I need help it has 8 questions PLEASE ASAP
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The best explanation for why the rotation is isometric is "<u>The rectangle did not change shape or size</u>". Hence the <u>second option</u> is the right choice.
There are four main categories of transformations:
- Translation (figure slides in any direction)
- Reflection (figure flips over a line)
- Rotation (figure turns about a fixed point)
- Dilation (the figure is enlarged or reduced)
A stiff transformation known as an isometry maintains perimeter and area while also preserving length and angle measurements. In other words, there is congruence between the preimage and the image. Translations, reflections, and rotations are therefore isometric, but dilations are not since the image and preimage are comparable, rather than congruent figures.
The transformation in the question will be isometric when the preimage of the rectangle before 360° rotation, will be congruent to the image after the rotation.
The congruency is best described by the option "The rectangle did not change shape or size", as that is the basis of congruency.
Thus, the best explanation for why the rotation is isometric is "<u>The rectangle did not change shape or size</u>". Hence the <u>second option</u> is the right choice.
Learn more about isometric transformations at
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The provided question is incomplete. For the complete question, refer to the attachment.
Based on the question when the ball is thrown vertically upwards by Melvin who stands at the edge of a cliff is modelled by an equation with variables t (representing the time in seconds) and h (representing the height in metres); h = t ( 8 - 5t )
a) After one second since the ball was released or in essence when t = 1, the height of the ball can be calculated by substituting for t in the equation thus solving for h.
when t = 1 ; h = ( 1 ) [ 8 - 5 ( 1 )]
h = 3
∴ at one second, the height of the ball was 3 metres
b) (i) when t = 5; h = ( 5 ) [ 8 - 5 ( 5 )]
h = - 85
b) (ii) after exactly 5 seconds after releasing, the height of the ball was -85 metres.
This suggests that the ball went upwards vertically, reached its highest point, and then fell downwards, passing the edge of the cliff where Marvin stood and goes down below the edge of the cliff (where it was released) by 85 metres, hence the - 85 metres.
c) assume that the top of the cliff is represented by h = 0; that is when the height of the ball is at the edge of the cliff, thus not above or below but at the height of 0.
when h = 0; (0) = t ( 8 - 5t)
0 = 8t - 5t²
by using the quadratic formula x = [ -b ± √(b² - 4ac) ] ÷ 2a where a = -5 b = 8 c = 0
Now t = [ -8 ± √(8² - 4(-5)(0)) ] ÷ 2(-5)
t = [ -8 ± √(64) ] ÷ (- 10)
⇒ t = (-8 + 8) ÷ (- 10) or t = (-8 - 8) ÷ (-10)
⇒ t = 0 & t = 1.6
∴ the ball has two points at which it is position is at the edge (when h = 0); this occurs when time is 0 seconds but also when time is 1.6 seconds. This coincides with part b when it was established that the path of the ball is that of a parabola with a maximum point.
a) After one second since the ball was released or in essence when t = 1, the height of the ball can be calculated by substituting for t in the equation thus solving for h.
when t = 1 ; h = ( 1 ) [ 8 - 5 ( 1 )]
h = 3
∴ at one second, the height of the ball was 3 metres
b) (i) when t = 5; h = ( 5 ) [ 8 - 5 ( 5 )]
h = - 85
b) (ii) after exactly 5 seconds after releasing, the height of the ball was -85 metres.
This suggests that the ball went upwards vertically, reached its highest point, and then fell downwards, passing the edge of the cliff where Marvin stood and goes down below the edge of the cliff (where it was released) by 85 metres, hence the - 85 metres.
c) assume that the top of the cliff is represented by h = 0; that is when the height of the ball is at the edge of the cliff, thus not above or below but at the height of 0.
when h = 0; (0) = t ( 8 - 5t)
0 = 8t - 5t²
by using the quadratic formula x = [ -b ± √(b² - 4ac) ] ÷ 2a where a = -5 b = 8 c = 0
Now t = [ -8 ± √(8² - 4(-5)(0)) ] ÷ 2(-5)
t = [ -8 ± √(64) ] ÷ (- 10)
⇒ t = (-8 + 8) ÷ (- 10) or t = (-8 - 8) ÷ (-10)
⇒ t = 0 & t = 1.6
∴ the ball has two points at which it is position is at the edge (when h = 0); this occurs when time is 0 seconds but also when time is 1.6 seconds. This coincides with part b when it was established that the path of the ball is that of a parabola with a maximum point.