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Nikitich [7]
3 years ago
11

HLPE QUICK PLEASE!! WILL GIVE BRAINLIEST!

Mathematics
2 answers:
alukav5142 [94]3 years ago
6 0

Answer:

C

Explanation not provided

Phoenix [80]3 years ago
4 0

Answer:

A.

Step-by-step explanation:

499.0 seconds which rounds to 500 seconds.

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Please help me with this question
Alex17521 [72]

Answer:

The answer is C.

Step-by-step explanation:

5*6=30

5 0
3 years ago
Read 2 more answers
(-8x + 1) – (8x – 1) =<br> ??
Bad White [126]

Answer:0

Step-by-step explanation:

6 0
3 years ago
A sequence is defined by
garik1379 [7]
I believe the answer is D.

3^2 is 9 and 4^2 is 16, and because 9/16 was the third term in the sequence, you raise both the numerator and the denominator by the number of the term minus 1.

So to find the seventh term, you raise the numerator and denominator by 7 - 1 = 6.

So the answer is 729/4096
5 0
3 years ago
Read 2 more answers
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
Ayúdenme porfavor necesito esto para mañana resuelto por el Sistema de Ecuaciones Lineales en
AlekseyPX

Deberias mirar videos en linea

7 0
3 years ago
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