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sergeinik [125]
3 years ago
15

. .06 cm = ________ mm

Mathematics
2 answers:
Alborosie3 years ago
5 0

Answer:

  • 0.6

Step-by-step explanation:

Hope it helps! :)

german3 years ago
5 0

Answer:

0.6

Step-by-step explanation:

hope this helps

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Help please!!! I’ll mark u as brainliest
sergeinik [125]

Answer:

Answer is below

Step-by-step explanation:

Yes, the graph would still be a function, but a different function than if the values were all different. This would cause two of the values to be the same and have no slope on the line. Before, the function was quadratic, but now it would be quadratic until the two values are equal.

7 0
4 years ago
Madison earned $80 last month pet sitting in her
xeze [42]

Answer:

26

Step-by-step explanation:

Subtract

4 0
3 years ago
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HELP ME PLEASEEEEEE.........................
serious [3.7K]
The answer would be 48.
All you would have to do is multiply the height (the straight line) and the length which is the 6.
4 0
3 years ago
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Given that (8,7) is on the graph of f(x), find the corresponding point for the function f(x)+4
ozzi

Answer:

( 8,11)

Step-by-step explanation:

When x = 8 the output is 7

The new function

f(x) +4

when x = 8

The output is f(8) +4= 7+4 = 11

( 8,11)

7 0
3 years ago
Read 2 more answers
A student walk 60m on a bearing
Katena32 [7]

Answer:

d = 234.6 m

Step-by-step explanation:

You can consider a system of coordinates with its origin at the beginning of the walk of the student.

When she start to walk, she is at (0,0)m. After her first walk, her coordinates are calculated by using the information about the incline and the distance that she traveled:

x_1=60cos28\°=52.97m\\\\y_1=60sin28\°=28.16m

she is at the coordinates (52.97 , 28.16)m.

Next, when she walks 180m to the east, her coordinates are:

(52.97+180 , 28.16)m = (232.97 , 28.16)m

To calculate the distance from the final point of the student to the starting point you use the Pythagoras generalization for the distance between two points:

d=\sqrt{(x-x_o)^2+(y-y_o)^2}\\\\x=232.97\\\\x_o=0\\\\y=28.16\\\\y_o=0\\\\d=\sqrt{(232.97-0)^2+(28.16-0)^2}m=234.6m

The displacement of the student on her complete trajectory was of 234.6m

8 0
3 years ago
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