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olchik [2.2K]
3 years ago
8

Tengo 5 manzanas y me regalan dos ​

Mathematics
1 answer:
cricket20 [7]3 years ago
3 0

Answer:

what?

Step-by-step explanation:

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Anon25 [30]

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kenny6666 [7]

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3 years ago
Please answer! this is an important exam that im taking right now.
Scorpion4ik [409]

Answer:

Required ordered pair is (0,0) for system of equation

Step-by-step explanation:

The given system of equation is

A). 9^{12x} = 9^{3y}

B). y^{4} - 9y^{2} x^{2} =16y

On simplifying the equation A

9^{12x} = 9^{3y}

Take log on both side,

(12x) (log9) = (3y) (log9)

4x=y

To find the solution of the system of the equation :

Replacing value of y=4x in equation B,

y^{4} - 9y^{2} x^{2} =16y

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256x^{4} - 144x^{2} x^{2} =64x

256x^{4} - 144x^{4}=64x

112x^{4}=64x

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x=0 and x=0.829

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8 0
4 years ago
What effect does doubling the length and width have on the perimeter
Paraphin [41]
The effect of doubling the length and width have on the perimeter would be 2x bigger
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3 years ago
Reduce this algebraic fraction 27c^4d^5/9c^7d^4 ...?
Naddika [18.5K]
The answer is \frac{3d}{ c^{3}}

The fraction is: \frac{27 c^{4} d^{5}}{9 c^{7}  d^{4} }
Let's rewrite it: \frac{27 c^{4} d^{5}}{9 c^{7} d^{4} }= \frac{27}{9}*\frac{c^{4}}{c^{7}}*\frac{d^{5}}{d^{4} }

Now, let's use the rule \frac{ x^{a} }{ x^{b}} =  x^{a-b} = \frac{1}{ x^{b-a} }:
\frac{27}{9}*\frac{c^{4}}{c^{7}}*\frac{d^{5}}{d^{4} }=3* \frac{1}{c^{7-4}}  * d^{5-4} =3* \frac{1}{ c^{3} } *d^{1}= \frac{3d}{ c^{3} }
5 0
3 years ago
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