<img src="https://tex.z-dn.net/?f=%282w%20%2B%201%29%20%7B%7D%5E%7B2%7D%20" id="TexFormula1" title="(2w + 1) {}^{2} " alt="(2w +

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zubka84 [21]

2 years ago

1) {}^{2} " align="absmiddle" class="latex-formula">

Mathematics

2 answers:

Lynna [10]2 years ago

7 0

Answer:

4w^2 +4w+1

this is the product.

Step-by-step explanation:

Alex787 [66]2 years ago

3 0

Answer:

4w^2+4w+1 is very correct

Step-by-step explanation:

so I just wanted to explain more

so you expand the question like this

(2w+1)(2w+1) then

further expand by 2w(2w+1)+1(2w+1) so you use what was in the first bracket to multiply the second bracket

then it will be 2w×2w= 4w^2 that how I got that part so you multiply what's in the bracket by what is outside of it = 4w^2 +2w+2w+1 = 4w^2 +4w not 4w^2 because it's not multiplication +1

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2x + 5y - 28 -2x - 5y = -18 Elimination

lidiya [134]

Answer:

no solution I believe

Step-by-step explanation:

b/c x and y cancel out so it cant be solved

6 0

3 years ago

Mike pays $20 per hour for algebra tutoring. The equation y = 20x relates the cost of tutoring (y) to the number of hours (x) he

kotykmax [81]

The unit rate in the equation is 20.

Option C) is the correct answer.

Step-by-step explanation:

Step 1 :

In the given equation y = 20x

where,

x ⇒ total number of hours for tutoring

20 ⇒ the cost per hour of tutoring

y ⇒ the total cost for total hours of tutoring

Step 2 :

Unit rate refers to the value per hour.

Here, the cost for 1 hour of tutoring is $20.

∴ The unit rate is $20 per hour of tutoring.

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3 years ago

VEEL

Andre45 [30]

Answer:

$a_n=-3(3)^{n-1}$ ; {-3,-9, -27,- 81, -243, ...}

$a_n=-3(-3)^{n-1}$ ; {-3, 9,-27, 81, -243, ...}

$a_n=3(\frac{1}{2})^{n-1}$ ; {3, 1.5, 0.75, 0.375, 0.1875, ...}

$a_n=243(\frac{1}{3})^{n-1}$ ; {243, 81, 27, 9, 3, ...}

Step-by-step explanation:

The first explicit equation is

$a_n=-3(3)^{n-1}$

At n=1,

$a_1=-3(3)^{1-1}=-3$

At n=2,

$a_2=-3(3)^{2-1}=-9$

At n=3,

$a_3=-3(3)^{3-1}=-27$

Therefore, the geometric sequence is {-3,-9, -27,- 81, -243, ...}.

The second explicit equation is

$a_n=-3(-3)^{n-1}$

At n=1,

$a_1=-3(-3)^{1-1}=-3$

At n=2,

$a_2=-3(-3)^{2-1}=9$

At n=3,

$a_3=-3(-3)^{3-1}=-27$

Therefore, the geometric sequence is {-3, 9,-27, 81, -243, ...}.

The third explicit equation is

$a_n=3(\frac{1}{2})^{n-1}$

At n=1,

$a_1=3(\frac{1}{2})^{1-1}=3$

At n=2,

$a_2=3(\frac{1}{2})^{2-1}=1.5$

At n=3,

$a_3=3(\frac{1}{2})^{3-1}=0.75$

Therefore, the geometric sequence is {3, 1.5, 0.75, 0.375, 0.1875, ...}.

The fourth explicit equation is

$a_n=243(\frac{1}{3})^{n-1}$

At n=1,

$a_1=243(\frac{1}{3})^{1-1}=243$

At n=2,

$a_2=243(\frac{1}{3})^{2-1}=81$

At n=3,

$a_3=243(\frac{1}{3})^{3-1}=27$

Therefore, the geometric sequence is {243, 81, 27, 9, 3, ...}.

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Answer:

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Step-by-step explanation:

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