3 years ago

Solve 5y^2+20=80, where y is the real number

Mathematics

1 answer:

soldi70 [24.7K]3 years ago

3 0

Answer:

2√3

Step-by-step explanation:

5y2+20=80

Step 1: Subtract 20 from both sides.

5y^2+20−20=80−20

5y^2=60

Step 2: Divide both sides by 5.

5y^2 /5 = 60/5

y2=12

Step 3: Take square root.

y=√12

y=2√3

You might be interested in

To find the interquartile range you need to ____the upper and lower quartile values.

Blababa [14]

Answer:

The correct options are: Interquartile ranges are not significantly impacted by outliers. Lower and upper quartiles are needed to find the interquartile range. The data values should be listed in order before trying to find the interquartile range. The option Subtract the lowest and highest values to find the interquartile range is incorrect because the difference between lowest and highest values will give us range. The option A small interquartile range means the data is spread far away from the median is incorrect because a small interquartile means data is nor spread far away from the median

8 0

2 years ago

Read 2 more answers

Which is a correct first step in solving 5 - 2x < 8x - 3?

charle [14.2K]

Answer:

5 < 10x-3

Step-by-step explanation:

5 - 2x < 8x - 3

Add 2x to each side

5 - 2x+2x < 8x - 3+2x

5 < 10x-3