I think K=1.123
you have to find by how much it's increasing, think of 6(1/2) times WHAT equals 7(1/3)
6(1/2)a=7(1/3)
a=1.123
That means each side will be the same times 1.123
x=4.492
you can round that to 4.49 or 4.5 or 5
Answer:
distribute the monomial into the polynomial and be careful of sign errors
Step-by-step explanation:
Let's solve this problem step-by-step.
STEP-BY-STEP SOLUTION:
In order to find the answer, we need to do as follows:
Average no. of people per km^2 = total no. of people ÷ total no. of people
Average no. of people per km^2 = 1, 080, 264, 388 ÷ 2, 973, 190
Average no. of people per km^2 = 363.335134317
Now we need to round the answer to the nearest whole number. In this case, we will be rounding down as displayed below:
Average no. of people per km^2 = 363 people
ANSWER:
Therefore, the answer is:
There are an average of 363 people per km^2 of land in India.
Please mark as brainliest if you found this helpful! :)
Thank you <3
Answer:
30/10 and 5/11
Step-by-step explanation:
<u>Answer</u><u> </u><u>:</u><u>-</u>
9(3+√3) feet
<u>Step </u><u>by</u><u> step</u><u> explanation</u><u> </u><u>:</u><u>-</u>
A triangle is given to us. In which one angle is 30° and length of one side is 18ft ( hypontenuse) .So here we can use trignometric Ratios to find values of rest sides. Let's lable the figure as ∆ABC .
Now here the other angle will be = (90°-30°)=60° .
<u>In ∆ABC , </u>
=> sin 30 ° = AB / AC
=> 1/2 = AB / 18ft
=> AB = 18ft/2
=> AB = 9ft .
<u>Again</u><u> </u><u>In</u><u> </u><u>∆</u><u> </u><u>ABC</u><u> </u><u>,</u><u> </u>
=> cos 30° = BC / AC
=> √3/2 = BC / 18ft
=> BC = 18 * √3/2 ft
=> BC = 9√3 ft .
Hence the perimeter will be equal to the sum of all sides = ( 18 + 9 + 9√3 ) ft = 27 + 9√3 ft = 9(3+√3) ft .
<h3>
<u>Hence </u><u>the</u><u> </u><u>perim</u><u>eter</u><u> of</u><u> the</u><u> </u><u>triangular</u><u> </u><u>pathway</u><u> </u><u>shown</u><u> </u><u>is</u><u> </u><u>9</u><u> </u><u>(</u><u> </u><u>3</u><u> </u><u>+</u><u> </u><u>√</u><u>3</u><u> </u><u>)</u><u> </u><u>ft</u><u> </u><u>.</u></h3>