If the distance is 45 miles and speed of both cyclist is 14 and 16 miles per hour then they will take time of 1.5 hour to meet.
Given First cyclist is riding at 14 miles per hour and second at 16 miles per hour. The distance is 45 miles.
We know that speed is the distance covered by an object in a particular period of time.
Speed=distance/time.
It is expressed as kilometers per hour or miles per hour, etc.
If both riders are riding towards each other then the speed will be 16+14 =30 miles per hour.
Distance=45 miles.
Time =distance/speed
=45/30
=1.5
Hence if first cyclist is riding at 14 miles per hour and second is riding at 14 miles per hour and the distance is 45 miles then they will meet after 1.5 hours.
Learn more about speed at brainly.com/question/4931057
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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,
![\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)
Yes, it should be taken into account
Answer:
m<U = 38degrees
Step-by-step explanation:
From the given diagram, <B = <U since both triangles are similar, hence;
Hence;
2y+2 = 3y-16
2y - 3y = -16 - 2
-y = -18
y = 18
Get m<U
m<U = 3y-16
m<U = 3(18) -16
m<U = 54 - 16
m<U = 38degrees
Answer:






Step-by-step explanation:
Given

See attachment for proper table
Required
Complete the table
Experimental probability is calculated as:

We use the above formula when the frequency is known.
For result of roll 2, 4 and 6
The frequencies are 13, 29 and 6, respectively
So, we have:



When the frequency is to be calculated, we use:


For result of roll 3 and 5
The probabilities are 0.144 and 0.296, respectively
So, we have:


For roll of 1 where the frequency and the probability are not known, we use:

So:
Frequency(1) added to others must equal 125
This gives:


Collect like terms


The probability is then calculated as:


So, the complete table is:





