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scoundrel [369]
3 years ago
15

Which expression is equivalent to 9x^5y^16/45x^5y^4

Mathematics
2 answers:
katrin2010 [14]3 years ago
5 0

Answer:

An equivalent expression to the given expression  \frac{9x^5y^{16}}{45x^5y^4} is \frac{y^{12}}{5}

Step-by-step explanation:

Given :  Expression \frac{9x^5y^{16}}{45x^5y^4}

We have to find an equivalent expression to the given expression  \frac{9x^5y^{16}}{45x^5y^4}

Consider the given expression  \frac{9x^5y^{16}}{45x^5y^4}

Cancel the common factor 9, we have,

=\frac{x^5y^{16}}{5x^5y^4}

Cancel the common factor x^5, we have,

=\frac{y^{16}}{5y^4}

\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}\:=\:x^{a-b}

We have, \frac{y^{16}}{y^4}=y^{16-4}=y^{12}

Thus, we get,

=\frac{y^{12}}{5}

Thus, An equivalent expression to the given expression  \frac{9x^5y^{16}}{45x^5y^4} is \frac{y^{12}}{5}

Pie3 years ago
4 0
Not sure i know what equivalent but i know the answer which is y^12/5.
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Two consecutive odd integers have a sum of 32. Find the integers.
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Let
x = first consecutive odd
x + 2 = second consecutive odd

Based on the problem, we equate
x + (x + 2) = 32
Solving for x,
2x + 2 = 32
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2x = 30
x = 30/2
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Therefore, the integers are 15 and 17.
6 0
4 years ago
3 more than 6 times a number
Snowcat [4.5K]

Answer:

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4 0
3 years ago
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Mike is three years older than his wife Karen.
kkurt [141]

Answer:

1. Karen= g^5 (not sure)
2. Mike = g^5 + 3 (not sure)

3. Karen = 25

Mike = 28

4. Karen = 35

Mike = 38

Step-by-step explanation:

1. if george age was g

karen age= g x 5 = g5

2. mike age= gx5 + 3

3. g = 5

karen age= 5 x 5 = 25

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8 0
2 years ago
Scores on an exam follow an approximately Normal distribution with a mean of 76.4 and a standard deviation of 6.1 points. What p
klasskru [66]

Answer:

99.89% of students scored below 95 points.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 76.4, \sigma = 6.1

What percent of students scored below 95 points?

This is the pvalue of Z when X = 95. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{95 - 76.4}{6.1}

Z = 3.05

Z = 3.05 has a pvalue of 0.9989.

99.89% of students scored below 95 points.

5 0
3 years ago
Need help on the middle question
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3 years ago
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