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nordsb [41]
3 years ago
5

For f(x)=3x+1 and g(x)=x^2-6, find (f o g)(x)

Mathematics
1 answer:
Eddi Din [679]3 years ago
8 0
The answer is:  " <span>3x² − 17 " .
______________________________________</span>
f(x) = 3x + 1 ;
________________________________________________
If g(x) = x² − 6 ;

then:   (f o g)(x)  = f(x² − 6) = 3x + 1 ;

=  3(x² − 6) + 1  =  3x² − 18 + 1 ;

=  3x² − 17  .
_______________________________________________
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&lt;&gt;) Melinda
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Answer:

The statement "500 divided by 50 is 10" is a true statement

Step-by-step explanation:

In mathematics, we have a term that we refer to as Place Value.

Place Value in mathematics can be defined as the value that a digit has based on its position or place in a number.

Examples of place value is:

Thousands represented by Th

Hundreds represented by H

Tens represented by T

Units represented by U

In the above question,

500 ÷ 50 gives us 10.

This is true because, 500 as a number, the Digit 5 has a place value of hundreds

50 as a number , the digit 5 has a place values of Tens.

10 as a number, the digit 1 has a place value of Tens.

In mathematics, the multiplication of 2 numbers in a place value of Tens always gives us a place value of hundreds.

For example, 10 × 50 = 500

Likewise, when we divide, a number with a place value of hundreds by a number with a place value of tens, we have a number with a place value of tens.

For example: 500 ÷ 50 = 10.

Therefore, the statement "500 divided by 50 is 10" is a true statement

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Read 2 more answers
How do I go about solving (27x^3/8y^9)^5/3? And what is the role of the numerator and denominator?
MrRissso [65]
\left( \frac{27x^3}{8y^9}\right)^ \frac{5}{3}  \\\\\\ =\left( \frac{(3x)^3}{(2y^3)^3}\right)^ \frac{5}{3} \\\\\\ =  \frac{(3x)^{3 \times  \frac{5}{3} }}{(2y^3)^{3 \times  \frac{5}{3} }} \\\\\\ =\frac{(3x)^5}{(2y^3)^{5 }} \\\\\\ =\frac{243x^5}{32y^{15}}

Now, If the exponent was negative like you asked....

\left( \frac{27x^3}{8y^9}\right)^ {-\frac{5}{3}} \\\\\\ =\left( \frac{8y^9}{27x^3}\right)^ {\frac{5}{3}}\\\\\\ =\left( \frac{(2y^3)^3}{(3x)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(2y^3)^{3 \times \frac{5}{3} }}{(3x)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(2y^3)^{5 }}{(3x)^5} \\\\\\ =\frac{32y^{15}}{243x^5}

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Step-by-step explanation:

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