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Ann [662]
4 years ago
10

7: The point B (-8,5) is dilated from the origin with a scale factor of 12 . ​What are the coordinates of B'? A (−6,3) B (−4,2.5

) C (−16,10) D (−10,7)
Mathematics
1 answer:
SSSSS [86.1K]4 years ago
6 0

Answer:

The correct option is option B.

The coordinate of B' is (-4,2.5).

Step-by-step explanation:

Formula of dilation:

Image = (pre-image)×(scale factor)

Scale factor:

The amount by which the shrinks or grows is called " scale factor".

  • If the scale factor less than 1 then the shape shrinks.
  • If the sale factor equal to 1 then the shape remains same.
  • If the scale factor greater than 1 then the shape enlarged.

Given point is B(-8,5)

Scale factor is \frac12.

The co-ordinate of image B is (-8\times \frac12,5\times \frac12)

                                                 =(-4,2.5)

The coordinate of B' is (-4,2.5).

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puteri [66]

Answer:

A) t = 1 second to reach the highest point

B) h = 96 feet

C) t = 3.449 seconds to hit the ground

Step-by-step explanation:

We notice that the equation that describes the position of the object in terms of the time, is a parabola (quadratic) with negative leading coefficient. therefore this is a parabola with arms pointing down, and with vertex at the top. It is at that top vertex that the maximum altitude of the object is reached.

We need therefore to find the position of the vertex (time "t" is the horizontal coordinate and height "h" is the vertical coordinate).

We know that the vertex (maximum of a parabola of this type - normally described by y=ax^2+bx+c) is given by x_{vertex} = -\frac{b}{2*a}. Therefore in our case, understanding that time "t" is equivalent to the variable "x", and eight "h" is equivalent to the variable "y", that "-16" is a, and "32" is b, we have that the maximum occurs for:

t_{max}=-\frac{32}{2*(-16)} = \frac{-32}{-32} = 1

which means at one second from the launching.

We then can find the answer to part B by replacing t with "1" second in the formula for the height "h":

h(1)=-16*(1)^2+32(1)+80=-16+32+80=96

That is : a height of 96 feet.

To find the answer for part C: the time to hit the ground, we solve for the variable "t" in the given expression for the height, by setting the height to zero (object touching the ground) and using the quadratic formula:

0=-16t^2+32t+80\\t=\frac{-32+/-\sqrt{32^2-4*(-16)*80} }{2*(-16)} \\ t=\frac{-32+/-\sqrt{6144} }{-32}

which gives us two solutions: t = -1.449 seconds, and t = 3.449 seconds

Since the negative time does not have physical meaning in our case (time before the object was launched), we adopt the second answer: t = 3.449 seconds

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