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enyata [817]
3 years ago
15

A translation is applied to the square formed by the points A(−3, −4) , B(−3, 5) , C(6, 5) , and D(6, −4) . The image is the squ

are that has vertices ​ A′(−3, −6) ​, ​ B′(−3, 3) ​, C′(6, 3) and D′(6, −6) .
Select the phrase from the drop-down menu to correctly describe the translation.
The square was translated

A. 2 units down
B. 10 units down
C. 10 units to the left
D. 2 units to the left
Mathematics
2 answers:
likoan [24]3 years ago
4 0
All of the x values are the same so it is not moving left or right.
The y values are changing by being 2 less than the original. 
If you subtract 2 from each y value you get the new set of ordered pairs.
It is moving 2 units down 
Letter A
Pepsi [2]3 years ago
3 0
X-coordinate is fixed. SO no left or right motion

Y coordinates are 2 less.
So It is moving 2 units down
Answer is A
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Endpoint given the Midpoint: * The midpoint of RP is M(2, 4). If one of the end points is R(-1,7), find the coordinates of the o
marin [14]

Answer:

The coordinates of other point are: (5,1)

Step-by-step explanation:

Given coordinates are:

M(x,y) = (2,4)

R(x_1,x_2) = (-1,7)

We have to find the coordinates of other point (x2,y2)

The formula for mid-point is given by:

M(x,y) = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})

Putting the values we get

(2,4) = (\frac{-1+x_2}{2}, \frac{7+y_2}{2})

Putting respective elements equal

\frac{-1+x_2}{2} = 2\\-1+x_2 = 2*2\\-1+x_2 = 4\\x_2 = 4+1\\x_2 = 5\\And\\\frac{7+y_2}{2} = 4\\7+y_2 = 4*2\\7+y_2 = 8\\y_2 = 8-7\\y_2 = 1

Hence,

The coordinates of other point are: (5,1)

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9. Calcule el valor de la fuerza que experimenta un ascensor cuando levanta 150 kg en los siguientes casos: a) Cuando asciende c
GenaCL600 [577]

Answer:

a) La fuerza neta que experimenta un ascensor cuando asciende con una aceleración de 4 metros por segundo al cuadrado es 600 newtons hacia arriba.

b) La fuerza neta que experimenta un ascensor cuando desciende con una aceleración de 6 metros por segundo al cuadrado es 900 newtons hacia abajo.

c) Por la fuerza de tensión sobre el cable del ascensor.

Step-by-step explanation:

a) La fuerza neta experimentada por el ascensor (F), en newtons, es:

F = m\cdot a (1)

Donde:

m - Masa, en kilogramos.

a - Aceleración, en metros por segundo cuadrado.

Si sabemos que m = 150\,kg y a = 4\,\frac{m}{s^{2}}, entonces la fuerza neta del ascensor es:

F = m\cdot a

F = 600\,N

La fuerza neta que experimenta un ascensor cuando asciende con una aceleración de 4 metros por segundo al cuadrado es 600 newtons hacia arriba.

b) Si sabemos que m = 150\,kg y a = -6\,\frac{m}{s^{2}}, entonces la fuerza neta del ascensor es:

F = m\cdot a

F = -900\,N

La fuerza neta que experimenta un ascensor cuando desciende con una aceleración de 6 metros por segundo al cuadrado es 900 newtons hacia abajo.

c) De manera simplificada, el ascensor experimenta dos fuerzas que definen la fuerza y aceleración netas: (i) La fuerza de tensión sobre el cable que traslada el ascensor y el peso total del ascensor, opuesta a la anterior y en función de la aceleración gravitacional. Puesto que la masa no varía en ningún caso, se concluye que el peso es constante, entonces la diferencia de valores se debe a la fuerza por tensión del cable. En el descenso, es fuerza es menor que en el ascenso.

8 0
3 years ago
URGENT ANSWER FAST PLEASE
xeze [42]
410.3ft^3. is correct





volume of the sphere with 3ft radius :
\frac{4  \times  \pi \times 3 \times 3 \times 3}{3}  = 113




volume of the sphere with 5ft radius:
\frac{4 \times \pi \times 5 \times 5 \times 5}{3}  = \frac{1570}{3}   = 523.3



difference
523.3 - 113 = 410.3





good luck
7 0
3 years ago
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