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vazorg [7]
3 years ago
12

Mr Carrillo's family bought lunch from Culvers. They also bought some sodas from

Mathematics
1 answer:
Svetradugi [14.3K]3 years ago
3 0
Tffrrrghdsdgggggdstdsd
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A farmer wants to use 200 feet of fencing for a chicken pen he wants the width of the pen to be 20 feet what is the length of th
Dominik [7]
The length should be 80 feet.
This is because the width is 20 since there are two sides for the width in pen, we have to add 20+20.  This would equal 40. Then we need to subtract 40 from 200. 200-40 would be 160. Then since there are two sides for the length, we need to split that in half. 160÷2 would be 80. To double check, you can do 80+80, and that would still be 160. So the length of one side would be 80. But of course, this whole thing would only be right if the pen has four sides(but since I have seen many chicken pens, there were all rectangular)

7 0
3 years ago
Read 2 more answers
Answer the question.
DIA [1.3K]
A. or 27 is the answer
5 0
3 years ago
Given m//n find the value of x
Alecsey [184]

Answer:

x=151

--------------------------

3 0
3 years ago
rick jogged the same distance on tuesday snd friday, and 8 miles on sunday for a total of 20 miles for the week. solve to find t
saveliy_v [14]

Ricky jogged 6 miles on tuesday and 6 miles on friday

<em><u>Solution:</u></em>

Given that,

Rick jogged the same distance on tuesday and friday

Let "x" be the distance jooged on each tuesday and friday

He also jogged for 8 miles on sunday

Total of 20 miles for the week

Therefore, we frame a equation as,

total distance jogged = miles jogged on tuesday + miles jogged on friday + miles jogged on sunday

20 = x + x + 8

20 = 2x + 8

2x = 20 - 8

2x = 12

x = 6

Thus Ricky jogged 6 miles on tuesday and 6 miles on friday

3 0
3 years ago
Tan(A+45°) - tan(A-45°) =2sec2A​
weeeeeb [17]

Answer:

\tan(A + 45 \degree)  -  \tan(A - 45 \degree)  \\   \\  = \frac{ \tan(A)  +  \tan(45 \degree) }{1 -  \tan(A)  \tan(45 \degree) }  -  \frac{ \tan(A)  -  \tan(45 \degree) }{1 +  \tan(A)  \tan(45 \degree) }  \\  \\

but tan 45° is 1;

=  \frac{ \tan(A) }{1 -  \tan(A) }  -  \frac{ \tan(A) }{1 +  \tan(A) }  \\  \\  \frac{\tan(A)  + { \tan }^{3}  A   + \tan(A)  -   { \tan}^{3}  A}{1 - { \tan }^{2} A }  \\  \\  =  (\frac{2  \tan(A) }{1 -  { \tan }^{2} A})  \\   \\  =  \frac{2 \sin(A) }{ \cos(A)  }  \times  { \sec }^{2} A \\  \\  =  \frac{2}{ \cos(2A) }  \\  \\  = 2 \sec(2A)

8 0
3 years ago
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