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Korolek [52]
2 years ago
12

A rectangular field has an area of 162 m² the width of this field is 9m what is the perimeter of this field

Mathematics
1 answer:
Andru [333]2 years ago
3 0

Answer:

54 m

Step-by-step explanation:

Alrighty, what have we here: an area of 162 m² and a width of 9m.

So: A=162m², W=9m, let's draw this: (see the screenshot attached)

Using the formula for area (Area = width x length) we see that:

162m^{2} =9m*L

Thus: (equating to L by dividing both sides by 9m)

⇒ \frac{162m^{2} }{9m}=18m=L

Now we have all of our dimensions, we can use the formula for the perimeter to calculate it:

perimeter=2W+2L=2*9m+2*18m=18m+36m=54m

And there we have our answer.

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<h2><u>question 1</u></h2>

n^{2} - 20n -96 = 0

use product and sum method

product = -96

sum = -20

numbers needed = ( -24 , 4)

n - 24 = 0

n + 4 = 0

hence <u>n = 24 and n = -4 </u>

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<h2><u>Question 2 </u></h2>

<u />x^{2} + 12 x = 48<u />

in the form ax^{2}  +bx +c = 0

= x^{2} +12x - 48

make use of the formula :

\frac{-b+-\sqrt{b^{2} -4ac} }{2a}

replace values to make 2 equations :

1.\frac{-12+\sqrt{12^{2} -4*1*-48} }{2*1} = 3.17

2.\frac{-12-\sqrt{12^{2} -4*1*-48} }{2*1} = -15.2

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<h2><u>Question 3 </u></h2>

<u />x^{2} -14x+40=0<u />

use product and sum method

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<h2><u>Question 4 </u></h2>

<u />5b^{2} -20b-18 = 7<u />

in the form ax^{2}  +bx +c = 0

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= b^{2} -4b-5 =0\\

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product = -5

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