference between any 2 consecutive numbers are the same: (x+6) - (x+4) = (x+4) - (x+2) = 2 and for the second 21-20 = 20-19 = ... = 17-16 = 1.
2)why isn't the least common multiple of 12 &15=90
isn't this how you do it:
12=2x2x3
15=3x5
So, 3x5x2x3=90 but they say its 60
i did it the exact same way as was shown in the example--> the least common multiple of 6,8,10
6=2x3
8=2x2x2
10=2x5 so, 2x2x2x3x5=120
note: i know to get the least common multiple you look at the multiples of the larger # until you reach a # that's common between them but why isn't the above method working?
First of all you should have 180 with that method, not 90. Notice that you have a shared factor of 3 in both the 12 and the 15. Say I started you off with 180 and had you find the factors of it. You would say 180 = 2x2x3x3x5 (I don't know why you removed the 2 from the equation since it should stay there). I could then ask you to use some of those factors to make 12 and 15. You'd say 'okay, let me do 2x2x3 = 12 and 3x5 = 15.' But notice how you used the 3 in both of those. Did you really need to use both 3s that were in the equation or could you have used the same one both times? The answer is the latter. That's why you should come up with 60 instead of 180. Because both 12 and 15 use a factor of 3, so when you just take all their factors and multiply them together you are doubling it up. If instead you take all unique factors of each and then look at all the shared factors and count them only once, you'll get your LCD. In this case you'd have 2x2x3x5 = 60.
For the second example of 6, 8, 10... you have 2x3=6, 2x2x2=8, and 2x5=10. So the unique factors are 3, 2, 2, and 5 (yes, the two 2s are unique since the 6 and the 10 only have ONE factor of 2 while 8 has THREE, so two of those are unique to the 8). The shared factor is a 2. So you do 2x2x2x3x5 = 120.
3)log(base2)(x-16)=log(base4)(x-4) what's the value of x?
ansr:20
my question is about the first step of the correct ansr explanation which is to put everything in terms of log(base2)
so becomes: log(base2)(x-16)=log(base2)(x-4)/log(base2)(4)
how did log(base4)(x-4) become log(base2)(x-4)/log(base2)(4)
what property of logs did they use, I looked at the properties of logs but couldn't figure out which one they used
It's a property where [log(base a) y] = [log(base x) y] / [log(base x) a]. You can use any new base.
Statement: Let a^x = y and so log(a)y 'log base a of y' = x. I want to show that for some other base b, log(a)y = log(b)y / log(b)a.
Proof: Start with a^x = y. Take the log of both sides, base b. We get log(b)a^x = log(b)y. Simplify the left side using the exponent rule. We get x*log(b)a = log(b)y. Now divide: x = log(b)y / log(b)a. Finally we know that x = log(a)y, so we have log(a)y = log(b)y / log(b)a. QED.
4)set S is the set of all points(x,y) in the coordinate plane such that x and y are both integers with absolute value less than 4. If one of these points is chosen at random, what is the probability that this point will be 2 units or less from the origin? ansr: 0.265
their ansr:
tot desired/tot possible=13/49=.265
total desired= (0,-2)(0,-1)(0,0)(0,1)(0,2)(-2,0)(-1,0)(1,0)(2,0)(-1,-1)(-1,1)
(1,-1)(1,1)
but what about these points: (2,1)(-2,1)(-2,-1)(2,-1) they are also 2 units or less from the origin why aren't they included with the desired values?
Use the distance formula; they aren't. They aren't referring to either x or y, they are referring to the point itself and its distance from (0,0). If I gave you point (3,4) then the distance from the origin = 5 which is obviously bigger than both 3 and 4.
5)((n+2)!-(n+1)!)/n!=? ansr: (n+1)^2
Their solution: (n+2)(n+1)-(n+1)= (n+1)^2 which is only solving the numerator, where did the denominator go??
Because the exclamation point is the symbol for factorial. The term n! = n*(n-1)*(n-2)*...*2*1. So when you write all that out you have (n+2)(n+1)👎(n-1)...(2)(1) - (n+1)👎...(2)(1) all over 👎(n-1)...(2)(1). Notice how all terms have 👎(n-1)...(2)(1) in them. So they all cancel out. There goes your entire denominator. You are left with (n+2)(n+1) - (n+1) in the numerator which = (n+1)(n+2 - 1) = (n+1)^2.
6)(fig2) if the area of the triangle is 15what is the length of AC? Ansr=6.2
This is how I did it:
Area=1/2 base(hight)
15=1/2 (CB)(AC)
30= (CB)(AC)
30/CB=AC
tan38=CB/(30/CB)
tan38=(CB)^2/30
CB=sqrt(30tan38)
CB=23.4
Therefore AC=30/23.4=1.2 which is not 6.2 what's wrong with this method, I know there is a shorter way but why is the above one wrong?
23.4 is 30tan38. You didn't square root it. Otherwise you are correct.
7)(fig1) what is the degree measure, to the nearest integer of angle ABO?
Ansr=40
In the process of attempting to understand their ansr, why is the length of OB=sqrt(2^2+3^2)I can't figure out which sides of which triangle are they using to come up with OB cuz if they're using the that main triangle already drawn then sqrt(2^2+3^2) should be the length of the hypotenuse not the length of OB?
They are using the triangle on the x-y plane (the 'floor' of the drawing). So the distance formula is the same as if you were looking at a 2D graph with point = (2,3) and the origin. That is the distance of OB = sqrt(13). The distance of OA = 3. You want angle ABO so tan(ABO) = OA / OB = 3/sqrt(13). Answer = 39.76 which is approximately 40 degrees.
