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torisob [31]
3 years ago
14

A rectangular garden has a total area of 48 sq yards. Draw and label two possible rectangular gardens with different side length

s that have the same area.
Mathematics
1 answer:
Pie3 years ago
6 0
                                  6yd
8yd                                                                                              a =48sq yd                                                                                                    (6×8=48)


                                         4yd 
12yd                                                                                      a=48sq yd                                                                                               (4×12=48)
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Please help it's urgent.
arlik [135]

Answer:

It is the second triangle.

Step-by-step explanation:

This is because the area of the square is given as 25^2 as the perpendicular, 144^2 as the base, and the hypotenuse is 169^2.

Length of the perpendicular = 5^2 (Because 25^2 / 5^2)

Length of the base = 12^2

Length of the hypotenuse = 13^2

So, the Pythagoras theorem is,

=> H^{2} = P^{2} + B^{2}

=> 13^{2} =12^{2} +5^{2}

This is true, because, 13*13 = 12*12 + 5*5

=> 169 = 144 + 25

=> 169 = 169

If my answer helped, kindly mark me as the brainliest!!

Thanks!!

3 0
2 years ago
write an inequality for the description of: ten times a number increased by four is no more than twenty five
Xelga [282]
10x +4 </ 25 (the </ is a less than or equal to sign)
3 0
3 years ago
How may shells does Joshua have?
Tpy6a [65]
24, since 54-30 is 24, and 30+24 is 54.
5 0
3 years ago
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5) Find the LCM of 12 and 32​
katrin [286]

Answer:

96

Step-by-step explanation:

Find the prime factorization of 12

12 = 2 × 2 × 3

Find the prime factorization of 32

32 = 2 × 2 × 2 × 2 × 2

Multiply each factor the greater number of times it occurs in steps i) or ii) above to find the LCM:

LCM = 2 × 2 × 2 × 2 × 2 × 3

LCM = 96

hope it helps ,pls mark me as brainliest

5 0
2 years ago
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4x + 6 = 8x – 10 graphing
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Are you graphing or trying to find the answer
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