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Svet_ta [14]
3 years ago
10

28.40 less than 28.400

Mathematics
2 answers:
Strike441 [17]3 years ago
3 0
The answer is 0. Even thought the second number has an extra zero at the end, they are the same value.
tamaranim1 [39]3 years ago
3 0
28.40 and 28.400 are the same thing, just like how 3 and 003 are the same.
So the answer is 0
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What is the volume of a rectangular prism with length 12 in., height 16 in., and width 13 in.?
Lilit [14]

Answer:

2496in3

Step-by-step explanation:

12x16x13=2496in3

7 0
3 years ago
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In the United States, we measure milk by the gallon. If we were using the metric system, what metric unit would we use
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The metric unit used for milk is the liter, L.

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2 years ago
What is the recursive rule for the sequence an=5(-2)n-1
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Multiplying an odd number of negative factors equals a negative

-5* 2n -1

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3 years ago
Lloyd was paid a 6% commission on sales of $95,292. How much money was he paid in commissions?
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4 0
3 years ago
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
3 years ago
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