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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:


Step-by-step explanation:

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Based on the rule of multiplication:

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Let us assume uniform width = x cm wide.

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Therefore,  length of rectangle made by frame = 50 + x+x = (2x+50) cm.

And width of rectangle made by frame = 70+x+x = (2x+70) cm.

We know, the area of rectangular portrait = 50 × 70 = 3500 cm^2.

Total area of the rectangle made by frame would be =  (2x+50) * (2x+70)

We know,

Actual area of frame = Area of rectangle made by frame -  area of rectangular portrait.

We also given "the area of the frame is the same as the area of the portrait."

We can setup an equation now,

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Subtracting 3500 from both sides, we get

3500-3500 = (2x+50) * (2x+70) - 3500-3500.

0 = (2x+50) * (2x+70) -7000.

FOIL (2x+50) * (2x+70), we get

0 = 2x*2x +2x*70 + 50*2x +50*70 - 7000.

0 = 4x^2 +140x +100x +3500 -7000.

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Dividing whole equation by 4, we get

x^2 +60x - 875 =0

Applying quadratic formula =\frac{-b\pm \sqrt{b^2-4ac}}{2a}, we get

=\frac{-60\pm \sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}

x=\frac{-60+\sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}=5\left(\sqrt{71}-6\right)

x=\frac{-60-\sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}:\quad -5\left(6+\sqrt{71}\right)

We cant take negative value.

So, x=5\left(\sqrt{71}-6\right)=12.13

We could take approximately 12 cm.



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