Ericka observed the total variation in the water level over 6.2 years to be −13.64 mm. She compared this to the average annual trend, which shows the water level rising 1.8 mm/year. Remember, she used 6.2 years as her time period. What is the difference between how much average water levels rose and how much the water level fell in the part of the river she observed?

2 & 7

step-by-step explanation:

he climbs 300 ft/hr at a constant rate. the peak is at 1,500 ft.

when going up, he reaches 600 ft in 2 hours. remember, that he is going in a constant rate. divide 600 with 300:

600/300 = 2

assuming that he takes the same route back (meaning that it's a 1500 ft route back, you can solve.

note also that you are solving when he gets back to 600 ft after he reaches the peak.

first, find the amount of hours it takes to get up the peak. divide the total amount (1500), with the ft/her (300) to get the amount of hours (x):

1500/300 = x

x = 5

it takes jamie 5 hours to get up. now, he is returning down, and reaches 600 (his current destination). remember that he climbs at a constant 300 ft/hr. divide 600 with 300:

600/300 = x

x = 2

now add the two numbers together.

x = 5 + 2 = 7

jamie takes 7 hours to go up the mountain and reach his current position when descending.

(total amount/constant rate) = (time).

remember that it is a constant rate, so in total, when he gets back down the mountain, it will take the same amount of time as when he goes up. the total amount for the round trip is: 5 + 5 = 10.

~

im with everythinglos so yeahhhhh

step-by-step explanation:

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.

3+3+3+2*2*2+29*12 = 365

step-by-step explanation: