To find the rate, first you divide 180mi/3hr to get 60mph. On the first day they were traveling at a rate of 60mph. It says that on the second day, they drove at the same rate and drove 300 miles. This means you divide 300mi/60mph to get 5hr. They drove 5hr on the second day. And finally, the question asks how long they drove total, so you add 3hr+5hr to get 8hr total.
Answer is 8hr.
Answer:
Snickerdoodles are 0.22$ each and chocolate chip are 0.43$ each
Step-by-step explanation:
using the information in the problem, you can make 2 separate equations: 8x+12y=6.92 and 5x+5y=3.25, let x= the cost of a snickerdoodle and y= the cost of a chocolate chip cookie.
With the 2 equations, we can isolate the y variable of the first equation and simplify which will give us y= -2/3x +6.92/12.
You can now take that equation and plug it into equation number 2:
5x+5(-2/3x + 6.92/12)=3.25.
solve that equation by multiplying 5 through the brackets: 5x-10/3x+34.6/12=3.25
add common terms: 5/3x=4.4/12
find the value of x, you will get x=0.22
plug the value of x (0.22) into one of your equations and solve for the y-value
Answer:
D. 
Step-by-step explanation:
Let be ![y = \Sigma_{i = 0}^{4} \left[5\cdot \left(\frac{1}{3} \right)^{i}\right]](https://tex.z-dn.net/?f=y%20%3D%20%5CSigma_%7Bi%20%3D%200%7D%5E%7B4%7D%20%5Cleft%5B5%5Ccdot%20%5Cleft%28%5Cfrac%7B1%7D%7B3%7D%20%5Cright%29%5E%7Bi%7D%5Cright%5D)
. To find the equivalent expression we must use the following property:
![y = c\cdot \Sigma_{i = 0}^{n} p(x) = \Sigma_{i=0}^{n} [c\cdot p(x)]](https://tex.z-dn.net/?f=y%20%3D%20c%5Ccdot%20%5CSigma_%7Bi%20%3D%200%7D%5E%7Bn%7D%20p%28x%29%20%3D%20%5CSigma_%7Bi%3D0%7D%5E%7Bn%7D%20%5Bc%5Ccdot%20p%28x%29%5D)
(1)
Based on this fact, we find the following equivalence:
Hence, the correct answer is D.
The answer is 2-7=5 is the answer
Step-by-step explanation: A line, ray, or line segment (referred to as segment) that is perpendicular to a given segment at its midpoint is called a perpendicular bisector. ... In the diagram above, RS is the perpendicular bisector of PQ, since RS is perpendicular to PQ and PS≅QS. Additionally, since PS≅QS, point S is the midpoint of PQ.
