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barxatty [35]
2 years ago
14

Caleb draws 14 dogs on each of 4 posters.He draws 18 cats on each of 6 other posters. If he draws 5 more dogs on each poster wit

h dogs, how many dogs ad cats does he draws?
Mathematics
1 answer:
MatroZZZ [7]2 years ago
5 0
Caleb draws 14 dogs on each of 4 posters.
Then,
Number of dogs drawn by caleb in the first instance = (14 * 4) dogs
                                                                                    = 56 dogs
Celeb drew 18 cats on each of 6 other posters.
Then
Number of cats drawn by Caleb in the first instance = (18 * 6) cats
                                                                                   = 108 cats
Again Caleb draws 5 more dogs on each poster with dogs.
Then,
The number of Dogs drwn in the second instance = (5 * 4) dogs
                                                                                 = 20 dogs
So, the total number of dogs drawn by Caleb = ( 56 + 20) dogs
                                                                         = 76 dogs
Then the total number of cats and dogs drawn by Caleb = 108 + 76
                                                                                           = 184
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Answer:

Sadie's error is " she made error in step 2 =\sqrt{3^2\times 6\times a^2\times a^5\times b^2\times b}  where a\geqslant 0"

Because she made error in splitting the powers to simplify the square root

<h3>Therefore the correct answer for Sadie's expression is 3ab\sqrt{6ab} where a\geqslant 0</h3>

Step-by-step explanation:

Given that " Sadie simplified the expression StartRoot 54 a Superscript 7 b cubed EndRoot, where a greater-than-or-equal-to 0, "

It can be written as \sqrt{54a^7b^3} where a\geqslant 0

The given expression is \sqrt{54a^7b^3} where a\geqslant 0

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Sadie's steps are

\sqrt{54a^7b^3}  where a\geqslant 0

=\sqrt{3^2\times 6\times a^2\times a^5\times b^2\times b}

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<h3>\sqrt{54a^7b^3}=3ab\sqrt{6a^5b} where a\geqslant 0</h3><h3><u>Now corrected steps are</u></h3>

\sqrt{54a^7b^3}  where a≥0

=\sqrt{(9\times 6)(a^{6+1})(b^{2+1})

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=\sqrt{3^2\times 6\times ((a^3)^2.a)(b^2.b) (by using the identity a^{mn}=(a^m)^n )

=3ab\sqrt{6ab}

Therefore \sqrt{54a^7b^3}=3ab\sqrt{6ab}  where a\geqslant 0

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Sadie's error is " she made error in step 2 =\sqrt{3^2\times 6\times a^2\times a^5\times b^2\times b} " where a\geqslant 0

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<h3>Therefore the correct answer for Sadie's expression is 3ab\sqrt{6ab} where a\geqslant 0</h3>
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