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Ivahew [28]
3 years ago
4

When a photo is reduced or enlarged it's length to width ratio usually remains the same .Aurelian wants to enlarge a 4 in by 6 i

n photo so that it has a width of 15 inches

Mathematics
1 answer:
Ilia_Sergeevich [38]3 years ago
8 0
Length:   4      2     10
Width:     6      3     15

You can tell because 15 isn't divisible by 6, but it is divisible by 3, which goes into 6.

Divide 6 by 2 to get 3.

To keep the ratio, we must also divide 4 by 2 to get 2.

Then, multiply 3 by 5 to get 15 for the width.

You must also multiply 2 by 5 to get the length of 10.

The picture's dimensions will be 10"x15".
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State the property.<br>4(2+3)=(2+3)4​
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The order of the expression is different.

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(12/21)... what is this fraction simplified
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\frac{12}{21} \\ \\ GCF = 3 \\ \\ 12 \div 3 = 4 \\ \\ 21 \div 3 = 7 \\ \\  \frac{4}{7} \\ \\ Answer: \fbox {4/7}
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A rectangular picture frame measures 63 cm by 63 cm. What is the perimeter or distance around the picture frame in meters? m
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6 0
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Sea un cuadrado de 2 pulgadas de lado uniendo los puntos medios se obtiene otro cuadrado inscrito en el anterior si repetimos es
Ne4ueva [31]

Answer:

1) La serie geométrica formada es

4, 2, 1,..., ∞

2) La suma al infinito de las áreas de los cuadrados es 8 in.²

Step-by-step explanation:

1) El área del primer cuadrado, a₁ = 2² = 4 pulgadas²

El área del siguiente cuadrado, a₂ = (√ (1² + 1²)) ² = (√2) ² = 2 pulg²

El área del siguiente cuadrado, a₃ = ((√ (2) / 2) ² + (√ (2) / 2) ²) = 1 pulg²

Por lo tanto, la razón común, r = a₂ / a₁ = 2/4 = a₃ / a₂ = 1/2

Las áreas de los cuadrados progresivos forman una progresión geométrica como sigue;

4, 4×(1/2), 4 ×(1/2)²,...,4×(1/2)^{\infty}

De donde obtenemos la serie geométrica formada de la siguiente manera;

4, 2, 1,..., ∞

2) La suma de 'n' términos de una progresión geométrica hasta el infinito para -1 <r <1 se da como sigue;

S_{\infty} = \dfrac{a}{1 - r}

Por lo tanto, la suma de las áreas de los cuadrados hasta el infinito se obtiene sustituyendo los valores de 'a' y 'r' en la ecuación anterior de la siguiente manera;

La \ suma \ al \ infinito \ del \ cuadrado \ S_{\infty}  = \dfrac{4 \ in.^2}{1 - \dfrac{1}{2} } = \dfrac{4 \ in.^2}{\left(\dfrac{1}{2} \right)} = 2 \times 4 \ in.^2= 8 \ in.^2

La suma al infinito de las áreas de los cuadrados, S_{\infty} = 8 in.²

7 0
2 years ago
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