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uranmaximum [27]
3 years ago
10

Solve (1/81)^x*1/243=(1/9)^−3−1 by rewriting each side with a common base.

Mathematics
1 answer:
elena55 [62]3 years ago
8 0

Answer:

x=(243)log_{\frac{1}{81}}[(\frac{1}{81})-1]

Step-by-step explanation:

you have the following formula:

(\frac{1}{81})^{\frac{x}{243}}=(\frac{1}{9})^{-3}-1

To solve this equation you use the following properties:

log_aa^x=x

Thne, by using this propwerty in the equation (1) you obtain for x

log_{(\frac{1}{81})}(\frac{1}{81})^{\frac{x}{243}}=log_{\frac{1}{81}}[(\frac{1}{81})-1]\\\\\frac{x}{243}=log_{\frac{1}{81}}[(\frac{1}{81})-1]\\\\x=(243)log_{\frac{1}{81}}[(\frac{1}{81})-1]

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The formula for the standard error is this:
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Now 2/3 of 10 :

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Select all the numbers that are common multiples of 4 and 6.
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12, 24, and 60.

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3. At noon Lan is in his blue mini-van 300 km north of Makenna in her Bugatti. Lan heads south
LiRa [457]

Answer:

The rate at which the distance between them is changing at 2:00 p.m. is approximately 1.92 km/h

Step-by-step explanation:

At noon the location of Lan = 300 km north of Makenna

Lan's direction = South

Lan's speed = 60 km/h

Makenna's direction and speed = West at 75 km/h

The distance  Lan has traveled at 2:00 PM = 2 h × 60 km/h = 120 km

The distance north between Lan and Makenna at 2:00 p.m = 300 km - 120 km = 180 km

The distance West Makenna has traveled at 2:00 p.m. = 2 h × 75 km/h = 150 km

Let 's' represent the distance between them, let 'y' represent the Lan's position north of Makenna at 2:00 p.m., and let 'x' represent Makenna's position west from Lan at 2:00 p.m.

By Pythagoras' theorem, we have;

s² = x² + y²

The distance between them at 2:00 p.m. s = √(180² + 150²) = 30·√61

ds²/dt = dx²/dt + dy²/dt

2·s·ds/dt = 2·x·dx/dt + 2·y·dy/dt

2×30·√61 × ds/dt = 2×150×75 + 2×180×(-60) = 900

ds/dt = 900/(2×30·√61) ≈ 1.92

The rate at which the distance between them is changing at 2:00 p.m. ds/dt ≈ 1.92 km/h

5 0
3 years ago
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