Answer:
1. 195=6.50·c
2. 30
Step-by-step explanation:
195=6.5·c
Divide both sides by 6.5
30=c
This question is asking for a list of fractions, that when you pull out common factors, they will all simplify back to the 8/5 fraction.
So if 2 is the common factor, you multiply the numerator and denominator by 2.
2: (8*2)/(5*2)= 16/10
3: (8*3)/5*3)= 24/15
4: (8*4)/(5*4)= 32/20
5: (8*5)/(5*5)= 40/25
10: (8*10)/5*10)= 80/50
Eloise's Potential List:
16/10, 24/15, 32/20, 40/25, 80/50
If you simplify any of these fractions above, you will get 8/5 again.
To work backwards:
80/50: 10 goes into 80, 10 goes into 50= 8/5
16/10: 2 goes into 16, 2 goes into 10= 8/5
Hope this helps! :)
Step-by-step explanation:
I am not fully sure what your teacher is aiming for. it friends very much on what you were just discussing in class (which I don't know).
but the first thing coming to mind is a minus sign ("-"). squaring a negative number removed the minus and makes the result equal to squaring the same positive number.
just for the undoing the 1/2 :
that is, because a fraction as exponent specifies in its denominator the root to be calculated for the basic value or expression.
so, 1/2 means square root. and yes, square is the inverse function of a square root, and it "undoes" the square root.
in exponent calculation it just means that for exponent 1 to the power of exponent 2 we simply multiply both exponents. and so, 1/2 × 2 = 1
FYI - the numerator still represents an original "to the power of" operation.
so, e.g. 3/2 would mean put the basis to the power of 3 and then do the square root of that result. or the other way around. these operations are commutative (the sequence does not matter).
1. 7b + 10
2. 13x
3. 4v
4. 2 + 2k
5. 12n
6. 0
7. 8 + 12x
8. 6v + 1
9. 7n
10. 7p
11. 1
12. 18 + 5n
13. 10x + 84
14. 18p - 40
That'd be true only if the value of "s" is the exact same one for both
namely if sec(s) = cos(s)
then solving for "s"
thus
![\bf sec(s)=cos(s)\qquad but\implies sec(\theta)=\cfrac{1}{cos(\theta)} \\\\\\ thus\cfrac{1}{cos(s)}=cos(s)\implies 1=cos^2(s)\implies \pm \sqrt{1}=cos(s) \\\\\\ \pm 1=cos(s)\impliedby \textit{now taking }cos^{-1}\textit{ to both sides} \\\\\\ cos^{-1}(\pm 1)=cos^{-1}[cos(s)]\implies cos^{-1}(\pm 1)=\measuredangle s](https://tex.z-dn.net/?f=%5Cbf%20sec%28s%29%3Dcos%28s%29%5Cqquad%20but%5Cimplies%20sec%28%5Ctheta%29%3D%5Ccfrac%7B1%7D%7Bcos%28%5Ctheta%29%7D%0A%5C%5C%5C%5C%5C%5C%0Athus%5Ccfrac%7B1%7D%7Bcos%28s%29%7D%3Dcos%28s%29%5Cimplies%201%3Dcos%5E2%28s%29%5Cimplies%20%5Cpm%20%5Csqrt%7B1%7D%3Dcos%28s%29%0A%5C%5C%5C%5C%5C%5C%0A%5Cpm%201%3Dcos%28s%29%5Cimpliedby%20%5Ctextit%7Bnow%20taking%20%7Dcos%5E%7B-1%7D%5Ctextit%7B%20to%20both%20sides%7D%0A%5C%5C%5C%5C%5C%5C%0Acos%5E%7B-1%7D%28%5Cpm%201%29%3Dcos%5E%7B-1%7D%5Bcos%28s%29%5D%5Cimplies%20cos%5E%7B-1%7D%28%5Cpm%201%29%3D%5Cmeasuredangle%20s)
