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Kryger [21]
3 years ago
11

What is (0,-6) reflected across the y-axis

Mathematics
2 answers:
mr_godi [17]3 years ago
5 0
(0,-6) reflected across the y axis is (0,6)
Nesterboy [21]3 years ago
3 0
Same thing (0, -6) if I’m not mistaken
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Tire a bycicle tire has a radius of 15 inches. What is the circumference of the tire? Round to the nearest tenth.
wlad13 [49]
94.2 is the circumference
3 0
3 years ago
Read 2 more answers
What is the percent of 840 is 546
lesantik [10]
K so its an equation      546         x
                                   over   = over
                                   840       100
when u do cross multiplication it will be 840x=546 * 100
                                                       x is about 64%
Hope this is helpful
4 0
3 years ago
The angle of elevation to the top of a skyscraper is measured to be 2° from a point on the ground 1 mile from the building. How
Nadusha1986 [10]

See picture below for a geometric view.


Let x = height of building


We use basic trigonometry to find x.


The tangent function = opposite side of right triangle OVER adjacent side of right triangle.


tan(2°) = x/1


Solve for x.


After multiplying both sides by 1 (here 1 represents 1 mile), we get

tan(2°) = x.


We now use the calculator to find x.


In fact, you can take it from here. Use your calculator.


3 0
3 years ago
If f(x) = -5x +5 and g(x) +3x +8, find g(f(x))
4vir4ik [10]

Answer:

Step-by-step explanation:

g(f(x)) =  g(f(x)) +3(-5x +5 ) +8

= -15x + 15 + 8

= -15x + 23

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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