Answer:
Let x be the number of regular health bars you buy and y the number of strawberry health bars you buy. Then:
0.75x+1.25y=3.75
x+y>=3
Step-by-step explanation:
For the first equation, we have to assume that you will spend all of your money, otherwise it becomes an inequation. The money you spend on regular bars is 0.75x dollars and the money you spend on strawberry bars is 1.25y, so if you spend your 3.75 dollars on the bars, then 0.75x+1.25y=3.75.
For the second, you will always buy x+y health bars, regular and strawberry. There isn't enough information to make this into a equation, the only thing we can deduce is the inequation x+y>=3.
If we also assume that x and y are integers (we can't buy half-bars or one-fourth of a bar) then the minimum number of bars we can buy is 3 (3 strawberry bars) and the maximum is 5 bars (5 regular bars). x+y must be an integer too, so the possibilities for the second equation are x+y=3, x+y=4 and x+y=5. There is a finite number of solutions in any case.
Answer:
The population of bacteria is increasing.
Step-by-step explanation:
The function modelling the population of bacteria in the given culture is an exponential function; We have a base 1.03 and an exponent t. An exponential function is said to be increasing if the base is strictly greater than 1, this implies that the population of the bacteria is increasing as t increases from 0 to infinity.
Answer:
![\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E6y%5E4%7D%7B81x%5E2y%5E8%7D%7D%5Crightarrow%5Cfrac%7B2x%7D%7B3y%7D%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B81x%5E2y%5E%7B10%7D%7D%7B81x%5E6y%5E6%7D%7D%20%5Crightarrow%5Cfrac%7B3y%7D%7B2x%7D%5C%5C%5Csqrt%5B3%5D%7B%5Cfrac%7B64x%5E8y%5E7%7D%7B125x%5E2y%5E%7B10%7D%7D%7D%5Crightarrow%5Cfrac%7B4x%5E2%7D%7B5y%7D%5C%5C%5Csqrt%5B5%5D%7B%5Cfrac%7B243x%5E%7B17%7Dy%5E%7B16%7D%7D%7B32x%5E7y%5E%7B21%7D%7D%7D%5Crightarrow%5Cfrac%7B3x%5E2%7D%7B2y%7D%5C%5C%5Csqrt%5B5%5D%7B%5Cfrac%7B32x%5E%7B12%7Dy%5E%7B15%7D%7D%7B243x%5E7y%5E%7B10%7D%7D%7D%20%5Crightarrow%5Cfrac%7B2xy%7D%7B3%7D%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E%7B10%7Dy%5E%7B9%7D%7D%7B256x%5E2y%5E%7B17%7D%7D%7D%5Crightarrow%5Cfrac%7Bx%7D%7B2y%7D)
Step-by-step explanation:
![\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}} =\sqrt[4]{\frac{(2^4)(x^{6-2})(y^{4-8})}{(3^4)}} =\sqrt[4]{\frac{2^4x^4y^{-4}}{3^4}} =\frac{2xy^{-1}}{3}=\frac{2x}{3y}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E6y%5E4%7D%7B81x%5E2y%5E8%7D%7D%20%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B%282%5E4%29%28x%5E%7B6-2%7D%29%28y%5E%7B4-8%7D%29%7D%7B%283%5E4%29%7D%7D%20%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B2%5E4x%5E4y%5E%7B-4%7D%7D%7B3%5E4%7D%7D%20%3D%5Cfrac%7B2xy%5E%7B-1%7D%7D%7B3%7D%3D%5Cfrac%7B2x%7D%7B3y%7D)
