3 years ago

Solve for X Please and thank you

Mathematics

2 answers:

Harlamova29_29 [7]3 years ago

6 0

-11<2x-1<7 First add 1 to all 3 sides.

-10<2x<8 Then divide by 2 on all 3 sides

-5<x<4 This is your answer

siniylev [52]3 years ago

5 0

Answer:

Step-by-step explanation:

-5<x<4. The 3rd answer.

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when a relation is drawn on an xy-plane what is another name for the set of all x values from its graph?

Galina-37 [17]

Another name for the set of all x-values for a relation is: domain.

<h3>What is the Domain of a Relation?</h3>

The domain of a relation can be defined as all the x-values that corresponds to the y-values that are plotted on a graph.

They are referred to as the input or the domain of the relation. Therefore, another name for the set of all x-values plotted for a relation on a graph is: domain.

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brainly.com/question/26098895

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It's a question from real and complex numbers which I can't solve. so someone PLZ HeLp

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Answer:

$\frac{1}{5}$

Step-by-step explanation:

Using the rules of exponents

$a^{m}$ × $a^{n}$ = $a^{(m+n)}$ , $\frac{a^{m} }{a^{n} }$ = $a^{(m-n)}$ , $(a^m)^{n}$ = $a^{mn}$

Simplifying the product of the first 2 terms

$\frac{a^{p^2+pq} }{a^{pq+q^2} }$ × $\frac{a^{q^2+qr} }{a^{qr+r^2} }$

= $a^{p^2-q^2}$ × $a^{q^2-r^2}$

= $a^{p^2-r^2}$

Simplifying the third term

5( $(a^p+r)^{p-r}$

= 5 $a^{(p+r)(p-r)}$ = 5 $a^{(p^2-r^2)}$

Performing the division, that is

$\frac{a^{(p^2-r^2)} }{5a^{(p^2-r^2)} }$ ← cancel $a^{(p^2-r^2)}$ on numerator/ denominator leaves

= $\frac{1}{5}$

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