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Pavlova-9 [17]
3 years ago
14

How long is the coastline of canada

Mathematics
1 answer:
netineya [11]3 years ago
8 0
All you need to do is add 12383 and 113211 to get 125594
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In a particular​ year, a total of 40 comma 285 students studied in two of the most popular host countries when traveling abroad.
klasskru [66]

Answer: 2nd most popular country =  15895 students

most popular country = 24390 students

Step-by-step explanation:

Total students: 40,285

2nd most popular country = x students

most popular country = x + 8,495 students

x + x + 8495 = 40285

2x + 8495 = 40285

2x = 40285 - 8495

2x = 31790

x = 15895

2nd most popular country = x = 15895 students

most popular country = x + 8,495 = 15895 + 8,495 = 24390 students

6 0
3 years ago
PLEASE HELP PLEASE PLEASE
Phantasy [73]
-2(-2)^4 + 5(-2)^2 - 4

-2(-2)= 4. 4^4= 16

5(-2)= -10. -10^2= 100 - 4= 96

96 + 16= 112

Answer: 112
8 0
4 years ago
13 cm<br> Find x.<br><br> 24 degrees
AleksandrR [38]
X is 66 degrees, you just subtract 24 to 90 and it gives you 66.
5 0
3 years ago
−5x 5 y 5 (−6xy 3 ) simplify
Readme [11.4K]

Answer:

|-2| + 2

Step-by-step explanation:

-4 / 2-5<em>i</em>

cos (<em>x</em>) / 1 - sin squared (<em>x</em>)

<u>|-2| + 2</u>

8 0
2 years ago
Read 2 more answers
How to work it out logically? Question a !!!
Arturiano [62]

Answer:

Fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.

Step-by-step explanation:

Given the figure with dimensions. we have to find the area of given figure.

Area of figure=ar(1)+ar(2)+ar(3)

Area of region 1 = ar(ANGI)+ar(AIB)

                          =L\times B+\frac{1}{2}\times base\times height\\\\=[1500\times (5000-2000-1500)]+\frac{1}{2}\times (3000-1500)\times (5000-2000-1500)\\\\=3375000m^2=337.5ha

Area of region 2 = ar(DHBC)

                       =2000\times1500\\\\=3000000m^2=300ha

Area of region 3 = ar(GFEH)

                             (2000+1500)\times 1000\\\\=3500000m^2=350ha

Hence, Area of figure=ar(1)+ar(2)+ar(3)=337.5ha+300ha+350ha

                                                 =987.5 ha

Now, we have to do straight-line fencing such that area become half and cost of fencing is minimum.

Let the fencing be done through x m downward from B which divides the two into equal area.

⇒ Area of upper part above fencing=Area of lower part below fencing

⇒ar(ANGB)+ar(GKLB)=ar(KLCM)+ar(MDCF)\\\\337500+3000x=(3500-x)\times 1000+2000(1500-x)\\\\3375000+3000x=3500000-1000x+3000000-2000x\\\\6000x=315000\\\\x=\frac{315000}{6000}=520.8m

Hence, fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.

7 0
3 years ago
Read 2 more answers
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