Answer:
1200 miles.
Step-by-step explanation:
Let d represent one way distance.
We have been given that a plane averaged 800 mph on a trip going east, but only 400 mph on the return trip.
The time taken while going east would be
.
The time taken while returning back would be
.
Since the total flying time in both directions was 4.5 hr, so we will equate sum of both times with 4.5 as:
![\frac{d}{800}+\frac{d}{400}=4.5](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7B800%7D%2B%5Cfrac%7Bd%7D%7B400%7D%3D4.5)
Make a common denominator:
![\frac{d}{800}+\frac{d*2}{400*2}=4.5](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7B800%7D%2B%5Cfrac%7Bd%2A2%7D%7B400%2A2%7D%3D4.5)
![\frac{d}{800}+\frac{2d}{800}=4.5](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7B800%7D%2B%5Cfrac%7B2d%7D%7B800%7D%3D4.5)
![\frac{d+2d}{800}=4.5](https://tex.z-dn.net/?f=%5Cfrac%7Bd%2B2d%7D%7B800%7D%3D4.5)
![\frac{3d}{800}=4.5](https://tex.z-dn.net/?f=%5Cfrac%7B3d%7D%7B800%7D%3D4.5)
![\frac{3d}{800}\times \frac{800}{3}=4.5\times \frac{800}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B3d%7D%7B800%7D%5Ctimes%20%5Cfrac%7B800%7D%7B3%7D%3D4.5%5Ctimes%20%5Cfrac%7B800%7D%7B3%7D)
![d=1.5\times \frac{800}{1}](https://tex.z-dn.net/?f=d%3D1.5%5Ctimes%20%5Cfrac%7B800%7D%7B1%7D)
![d=1200](https://tex.z-dn.net/?f=d%3D1200)
Therefore, the one-way distance was 1200 miles.
Answer:
n = 8
Step-by-step explanation:
-4 + 8 = n/2
4 = n/2
4 x 2 = n
8 = n
Answer:
![ab=\gcd(a,b)\cdot \text{lcm}(a,b)](https://tex.z-dn.net/?f=ab%3D%5Cgcd%28a%2Cb%29%5Ccdot%20%5Ctext%7Blcm%7D%28a%2Cb%29)
Step-by-step explanation:
Using the hint, write a and b in the following prime factorization:
where
for i ≠ j.
Then by the formulae for gcd(a,b) and lcm(a,b) we know that:
![\gcd(a,b)=p_1^{\min(x_1,y_1)}p_2^{\min (x_2,y_2)} \cdots p_t^{\min(x_t,y_t)}](https://tex.z-dn.net/?f=%5Cgcd%28a%2Cb%29%3Dp_1%5E%7B%5Cmin%28x_1%2Cy_1%29%7Dp_2%5E%7B%5Cmin%20%28x_2%2Cy_2%29%7D%20%5Ccdots%20p_t%5E%7B%5Cmin%28x_t%2Cy_t%29%7D)
![\text{lcm}(a,b)= q\cdot r\cdot p_1^{\max(x_1,y_1)}p_2^{\max(x_2,y_2)}\cdots p_t^{\max(x_t,y_t)}](https://tex.z-dn.net/?f=%20%5Ctext%7Blcm%7D%28a%2Cb%29%3D%20q%5Ccdot%20r%5Ccdot%20p_1%5E%7B%5Cmax%28x_1%2Cy_1%29%7Dp_2%5E%7B%5Cmax%28x_2%2Cy_2%29%7D%5Ccdots%20p_t%5E%7B%5Cmax%28x_t%2Cy_t%29%7D)
Note that the expression
for all i, since if the minimum is, <em>without loss of generality</em>,
, then the maximum must be
, and viceversa. Then, it is straightforward to verify that when we multiply gcd(a, b) and lcm(a, b) its prime factorization matches the prime factorization of ab, and so we can see the equaility holds:
![\gcd(a,b)\cdot \text{lcm}(a,b)=ab.](https://tex.z-dn.net/?f=%5Cgcd%28a%2Cb%29%5Ccdot%20%5Ctext%7Blcm%7D%28a%2Cb%29%3Dab.)
The expression to represent the number of minutes Andrew ran is x = 4y - 4
<em><u>Solution:</u></em>
Given that, Andrew ran 4 minutes less than 4 times as many minutes as Chris ran
Let "y" be the number of minutes Chris ran
Let "x" be the number of minutes Andrew ran
From given,
Andrew ran 4 minutes less than 4 times as many minutes as Chris ran
Which means,
Number of minutes Andrew ran = 4 times number of minutes Chris ran - 4
![x = 4y - 4](https://tex.z-dn.net/?f=x%20%3D%204y%20-%204)
Thus the expression to represent the number of minutes Andrew ran is found
Answer:
repeating-- terminating-- terminating-- repeating.
Step-by-step explanation: