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Vladimir [108]
3 years ago
13

How will the graph of the function f(x) = 3* translate when the function is changed to f(x) = 3(x-2)?

Mathematics
1 answer:
Lelechka [254]3 years ago
6 0

Answer: B

Step-by-step explanation:

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This is definitely<em><u> </u></em><em><u>FALSE</u></em>

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Three sides of a fence and an existing wall form a rectangular enclosure. The total length of a fence used for the three sides i
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<span>Let the wall's length be y. So the fence parallel is also y.

We form the equation;
2x + y = 240

Now the area A is x*y, which is given as 5500.

We have 2 equations with 2 variables. Solve them and get the answer.

xy = 5500
2x + y = 240

Y = 5500/x

2x + (5500/x) = 240
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Soon you will determine the right triangle connection the pythagorean theorem
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3 years ago
Let f(x)=x^3+3 and g(x) = 2x^2 + 2x +1. What is g(f(-2))?
emmasim [6.3K]

Answer:

61

Step-by-step explanation:

This composite function can be easily solved

First solve f(-2)

f(-2)=(-2)³+3

f(-2)=-8+3=-5

Now plug -5 for x in function g(x)

=2(5)²+2(5)+1

=50+10+1

=61

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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
3 years ago
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