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3 years ago

Solve for 1/3(8x-5)-4=-3

Mathematics

2 answers:

Tems11 [23]3 years ago

8 0

Answer: x = 1

Step-by-step explanation

irakobra [83]3 years ago

5 0

Let's solve your equation step-by-step.

13(8x−5)−4=−3

Step 1: Simplify both sides of the equation.

13(8x−5)−4=−3

(13)(8x)+(13)(−5)+−4=−3 (Distribute)

83x+−53+−4=−3

(83x)+(−53+−4)=−3 (Combine Like Terms)

83x+−173=−3

Step 2: Add 17/3 to both sides.

83x+−173+173=−3+173

83x=83

Step 3: Multiply both sides by 3/8.

(38)*(83x)=(38)*(83)

X = 1

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

max2010maxim [7]

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:

<em> 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</em>

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

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

8 0

3 years ago

What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

$(2 x+y)^{5}$

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}$

$$+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}$

Let us solve the term one by one.

$$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}$

$$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y$

$$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}$

$$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}$

$$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}$

$$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}$

Substitute these into the above expansion.

$(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}$

The coefficient of $x^{2} y^{3}$ is 40.

Option C is the correct answer.

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3 years ago

a boat is bringing a diver up at 25 feet per minute, the diver is currently at 200 feet, where was he 10 minutes ago ???

Sidana [21]

Answer:he was at -50 feet

Step-by-step explanation:

If he is going 25 feet per minute then you do 25x10 winch is 250 to see how far they have been total then since hes at 200 feet now you subtract 250 from 200 and you get your answer

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3 years ago

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Arlecino [84]

Answer:

linear and non-proportional

Step-by-step explanation:

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Tomtit [17]

Answer:

Verified

Step-by-step explanation:

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