What is a product of 2 numbers

Isosceles triangle LMN is graphed with vertices L(0, 1), M(3, 5), and N(6,1). What is the slope of side LN?

Vaselesa [24]

L(0, 1)
N(6, 1)

slope formula;
m = y2 - y1/ x2 - x1

m = 1 - 1/ 6 - 1

m = 0/5

m = 0

hope this helped, God bless!

9 0

4 years ago

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What is the fourth term in the expansion of (x + 2y)^8?

Hatshy [7]

Base on my calculation and the simplification of the form and finding its 4th power of the x+2y, i came up with a solution and finally found an answer of B. 448x^5y^3, I hope i answer your questions directly and please feel free to ask for more and ask for clarifications

4 0

4 years ago

Andrew and Kate have $30 to spend on dinner, tax, and gratuity at Mo’s Restaurant. Tax is 6%, and they will give a 15% tip on th

Jlenok [28]

<span>$21 times .06 = $1.26 in taxes and $21 times .15 = $3.15 in tip so you add $21+ 1.26 + 3.15 = $25.41

</span>

7 0

3 years ago

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an arts academy requires there to be 3 teachers for every 75 students and 6 tutors for every 72 students. How many students does

Zinaida [17]

Each teacher has 25 students. 75/3 would be 25. I hope this helped!

5 0

3 years ago

Simplify u^2+3u/u^2-9<br> A.u/u-3, =/ -3, and u=/3<br> 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