Answer:
- 1
Step-by-step explanation:
2x + 3 = x - 4
2x - x = - 4 + 3
x = - 1
The solution to the above equation is - 1.
<span>4p - 5 • (p + 6)
</span>-p - 30 = -1 • (p + 30)<span>
</span>
<span> -p - 30</span>
The car takes the time 5.5 hours i.e.5 hours and 30 minutes to travel 462 kilometres when the car travels at average speed of 84 kilometres per hour
Given the average speed of the car is 84 kilometres per hour and the distance to be travelled is 462 kilometres and asked to find the time taken for the journey of 462 kilometres
To calculate the time taken for the journey we use the below-mentioned formula
![speed=\frac{distance}{Time taken}](https://tex.z-dn.net/?f=speed%3D%5Cfrac%7Bdistance%7D%7BTime%20taken%7D)
(speed)×(Time taken)=distance
![Time taken=\frac{distance}{speed}](https://tex.z-dn.net/?f=Time%20taken%3D%5Cfrac%7Bdistance%7D%7Bspeed%7D)
Time taken=![\frac{462 kilometres}{84KMPH}](https://tex.z-dn.net/?f=%5Cfrac%7B462%20kilometres%7D%7B84KMPH%7D)
Time taken=5.5 hours i.e.5 hours and 30 minutes
As a result, the time spent on the car journey is 5.5 hours
Hence, It takes the time 5hours and 5 minutes to travel 462 kilometres
Learn more about time here:
brainly.com/question/28050940
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Answer:
- Decay rate, r = 0.014
- Initial Amount =120,000
![P(t)=120000(0.986)^t](https://tex.z-dn.net/?f=P%28t%29%3D120000%280.986%29%5Et)
- P(10)=104,220
Step-by-step explanation:
The exponential function for growth/decay is given as:
![P(t)=P_0(1 \pm r)^t, where:\\P_0$ is the Initial Population\\r is the growth/decay rate\\t is time](https://tex.z-dn.net/?f=P%28t%29%3DP_0%281%20%5Cpm%20r%29%5Et%2C%20where%3A%5C%5CP_0%24%20is%20the%20Initial%20Population%5C%5Cr%20is%20the%20growth%2Fdecay%20rate%5C%5Ct%20is%20time)
In this problem:
The city's initial population is 120,000 and it decreases by 1.4% per year.
- Since the population decreases, it is a Decay Problem.
- Decay rate, r=1.4% =0.014
- Initial Amount =120,000
Therefore, the function is:
![P(t)=120000(1 - 0.014)^t\\P(t)=120000(0.986)^t](https://tex.z-dn.net/?f=P%28t%29%3D120000%281%20-%200.014%29%5Et%5C%5CP%28t%29%3D120000%280.986%29%5Et)
When t=10 years
![P(10)=120000(0.986)^10\\=104219.8\\\approx 104220 $ (to the nearest whole number)](https://tex.z-dn.net/?f=P%2810%29%3D120000%280.986%29%5E10%5C%5C%3D104219.8%5C%5C%5Capprox%20104220%20%24%20%28to%20the%20nearest%20whole%20number%29)
Answer:
The equation you are given is a quadratic. The standard form of a quadratic is y = a(x-h)2 + k where (h,k) is the vertex of the graph, which is a parabola. Vertically moving the graph 4 units upward means that you are moving k +4 units.
y = a(x-h)2 + k standard form
y = 5x2 - 4 original equation
y = 5(x-0)2 - 4 re-written in standard form
h = 0 k = -4
Four (4) units up is k + 4--->-4 + 4 = 0.
Therefore, f(x) = 5x2 + 0--->f(x) = 5x2.
Step-by-step explanation:
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