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Nataliya [291]
3 years ago
11

I just need help with this table

Mathematics
1 answer:
Mashutka [201]3 years ago
6 0

Answer:

from the looks of it, all you have to do is, for f(x), is plug it in as an exponent. in order (top to bottom), it should be: 64, 2048, 4096, 8192.

g(x) is being squared and then multiplied, so it should be (from top to bottom): 720, 2420, 2880, 3380

Step-by-step explanation:

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Danielle can run one mile in 8.835 minutes Alex can run 1 mile and 9.75 minutes how much longer in minutes does it take for Alex
hichkok12 [17]

Answer:

0.915 minutes

Step-by-step explanation:

First, take the 9.75 minutes that it takes Alex and the 8.835 minutes it takes Danielle.

Second, line up the numbers annexing zeros as placeholders as needed.

Third, subtract them

9.750

-8.835

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0.915

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3 years ago
May someone help me plz? Thank you!
Rama09 [41]

Answer:

The answer is C!

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URGENT!!! PLEASE HELP
Molodets [167]

Answer: System 1 has an infinite number of solutions and System II has no solutions.

Step-by-step explanation:

If we look at system 1, we see that if we multiply the first equation by 2 and subtract the y value so that its on the same side as x, the equations are the same. That means that no matter the x or y value, it will be a solution.

If we look at system 2, however, we see that if we multiply the top equation by 3, the equations are the exact same except for the fact that the top equation has a -21. This means that no matter the x or y, there will never be a solution.

3 0
3 years ago
Solve this A tourist starts off from town A and travels for 50km on a bearing of N80°W to town B.At town B ,he continues for ano
VLD [36.1K]

Answer:

The point C is 12.68 km away from the point A on a bearing of S23.23°W.

Step-by-step explanation:

Given that AB is 50 km and BC is 40 km as shown in the figure.

From the figure, the length of x-component of AC = |AB sin 80° - BC cos 20°|

=|50 sin 80° - 40 cos 20°|=11.65 km

The length of y-component of AC = |AB cos 80° - BC sin 20°|

=|50 cos 80° - 40 sin 20°|= 5 km

tan\theta= 5/11.65

\theta=23.23°

AC= \sqrt{5^2+11.65^2}=12.68 km

Hence, the point C is 12.68 km away from the point A on a bearing of S23.23°W.

8 0
2 years ago
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