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tester [92]
2 years ago
5

1. ADEA has vertices D(8,4), E(2, 6), and F(3, 1). What are the vertices of the image after a dilation with a scale factor of 5

using the origin as the center of dilation? TO​
Mathematics
1 answer:
lubasha [3.4K]2 years ago
3 0

Answer:

The vertices of the image are D' (40, 20), E' (10, 30), F' (15, 5)

Step-by-step explanation:

If the point (x, y) dilated by a scale factor k using the origin as a center of dilation, then its image is (kx, ky)

Let us use this rule to solve our question

∵ The vertices of ΔDEF are D (8, 4), E (2, 6), and F (3, 1)

∵ ΔDEF dilated by a scale factor 5

∴ k = 5

∵ The origin as the center of dilation

→ That means, multiply the coordinates of every point by 5

∴ D' = (8 × 5, 4 × 5)

∴ D' = (40, 20)

∴ E' = (2 × 5, 6 × 5)

∴ E' = (10, 30)

∴ F' = (3 × 5, 1 × 5)

∴ D' = (15, 5)

∴ The vertices of the image are D' (40, 20), E' (10, 30), F' (15, 5)

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Step-by-step explanation:

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\frac{dT}{dt}=-k(T-T_{s})\\\frac{dT}{(T-T_{s})}=-k*dt\\\int\limits {\frac{1}{T-T_{s} } } \, dT=-\int\limits {k} \, dt\\ln(T-T_{s})=k*t+C\\e*ln(T-T_{s})=e^{k*t+C} \\T-T_{s}=e^{k*t}*e^{C}\\e^{C}=C\\T-T_{s}=e^{k*t}*C\\T=T_{s}+e^{k*t}*C

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145=66+124*e^{-k*t} \\145-66=124*^{-0.087*t}

ln(0.637)=ln*e(-0.087*t)\\-0.45=-0.087*t\\t=5.18minutes

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