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galina1969 [7]
3 years ago
13

Two more questions , 15 points.

Mathematics
1 answer:
Mumz [18]3 years ago
7 0

The formula that is useful for solving both of these problems is ...

(v_2)^2-(v_1)^2=2ad\qquad\text{a=acceleration, d=distance}\\\\ \text{where $v_1$ and $v_2$ are initial and final velocities}

9. Given v₁=15, a=9.8, d=10, find v₂.

... (v₂)² = 15² + 2·9.8·10

... v₂ = √421 ≈ 20.5 . . . . m/s

10. Given d=12 m when a=-9.8 m/s² and v₂=0, find d when a=0.17·(-9.8 m/s²).

The formula tells us that d=(v₁)²/(2a), which is to say that the distance is inversely proportional to the acceleration. If acceleration is 0.17 times that on earth, distance will be 1/0.17 ≈ 5.88 times that on earth.

(12 m)/0.17 ≈ 70.6 m

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We conclude that the original triangle ΔDEF is dilated by a scale factor of 5/2 to get the image triangle ΔD'E'F'.

Therefore, 'Cárl' is correct. Hence, option 'a' is correct.

Step-by-step explanation:

Checking the location of the vertices of the original triangle ΔDEF

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Checking the location of the dilated vertices of the image triangle ΔD'E'F'

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Notice that if we multiply the vertices of the original triangle ΔDEF by 5/2, we get the correct image vertices of image triangleΔD'E'F'

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E(2, 4) → E'(5/2 (2), 5/2(4))  → D'(5, 10)

F(0, 2) → E'(5/2 (0), 5/2(2))  → F'(0, 5)

Thus, we conclude that the original triangle ΔDEF is dilated by a scale factor of 5/2 to get the image triangle ΔD'E'F'.

Therefore, 'Cárl' is correct. Hence, option 'a' is correct.

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