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kifflom [539]
3 years ago
10

What is the largest possible area (in cm²) for a rectangle with a perimeter of 120 cm?

Mathematics
1 answer:
kompoz [17]3 years ago
8 0

Answer:

900 cm²

Step-by-step explanation:

Perimeter of a rectangle is given by

P=2(l+w)

Where l and w are length and width respectively.

Substituting P with 120 cm

120=2(l+w)

Divide both sides by 2

60=l+w

Making w the subject of formula

w=60-l

Area, A=lw

Substitute w with 60-l

A=l(60-l)=60l-l²

Since A=60l-l²

Get the first derivative of A with respect to l

0=60-2l

l=60/2=30

Therefore, w=60-l=60-30=30

The length abd width are both 30cm. This is a square and a square is also a type of rectangle since it has four sides.

Area=30*30=900cm²

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Find the fifth roots of 32(cos 280° + i sin 280°)
shusha [124]

Answer:

The fifth root is 2[cos(56°) + i sin(56°)]

Step-by-step explanation:

* To solve this problem we must revise De Moiver's rule

- In the complex number with polar form

∵ z = r(cosФ + i sinФ)

∴ z^n = r^n(cos(nФ) + i sin(nФ))

* In the problem

- The fifth root means z^(1/5)

- We can put 32 as a form a^n

∵ 32 = 2 × 2 × 2 × 2 × 2 = 2^5

∴ z = 2^5[cos(280°) + i sin(280°)]

* Lets find z^(1/5)

*z^{\frac{1}{5}}=[2^{5}]^{\frac{1}{5} } (cos(\frac{1}{5})(280)+isin(\frac{1}{5})(280)

*(2^{5})^{\frac{1}{5}}=2^{5.\frac{1}{5}}=2

∴ z^(1/5) = 2[cos(56) + i sin(56)]

* The fifth root of 32[cos(280°) + i sin(280°)] is 2[cos(56°) + i sin(56°)]

 

4 0
3 years ago
Scientists released 8 rabbits into a new habitat in year 0
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Answer:

A. f(x)= 8(2)^x

Step-by-step explanation:

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What is the equation of the line that passes through the point (6,-4) and has a slope of -\frac{1}{6}− 6 1 ​ ?
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Answer:

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Step-by-step explanation:

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7 0
3 years ago
If g=8 what is the value if the expression g/2+3​
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