![\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} =\sqrt[4]{\frac{(3^4)(x^{2-6})(y^{10-6})}{(2^4)}} =\sqrt[4]{\frac{3^4x^{-4}y^{4}}{2^4}} =\frac{3x^{-1}y^1}{3}=\frac{3y}{2x}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B81x%5E2y%5E%7B10%7D%7D%7B81x%5E6y%5E6%7D%7D%20%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B%283%5E4%29%28x%5E%7B2-6%7D%29%28y%5E%7B10-6%7D%29%7D%7B%282%5E4%29%7D%7D%20%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B3%5E4x%5E%7B-4%7Dy%5E%7B4%7D%7D%7B2%5E4%7D%7D%20%3D%5Cfrac%7B3x%5E%7B-1%7Dy%5E1%7D%7B3%7D%3D%5Cfrac%7B3y%7D%7B2x%7D)
![\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}} =\sqrt[3]{\frac{(4^3)(x^{8-2})(y^{7-10})}{(5^3)}} =\sqrt[3]{\frac{4^3x^6y^{-3}}{5^3}} =\frac{4x^2y^{-1}}{5}=\frac{4x^2}{5y}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Cfrac%7B64x%5E8y%5E7%7D%7B125x%5E2y%5E%7B10%7D%7D%7D%20%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%284%5E3%29%28x%5E%7B8-2%7D%29%28y%5E%7B7-10%7D%29%7D%7B%285%5E3%29%7D%7D%20%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B4%5E3x%5E6y%5E%7B-3%7D%7D%7B5%5E3%7D%7D%20%3D%5Cfrac%7B4x%5E2y%5E%7B-1%7D%7D%7B5%7D%3D%5Cfrac%7B4x%5E2%7D%7B5y%7D)
![\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}} =\sqrt[5]{\frac{(3^5)(x^{17-7})(y^{16-21})}{(2^5)}} =\sqrt[5]{\frac{3^5x^{10}y^{-5}}{2^5}} =\frac{3x^2y^{-1}}{2}=\frac{3x^2}{2y}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B%5Cfrac%7B243x%5E%7B17%7Dy%5E%7B16%7D%7D%7B32x%5E7y%5E%7B21%7D%7D%7D%20%3D%5Csqrt%5B5%5D%7B%5Cfrac%7B%283%5E5%29%28x%5E%7B17-7%7D%29%28y%5E%7B16-21%7D%29%7D%7B%282%5E5%29%7D%7D%20%3D%5Csqrt%5B5%5D%7B%5Cfrac%7B3%5E5x%5E%7B10%7Dy%5E%7B-5%7D%7D%7B2%5E5%7D%7D%20%3D%5Cfrac%7B3x%5E2y%5E%7B-1%7D%7D%7B2%7D%3D%5Cfrac%7B3x%5E2%7D%7B2y%7D)
![\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} =\sqrt[5]{\frac{(2^5)(x^{12-7})(y^{15-10})}{(3^5)}} =\sqrt[5]{\frac{2^5x^{5}y^{5}}{3^5}} =\frac{2x^1y^{1}}{3}=\frac{2xy}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B%5Cfrac%7B32x%5E%7B12%7Dy%5E%7B15%7D%7D%7B243x%5E7y%5E%7B10%7D%7D%7D%20%3D%5Csqrt%5B5%5D%7B%5Cfrac%7B%282%5E5%29%28x%5E%7B12-7%7D%29%28y%5E%7B15-10%7D%29%7D%7B%283%5E5%29%7D%7D%20%3D%5Csqrt%5B5%5D%7B%5Cfrac%7B2%5E5x%5E%7B5%7Dy%5E%7B5%7D%7D%7B3%5E5%7D%7D%20%3D%5Cfrac%7B2x%5E1y%5E%7B1%7D%7D%7B3%7D%3D%5Cfrac%7B2xy%7D%7B3%7D)
![\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}} =\sqrt[4]{\frac{(2^4)(x^{10-2})(y^{9-17})}{(4^4)}} =\sqrt[4]{\frac{2^4x^{8}y^{-8}}{4^4}} =\frac{2x^{1}y^{-1}}{4}=\frac{x}{2y}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E%7B10%7Dy%5E%7B9%7D%7D%7B256x%5E2y%5E%7B17%7D%7D%7D%20%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B%282%5E4%29%28x%5E%7B10-2%7D%29%28y%5E%7B9-17%7D%29%7D%7B%284%5E4%29%7D%7D%20%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B2%5E4x%5E%7B8%7Dy%5E%7B-8%7D%7D%7B4%5E4%7D%7D%20%3D%5Cfrac%7B2x%5E%7B1%7Dy%5E%7B-1%7D%7D%7B4%7D%3D%5Cfrac%7Bx%7D%7B2y%7D)
Thus,
<span>Representing in fraction for the best understanding
</span>
<span>
When the exponent is negative, the inversion is made, the numerator becomes the denominator, and the denominator becomes the numerator, and the change of signal from the exponent, which was negative, becomes positive.
</span>
Answer:
c.-1/6
Answer:
C. (3, 3)
Step-by-step explanation:
