1/6p - 4/5
Explanation: -2/3p needs to have the same denominator as 5/6p so they can combine. (Note, whatever sign is in front of a number determines if it is negative or positive. Your equation indicates that the bolded parts are negative and the other parts are positive: -2/3p + 1/5 - 1 + 5/6p. So what we are really doing in an equation like this is combining some numbers together)
-2/3p and 5/6p and the least common multiple of six, so we adjust the numbers to each have a denominator of 6. 5/6p is already there, so we need to adjust -2/3p.
We can multiply both the numerator and denominator by 2 to turn -2/3p into -4/6p. -4/6p has the same value as -2/3p, and now it also has the same denominator as 5/6p!
We then have to combine like terms, therefore combining -4/6p with 5/6p. When adding together fractions, the denominator does not change (this is why both numbers must have the same denominators). So our answer will be _/6p. So we combine -4 with 5, giving us one.
Therefore, -2/3p combined with 5/6p is 1/6p.
Now we have to combine our other set of like terms, 1/5 and -1. We can do this the same way that we combined the other numbers.
1/5 and -1 need the same denominator. This is simple because 1 can easily be figured out with any denominator, as long as the numerator and denominator are the same. This would make 1 = 5/5. But we need it negative, so it would be -5/5.
Now that we have common denominators, we can combine!
1/5 - 5/5
Remember what we said before, the solution will have the same denominator, so all we need to do is (in this case) subtract the numerators.
1 - 5 = 4
So that would be -4/5.
With the like terms combined, we just need to put our two combinations (1/6p and -4/5) together!
Our answer: 1/6p - 4/5
I hope that helps!
Answer:
First table
Step-by-step explanation:
The first table shows the relationship y=1/7x
The second table doesn't show a relationship due to (6,15)
The third table doesn't show a relationship due to (3,5)
The fourth table doesn't show a relationship due to (2,10)
So the first table is correct
Hmm, one way we can do this is by assigning numbers to each
A=4 and B=3
A>B because 4>3
so
A. 2(A+B)=2(4+3)=2(7)=14
B. A+B^2=4+3^2=4+9=13
C. A^2+B^2=4^2+3^2=16+9=25
D. A^2-B^2=4^2-3^2=16-9=7
the largest is 25 so C
(x - xo)^2 + (y - yo)^2 = r^2
[x - (-3)]^2 + [y - (2)]^2 = (1.5)^2
(x+3)^2 + (y-2)^2 = 2.25
Answer: option C
bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.
![\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}](https://tex.z-dn.net/?f=%5Cbf%20cos%5Cleft%5B%20sin%5E%7B-1%7D%5Cleft%28%20-%5Ccfrac%7B5%7D%7B13%7D%20%5Cright%29%20%5Cright%5D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bthen%20we%20can%20say%20that%7D~%5Chfill%20%7D%7Bsin%5E%7B-1%7D%5Cleft%28%20-%5Ccfrac%7B5%7D%7B13%7D%20%5Cright%29%5Cimplies%20%5Ctheta%20%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Btherefore%20then%7D~%5Chfill%20%7D%7Bsin%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B-5%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B13%7D%7D%7D%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%7D%7D)
![\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cpm%5Csqrt%7B13%5E2-%28-5%29%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B144%7D%3Da%5Cimplies%20%5Cpm%2012%3Da%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20cos%5Cleft%5B%20sin%5E%7B-1%7D%5Cleft%28%20-%5Ccfrac%7B5%7D%7B13%7D%20%5Cright%29%20%5Cright%5D%5Cimplies%20cos%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B%5Cpm%2012%7D%7D%7B13%7D)
le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.
