Answer:
Dana would go 18 miles in 90 minutes because she is going a mile for every five minutes, so if she bikes for another forty minutes she will have biked 8 miles. This means that you would multiply 8 miles (remember its a mile per 5 minutes) by five minutes and that would give you forty. Then just add that on to the 50 minutes she has already biked.
or do it this way
To get the answer you need to find out how many miles she bikes in 10minutes.
Divide 50minutes by 10 you get 5. Now you divide 10(miles) by 5 and you get 2. So it means that in 10minutes she bikes 2miles. Now you need to know how many miles she bikes in 90minutes so if she bikes 2 in 10 then because 90:10=9 you do 2*9=18. She can bike 18miles in 90minutes. :))
Step-by-step explanation:
if this was helpful mark me as brainless
Answer:
63% increase.
Step-by-step explanation:
You don't really need to do anything because it's probably a 63% increase as 101 is close to 100 already. If you'd like:


≈
63%
So we are given two points, say P1(4,7), P2(x,19).
Slope is given by
m=3=(y2-y1)/(x2-x1)=(19-7)/(x-4)
solve for x
3=(19-7)/(x-4)
cross multiply
3(x-4)=12
3x-12=12
3x=12+12=24
x=8
Check the picture below.
so then, the perimeter of that hexagon will just be the sum of all its 6 sides, or namely 3⅖ + 3⅖ + 3⅖ + 3⅖ + 3⅖ + 3⅖, or just 6( 3⅖ ).
![\bf \textit{area of a regular polygon}\\\\ A=\cfrac{1}{2}ap~~ \begin{cases} a=apothem\\ p=perimeter\\[-0.5em] \hrulefill\\ a=3\\ p=6\left(3\frac{2}{5} \right) \end{cases}\implies A=\cfrac{1}{2}(3)\left[ 6\left(3\frac{2}{5} \right) \right]\implies A=\cfrac{1}{2}(3)\left[ 6\left(\cfrac{17}{5} \right) \right] \\\\\\ A=\cfrac{1}{2}(3)\left(\cfrac{102}{5} \right)\implies A=\cfrac{1}{2}\left( \cfrac{306}{5} \right)\implies A=\cfrac{153}{5}\implies A=30\frac{3}{5}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20regular%20polygon%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7Dap~~%20%5Cbegin%7Bcases%7D%20a%3Dapothem%5C%5C%20p%3Dperimeter%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D3%5C%5C%20p%3D6%5Cleft%283%5Cfrac%7B2%7D%7B5%7D%20%5Cright%29%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%283%29%5Cleft%5B%206%5Cleft%283%5Cfrac%7B2%7D%7B5%7D%20%5Cright%29%20%5Cright%5D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%283%29%5Cleft%5B%206%5Cleft%28%5Ccfrac%7B17%7D%7B5%7D%20%5Cright%29%20%5Cright%5D%20%5C%5C%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7D%283%29%5Cleft%28%5Ccfrac%7B102%7D%7B5%7D%20%5Cright%29%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%5Cleft%28%20%5Ccfrac%7B306%7D%7B5%7D%20%5Cright%29%5Cimplies%20A%3D%5Ccfrac%7B153%7D%7B5%7D%5Cimplies%20A%3D30%5Cfrac%7B3%7D%7B5%7D)
Answer:
Height of shuttle at that instant = 1.17 miles
Step-by-step explanation:
Refer the given figure.
C is the position of rocket and B is the position of observer.
Here we need to find AC, that is the height of shuttle at that instant.
Using trigonometric rules on ΔABC, we have

Height of shuttle at that instant = AC = 1.17 miles