Use the cosine function.
cos x = 2/12
cos x = 0.167
use the inverse of cosine:
x = 80.405
The answer is C.
Hope this helps :)
that is not a function, because there are two outputs for three. two inputs can have one output, but one input cant have two outputs.
Answer:
<u>infinitely many solutions</u>
Step-by-step explanation:
The system of equations :
- 3x + 2y = 7
- -4.5x - 3y = -10.5
Multiplying Equation 1 times 3 and Equation 2 times 2 :
- 9x + 6y = 21
- -9x - 6y = -21
Putting the equations in standard form after simplifying :
- 6y = -9x + 21 ⇒ <u>y = -1.5x + 3.5</u>
- -6y = 9x - 21 ⇒ <u>y = -1.5x + 3.5</u>
<u />
As both equations are the same, the system will have <u>infinitely many solutions</u>.
To simplify
![\sqrt[4]{\dfrac{24x^6y}{128x^4y^5}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B24x%5E6y%7D%7B128x%5E4y%5E5%7D%7D)
we need to use the fact that
![\sqrt[4]{x^4}=|x|](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E4%7D%3D%7Cx%7C)
Why the absolute value? It's because
.
We start by rewriting as
![\sqrt[4]{\dfrac{2^23x^6y}{2^6x^4y^5}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B2%5E23x%5E6y%7D%7B2%5E6x%5E4y%5E5%7D%7D)
![\sqrt[4]{\dfrac{2^23x^4x^2y}{2^42^2x^4y^4y}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B2%5E23x%5E4x%5E2y%7D%7B2%5E42%5E2x%5E4y%5E4y%7D%7D)
Since
, we have
, and the above reduces to
![\sqrt[4]{\dfrac{3x^2y}{2^4y^4y}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B3x%5E2y%7D%7B2%5E4y%5E4y%7D%7D)
Then we pull out any 4th powers under the radical, and simplify everything we can:
![\dfrac1{\sqrt[4]{2^4y^4}}\sqrt[4]{\dfrac{3x^2y}{y}}](https://tex.z-dn.net/?f=%5Cdfrac1%7B%5Csqrt%5B4%5D%7B2%5E4y%5E4%7D%7D%5Csqrt%5B4%5D%7B%5Cdfrac%7B3x%5E2y%7D%7By%7D%7D)
![\dfrac1{|2y|}\sqrt[4]{3x^2}](https://tex.z-dn.net/?f=%5Cdfrac1%7B%7C2y%7C%7D%5Csqrt%5B4%5D%7B3x%5E2%7D)
where
allows us to write
, and this also means that
. So we end up with
![\dfrac{\sqrt[4]{3x^2}}{2y}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B4%5D%7B3x%5E2%7D%7D%7B2y%7D)
making the last option the correct answer.
Answer:
= 3.9 candy per pound
Step-by-step explanation:
This is a fraction equal to
7.8 candy ÷ 2 pounds
We want a unit rate where
1 is in the denominator,
so we divide top and bottom by 2
7.8 candy ÷ 2
________
2 pounds ÷ 2
=
3.9 candy
_______
1 pound
=
3.9 candy
_______
pound
= 3.9 candy per pound