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4 years ago

What is the expression in radical form?

Mathematics

1 answer:

hjlf4 years ago

8 0

Answer:

The answer is 1st point.

Step-by-step explanation:

You can seperate everything out and multiply together :

(5xy²z³)^2/5

= (5^2/5) × (x^2/5) × (y²^2/5) × (z³^2/5)

= (⁵√25) × (⁵√x²) × (⁵√y⁴) × (⁵√z⁶)

= ⁵√(25x²y⁴z⁶)

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Please help. Find the values of the variable in each figure. 17 5 11 x.

sweet-ann [11.9K]

15 i really hope this helps

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3 years ago

What’s the equation of the horizontal asymptote

Maksim231197 [3]

By inspection, the equation for the horizontal asymptote is y = -3.

6 0

3 years ago

∠1 and ∠2 are a linear pair. m∠1 = x - 29, and m∠2 = x + 61. Find the measure of each angle. A) ∠1 = 74; ∠2 = 106 B) ∠1 = 74; ∠2

Ne4ueva [31]

Answer:

Step-by-step explanation:

∠1 + ∠2 = 180 {linear pair}

x - 29 + x + 61 = 180 {add the like terms}

2x + 32 = 180 {subtract 32 from both sides}

2x + 32 - 32 = 180 - 32

2x = 148

x = 148/2

x = 74

∠1 = x - 29 = 74 - 29 = 45

∠2 = x + 61 = 74 + 61 = 135

8 0

4 years ago

3. Make a class frequency table using 2 rule of the following data. Find the Mean and make a Cumulative frequency Histogram [10]

sashaice [31]

Answer: Mean = 284

Step-by-step explanation:

Data:

125, 126, 127, 127, 137

201, 202, 207, 211, 215, 218, 224, 237, 261

301, 308, 310, 315, 330, 330, 360, 380, 393, 394

401, 401, 410, 421, 422, 424

Class Frequency Table:

$\boxed{\begin{array}{c|c|c}\underline{\text{Intervals}}&\underline{\text{\qquad Frequency\qquad}}&\underline{\text{Cumulative Frequency}}\\0-49&0&0\\50-99&0&0\\100-149&5&5\\150-199&0&5\\200-249&8&13\\250-299&1&14\\300-349&6&20\\350-399&4&24\\400-449&6&30\\450-499&0&30\end{array}}\\.\qquad \qquad \qquad \qquad \quad \ 30$

Mean:

The sum of the numbers divided by the quantity of numbers:

8518/30 = 283.933

Cumulative Frequency Histogram:

Create the graph based on the Interval and Cumulative Frequency of the class frequency table above. (see image)

3 0

4 years ago

Brian, a landscape architect, submitted a bid on each of three home landscaping projects. He estimates that the probabilities of

lesantik [10]

Answer:

a: 0.08, or 8% chance he wins all 3

b: 0.42, or 42% chance he wins 2

Step-by-step explanation:

P(win A) = 0.8

P(lose A) = 0.2

P(win B) = 0.5

P(lose B) = 0.5

P(win C) = 0.2

P(lose C) = 0.8

The situations are independent, so we multiply probabilities together.

To win all 3: P(win A)*P(win B)*P(win C) = 0.8*0.5*.02 = 0.08

To win 2 of the 3 there are 3 ways to do this. We add up the probabilities of the 3 situations...

P(win A)*P(win B)*(lose C) = 0.8*0.5*0.8 = 0.32

P(win A)*P(lose B)*P(win C) = 0.8*0.5*0.2 = 0.08

P(lose A)*P(win B)*P(win C) = 0.2*0.5*.02 = 0.02

0.32 + 0.08 + 0.02 = 0.42

5 0

3 years ago