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3 years ago

5.3.4 Journal: Two-Variable Systems Elimination I need urgent help

Mathematics

1 answer:

max2010maxim [7]3 years ago

8 0

Answer:

1) X stands for individual acts and y, group acts. 2) Each scenario describes a different period in minutes, but each one respecting their different amounts (individual and group acts). 3) $S=\left \{ 5,10 \right \}$

Step-by-step explanation:

Completing with what was found:

1) Here is a summary of the scenario your classmate presented for the talent show:Main show The main show will last two hours and will include twelve individual acts and six group acts.Final show The final show will last 30 minutes and will include the top four individual acts and the top group act.The equations he came up with are: 12x+ 6y= 120, 4x+ y= 30

1. What do x and y represent in this situation?

X stands for individual acts and y, group acts.

Besides that, In the system of equation, they represent the time for x, and the time for y.

2. Do you agree that your classmate set up the equations correctly? Explain why or why not.

Yes, that's right. Each scenario describes a different period in minutes, but each one respecting their different amounts (individual and group acts). Either for 120 minutes or 30 minutes length. And their sum totalizing the whole period.

3. Solving the system by Elimination

$\left\{\begin{matrix}12x+ 6y= 120\\ 4x+ y= 30\end{matrix}\right.\\\left\{\begin{matrix}12x+ 6y= 120\\ 4x+ y= 30\:*(-3)\end{matrix}\right.\\\left\{\begin{matrix}12x+ 6y= 120\\ -12x+ -3y= -90\end{matrix}\right.\\3y=30\Rightarrow y=10\\4x+(10)=30\Rightarrow 4x=20\Rightarrow x=5\\S=\left \{ 5,10 \right \}$

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Please help

ehidna [41]

Answer:

The length between the two points is RT = 1.3 units

Step-by-step explanation:

The coordinates of the point R and T are R(2,1.2) and T(2,2.5).

Now, by DISTANCE FORMULA:

The distance between two coordinates P and Q with coordinate P(a,b) and Q(c,d) is given as:

$PQ = \sqrt{(a-c)^2 + (b-d)^2}$

So, here the distance RT = $\sqrt{(2-2)^2 + (2.5 - 1.2)^2} = \sqrt{0^2 + (1.3)^2} = 1.3$

or, RT = 1.3 units

Hence, the length between the two points is RT = 1.3 units

6 0

3 years ago

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$