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WARRIOR [948]
3 years ago
8

Please help me I beg!

Mathematics
1 answer:
skad [1K]3 years ago
3 0

Answer:

38.9

Step-by-step explanation:

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Question four should be -7, question five should be 99, and question six should be 56.

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three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal
Nana76 [90]

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

=     x + x + (x + \frac{6}{5}) = \frac{27}{4}

=     3x + \frac{6}{5} = \frac{27}{4}

=     3x = \frac{27}{4} - \frac{6}{5}

( let's take the LCM of 4 and 5 = 20

=     3x = \frac{135}{20} - \frac{24}{20}

=     3x = \frac{111}{20}

=       x = \frac{111}{20} ÷ \frac{3}{1} = \frac{111}{20} × \frac{1}{3} = \frac{37}{20}

So, the weight of the equal bags are \frac{37}{20} and the weight of the third bag ( heavy one ) is \frac{37}{20} + \frac{6}{5} = \frac{37}{20} + \frac{24}{20} = \frac{61}{20}

1st bag =     \frac{37}{20} kg

2nd bag =  \frac{37}{20} kg

3rd bag =   \frac{61}{20} kg

Hope it helps...

7 0
3 years ago
PLEASE HELP ME me me me​
kykrilka [37]

Answer:

Andre has the correct answer. When simplified his answer is equivalent to the original equation.

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Convert from radians to degrees -7pi/2
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Π=180⁰
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Which equations below represent nonlinear functions?
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