Math helppp! PLEASE HELP!?

Question

Okay so i have 2 problems which really dont make sense to me! Can someone please solve it and try to explain it to me!


1.) A cow is tethered by a 50-meter rope to a rectangular barn. The dimensions of the barn are 60 m X 30 m. The rope is fastened to a hook on the barn that is 10 meters from the corner on the longest side of the barn. Over exactly how much ground can the cow graze?


2.) The weather during my vacation was as follows…
•It was cloudy on 13 different days, but it was never cloudy for an entire day
•Cloudy mornings were followed by clear afternoons
•Cloudy afternoons were preceded by clear mornings
•There were 11 clear mornings and 12 clear afternoons in all
How long was my vacation?

PLEASE HELPP ME! Solve and explainn!!

Okay guys im not cheating. I honestly needed help and its not like im gonna just copy all the work. I am going to go through the process myself and try to get an answer also.

Answer 1

The first one's pretty easy if you get the diagram correctly
Draw the rectangle ABCD (AB=CD=60) and put the cow at A
Then, with centre A and radius 50, draw an arc cutting the rectangle at points E (on AB) & F (on CD)
Then the area you need is (Area of sector AEF)+(Area of triangle AFD)
That should do it.

The second requires a little clear thinking
You can have entire clear days or combination of (clear & cloudy)
Let x = days of clear mornings & cloudy afternoons
Let y = days of cloudy mornings & clear afternoons
Then, the conditions mentioned above say,
x+y=13 (total different cloudy days)
11-x = 12-y (this means that we need clear mornings & clear afternoons in equal numbers after taking the numbers away to complete a cloudy day)
Solve the two equations. I think your vacations lasted for 18 days.

Answer 2

1) The area of the barn is 60m x 30m = 1800m^2
The length of the rope is 50 meter and the shorter side is 30 meter from a right angle triangle. Based on theorem P?, the maximum the cow can reach is at the 50th meter (30^2 + 40^2 = 50^ and 40 + 10 = 50)of the other longer side. Since the rope is fastened to a hook on the barn that is 10 meters from the corner on the longest side of the barn.
Hence the remaining length of the other longer side is 60 - 50 = 10 meter.
The side where the rope is fastened the cow could reach up to the conner of the barn. Hence the unreachable area form another right angle triangle which is 10 meter which the remaining length of the other longer side and the length of the shorter side which is 30 meter.
The area that is unreachable is 1/2 x 30 x 10 = 150m^2
The area of ground that could be graze by the cow is 1800 - 150 = 1650m^2.
2) Since
• It was never cloudy for an entire day
•Cloudy mornings were followed by clear afternoons
•Cloudy afternoons were preceded by clear mornings
•There were 11 clear mornings and 12 clear afternoons in all
11 + 12 = 23 days

Answer 3

1.A cow is tethered by a 50-meter rope to a rectangular barn. The dimensions of the barn are 60 m X 30 m. The rope is fastened to a hook on the barn that is 10 meters from the corner on the longest side of the barn. Over exactly how much ground can the cow graze?

50 meters is radius of a possible grazing circle limited by the barn.
so figure out area of half a 50 m radius circle
pi * r^2 = pi * (50 m)^2 = 2500pi m2
half of this are is obviously clear for the cow to graze,
1250pi m2

with the rope hanging up 10 m in on the 60 m side, the radius becomes 40m for a quarter of that circle
pi * (40 m)^2 = 1600pi m2, only 1/4 of this circle are is reached, so 400pi m2

now the 50m rope can stretch along the 30 m side, reducing the rope length to 10 m: (50 - 10 - 30) = 10 m
a circle area with radius 10m for only 1/4 of the circle
pi * (10m)^2 = 100pi m2, only 1/4 circle gives 25pi m2

so the grazing area is the sum of
(1250 + 400 + 25)pi m2 = 1675pi m2 = 5262 m2

I really liked Singh setup for no.2...stay with that

Answer 4

a million. via fact if u cases a destructive quantity via itself, the destructive will become a favorable. ( -a million X -a million = a million ) 2. 10 miles: 2X5=10 3. 5.2 X 10^11: 520000000000 take out the #'s that r no longer 0 and make beneficial the single in front is interior the single's locations. count quantity the different # locations and cases the # that's interior the single's place via 10^different#locations 4. 2.34 X 10^6: comparable as above, if u ignored a 0 desire ur in a position to bypass math and desire this helps! XD

Answer 5

1) depends on if it is an open barn (without walls)

2) cannot tell - how many daze were completely sunny?
could have been gone 60 daze - MY KINDA VACATION!!!

Answer 6

sorry but im ot goin to help yo chet

Answer 7

stop cheating and do your homework yourself lol