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2 years ago

Which expression has the same GCF as 15x2 - 21x?

Mathematics

2 answers:

Jlenok [28]2 years ago

7 0

Answer:First, find the greatest common factor of the coefficients of the given expression.

The greatest common factor of 15 and 21 is 3. Factor this out of the polynomial.

Next, find the greatest common factor of the variables.

The greatest common factor of x2 and x is x. Factor this out of the polynomial.

So, the greatest common factor of the expression 15x2 − 21x is 3x.

Now, consider the expression 15x2 − 21.

The greatest common factor of the coefficients 15 and 21 is 3. Only one term has a variable, so there is no common variable factor. Factor 3 out of the polynomial.

So, the greatest common factor of the expression 15x2 − 21 is 3.

Consider the expression 18x2 − 24x.

The greatest common factor of the coefficients 18 and 24 is 6, and the greatest common factor of the variables x2 and x is x. Factor these out of the polynomial.

So, the greatest common factor of the expression 18x2 − 24x is 6x.

Consider the expression 12x2 − 18x.

The greatest common factor of the coefficients 12 and 18 is 6, and the greatest common factor of the variables x2 and x is x. Factor these out of the polynomial.

So, the greatest common factor of the expression 18x2 − 24x is 6x.

Consider the expression 12x2 − 15x.

The greatest common factor of the coefficients 12 and 15 is 3, and the greatest common factor of the variables x2 and x is x. Factor these out of the polynomial.

So, the greatest common factor of the expression 12x2 − 15x is 3x.

Therefore, the expression 12x2 − 15x has the same GCF as 15x2 − 21x.

there you go

Mice21 [21]2 years ago

5 0

Answer:

$15x^2 +21 x$

Step-by-step explanation:

We have the following expression:

$15x^2 -21 x$

We can find the GCF with the following steps.

1. Find the GCF for the numerical part 15 and -21

The factors for 15 are : 1,3,5,15

The factors for -21 are -21,-7,-3,-1,1,3,7,21

And the common factors are 1,3

The GCF for the numerical part is 3*1 = 3

2. Find the GCF for the variable

The factors of x^2 are : x*x

The factors of x are: x

The common factors are x for this case, so then the GCF for the variable part is x

3. Multiply the two results from steps 1 and 2

GCf= 3*x = 3x

If we consider the new expression:

$15x^2 +21x$

We can find the GCF with the following steps.

1. Find the GCF for the numerical part 15 and -21

The factors for 15 are : 1,3,5,15

The factors for -21 are 1,3,7,21

And the common factors are 1,3

The GCF for the numerical part is 3*1 = 3

2. Find the GCF for the variable

The factors of x^2 are : x*x

The factors of x are: x

The common factors are x for this case, so then the GCF for the variable part is x

3. Multiply the two results from steps 1 and 2

GCf= 3*x = 3x

And as we can see both GCF are equal. So then we satisfy the condition required.

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Answer:

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Step-by-step explanation:

Present Value of a Growing Perpuity is calculated using the following formula

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2 years ago

Decrease £16870 by 3% , give your answer rounded 2 decimal places

lidiya [134]

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2 years ago

A random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a st

Orlov [11]

Answer:

95% confidence interval for the mean credit hours taken by a student each quarter is [14.915 hours , 15.485 hours].

Step-by-step explanation:

We are given that a random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a standard deviation of 2.3 credit hours.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample credit hours per quarter = 15.2 credit hours

s = sample standard deviation = 2.3 credit hours

n = sample of students = 250

$\mu$ = population mean credit hours per quarter

Here for constructing 95% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < $t_2_4_9$ < 1.96) = 0.95 {As the critical value of t at 249 degree of

freedom are -1.96 & 1.96 with P = 2.5%}

P(-1.96 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $15.2-1.96 \times {\frac{2.3}{\sqrt{250} } }$ , $15.2+1.96 \times {\frac{2.3}{\sqrt{250} } }$ ]

= [14.915 hours , 15.485 hours]

Therefore, 95% confidence interval for the mean credit hours taken by a student each quarter is [14.915 hours , 15.485 hours].

The interpretation of the above confidence interval is that we are 95% confident that the true mean credit hours taken by a student each quarter will be between 14.915 credit hours and 15.485 credit hours.

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