Convert 8.2 cm to m.

Can you form a triangle with these given side lengths? 5 , 5 , 5 PLEASE HELP ME!!!

I am Lyosha [343]

Answer:

8 0

3 years ago

Find the value of p . (√n)to the power p equal to n

Allushta [10]

Answer:

p=2

$\sqrt{x} * \sqrt{x} = x$

$\sqrt{n} *\sqrt{n} = n$

$\sqrt{n} ^{2} = n$

$\sqrt{n} ^{p} = n$

SO.... P = 2

Step-by-step explanation:

4 0

3 years ago

Can someone pls help me with this?

hoa [83]

Answer: It would be A because the base cost is three dollars with extra 150 per order

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Determine the domain and range of the graph.

nirvana33 [79]

Answer:

Domain [-4,4]

Range [-2,2]

Step-by-step explanation:

The domain is the x-values of the graph and the range in the y-values. When writing domain and range it should be from least to greatest. So to find the domain find the lowest x-value on the graph and then the highest. Next, do the same for y-values. Finally, either surround each value with parentheses or bracket, the difference is that brackets mean that value is included, while parentheses mean that value is not actually on the graph.

In this case, the lowest x-value is -4 and the highest is 4, both values are included as signified by the closed circles, therefore the domain is [-4,4]. The lowest y value is -2 and the highest is 2, both are included, therefore the range is [-2,2].

4 0

3 years ago

Simplify u^2+3u/u^2-9 A.u/u-3, =/ -3, and u=/3 B. u/u-3, u=/-3

VashaNatasha [74]

  The correct answer is: Answer choice: [A]:
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→  " $\frac{u}{u-3}$ " ; " { u $\neq$ ± 3 } " ;

  →  or, write as: " u / (u − 3) " ;  {" u ≠ 3 "}  AND: {" u ≠ -3 "} ;
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Explanation:
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We are asked to simplify:

   $\frac{(u^2+3u)}{(u^2-9)}$ ;

Note that the "numerator" —which is: "(u² + 3u)" — can be factored into:
  → " u(u + 3) " ;

And that the "denominator" —which is: "(u² − 9)" — can be factored into:
  → "(u − 3) (u + 3)" ;
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Let us rewrite as:
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→ $\frac{u(u+3)}{(u-3)(u+3)}$ ;

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→ We can simplify by "canceling out" BOTH the "(u + 3)" values; in BOTH the "numerator" AND the "denominator" ;  since:

" $\frac{(u+3)}{(u+3)} = 1$ " ;

→ And we have:
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→  " $\frac{u}{u-3}$ " ; that is: " u / (u − 3) " ; { u $\neq 3$ } .
and: { u $\neq-3$ } .

→ which is:  "Answer choice: [A] " .
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NOTE:  The "denominator" cannot equal "0" ; since one cannot "divide by "0" ;

and if the denominator is "(u − 3)" ; the denominator equals "0" when "u = -3" ; as such:

"u $\neq$ 3" ;

→ Note: To solve: "u + 3 = 0" ;

Subtract "3" from each side of the equation;

→ " u + 3 − 3 = 0 − 3 " ;

→ u = -3 (when the "denominator" equals "0") ;

→ As such: " u $\neq$ -3 " ;

Furthermore, consider the initial (unsimplified) given expression:

→   $\frac{(u^2+3u)}{(u^2-9)}$ ;

Note:  The denominator is: "(u²  − 9)" .

The "denominator" cannot be "0" ; because one cannot "divide" by "0" ;

As such, solve for values of "u" when the "denominator" equals "0" ; that is:
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→ " u² − 9 = 0 " ;

→ Add "9" to each side of the equation ;

→ u² − 9 + 9 = 0 + 9 ;

→ u² = 9 ;

Take the square root of each side of the equation;
to isolate "u" on one side of the equation; & to solve for ALL VALUES of "u" ;

→ √(u²) = √9 ;

→ | u | = 3 ;

→ " u = 3" ; AND; "u = -3 " ;

We already have: "u = -3" (a value at which the "denominator equals "0") ;

We now have "u = 3" ; as a value at which the "denominator equals "0");

→ As such: " u $\neq 3$ " ; "u $\neq$ -3 " ;

or, write as: " { u $\neq$ ± 3 } " .

_________________________________________________________

6 0

3 years ago