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Alex17521 [72]
3 years ago
15

The square shown is composed of smaller equally-sized squares. The shaded section has an area of 9/25 square inches. What is the

area of the large square.
Mathematics
1 answer:
barxatty [35]3 years ago
8 0

I believe this is a typo for this question.  The boxes shaded are 4 out of 25.


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The coordinates of point T are ​(​0,5​). The midpoint of ST is ​(​3,-5​). Find the coordinates of point S. (Type an ordered pair
Sholpan [36]
The coordinates are (6,-10)
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Nikita bought a flat for rupees 4987653. Round off the price of the flat to the nearest thousands. ​
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If you round to the nearest thousands then it would be 4988000.
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I need the answer ASAP please and thank you
o-na [289]

Answer:

-3

Step-by-step explanation:

We know that both the larger triangle is dialated by some factor.

To start divide the two given sides:

36/24=1.5

This means that the larger triangle is dialated by a factor of 1.5, meaning that all sides of the smaller triangle must be multipled by 1.5 to get the larger triangle.

By knowing that we can multiply 3x-7 by 1.5 and set it equal to 6x-6.

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4.5x-10.5=6x-6

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Escribe la posición del móvil si el diámetro de la trayectoria es de 10 m y la distancia recorrida es de 190 m
lions [1.4K]

Step-by-step explanation:

La posici´on de una part´ıcula que se mueve unidimensionalmente esta definida por la ecuaci´on:

x(t) = 2t

3 − 15t

2 + 24t + 4 donde 0x

0 y

0

t

0

se expresan en metros y segundos respectivamente. Determine:

a. ¿Cu´ando la velocidad es cero?

b. La posici´on y la distancia total recorrida cuando la aceleraci´on es cero.

Soluci´on:

a. Recordemos que:

v(t) =

dx

dt =

d

dt(2t

3 − 15t

2 + 24t + 4) = 6t

2 − 30t + 24

Sea t

0

el tiempo en que la velocidad se anula, entonces v(t

0

) = 0.

De este modo:

0 = v(t

0

) = 6(t

0

)

2 − 30(t

0

) + 24 = 6[(t

0

)

2 − 5(t

0

) + 4] = 6[(t

0

) − 4][(t

0

) − 1]

As´ı tenemos que:

t

0

1 = 4, t

0

2 = 1

De este modo, tenemos que la velocidad se anula al primer segundo y a los cuatro segundos.

b. Recordemos que:

a(t) =

dv

dt =

d

dt(6t

2 − 30t + 24) = 12t − 30

Ahora sea t

0

el instante en que la aceleraci´on se anula, entonces a(t

0

) = 0

Ahora:

0 = a(t

0

) = 12t

0 − 30

As´ı tenemos que: t

0 =

30

12 =

5

2

Por lo tanto, la posici´on en este instante es:

x(t

0

) = x

5

2

= 2

5

2

3 − 15

5

2

2 + 24

5

2

+ 4 = 125

4 − 3

125

4 + 60 + 4 = −2

125

4 + 64 = −

125

2 +

128

2 =

3

2

De este modo, la posici´on de la part´ıcula cuando la aceleraci´on es cero es de 3

2 metros.

Adem´as la distancia total recorrida esta dada por:

distancia = |x(t

0

) − x(0)| = |

3

2 − 4| =

5

2

Finalmente la distancia total recorrida es: 5

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The percent of decrease is 12.5 but if you need it rounded to the nearest whole number it is 13
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