$(2x+3)^2=11\to2x+3=\pm\sqrt{11}\ \ \ |-3\\\\2x=-\sqrt{11}-3\ \vee\ 2x=\sqrt{11}-3\ \ \ |:2\\\\x=\dfrac{-\sqrt{11}-3}{2}\ \vee\ x=\dfrac{\sqrt{11}-3}{2}$

Answer: C. and E.

7 0

4 years ago

Quick only 5 minutes to answer!

garik1379 [7]

Answer:

14 and 2/7 minutes, or 100/7 minutes

Step-by-step explanation:

To begin with, supposed the rate at which Tyler and Kiran are running are variables T and K. We can come up with the equations

30T= 4,200 and

30K= 6,300.

Where 30 is the number of minutes, multiplied by the rate at which they are running to equal the total distance ran. We can isolate the variables by dividing both sides of both equations by 30, resulting in Tyler's speed of 140 meters per minute and Kiran's speed of 210 meters per minute.

We can then plug these values into a final equation where x is the amount of time taken for Kiran to run 1 km more than Tyler

We end up with this: 210x= 140x+1000

Solve first by subtracting 140x from 210x to get 70x, and then divide both sides by 70. This isolates x and you get a final answer of 14 2/7 minutes, or 100/7 minutes

And sorry I couldn't get this to you in the 5 minutes you needed, but I hope this at least helps you understand the concept :')

7 0

3 years ago

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