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skad [1K]
3 years ago
11

Consider the equation 5x (2x + 1) - 3(2x + 1) = 0.

Mathematics
1 answer:
kondor19780726 [428]3 years ago
7 0

Answer:

It is equivalent to 15x – 3)(2x + 1) = 0.       False

It is equivalent to (5x – 3)(2x + 1) = 0        True

Either   x=-1/2   or    5/3

Step-by-step explanation:

Consider the equation 5x (2x + 1) - 3(2x + 1) = 0.

It is equivalent to 15x – 3)(2x + 1) = 0.       False

It is equivalent to (5x – 3)(2x + 1) = 0        True

When you take pull-out (2x+1), you will be left with (5x-3)

5x (2x + 1) - 3(2x + 1) = 0.

(2x+1)(5x-3)=0

Either 2x + 1 = 0                        

2x = -1    

Divide both-side of the equation by 2

2x/2 = -1/2                                    

  x= -1/2                                        

OR

5x-3 =0

5x = 3

Divide both-side of the equation by 5

5x/5 = 3/5

x = 5/3

Either   x=-1/2   or    5/3

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Sidana [21]

Answer:

x = 17.2

Step-by-step explanation:

since they are supplementary, when you add then together they equal 180

4x + 8 + 6x = 180

Combine like terms

10x + 8 = 180

subtract 8 from both sides

10x = 172

divide each side by 10

x = 17.2

hope this helps :)

7 0
3 years ago
Jasmyn uses 42% of a gift card to buy groceries if she spent $67.20 on groceries, how much was on the gift card to begin with
lapo4ka [179]

Answer:

The card was worth $160.

Step-by-step explanation:

It is given that:

Percent of gift card used = 42%

Cost of groceries = $67.20

Let,

x be the total amount on gift card.

42% of x = 67.20

\frac{42}{100}x=67.20

0.42x = 67.20

\frac{0.42x}{0.42}=\frac{67.20}{0.42}\\x=160

Therefore,

The card was worth $160.

4 0
3 years ago
State one form of the Law of Cosines and provide a trick for writing the other two forms and explain when Law of Cosines should
Yuki888 [10]

Solving for <em>Angles</em>

\displaystyle \frac{a^2 + b^2 - c^2}{2ab} = cos∠C \\ \frac{a^2 - b^2 + c^2}{2ac} = cos∠B \\ \frac{-a^2 + b^2 + c^2}{2bc} = cos∠A

* Do not forget to use the <em>inverse</em> function towards the end, or elce you will throw your answer off!

Solving for <em>Edges</em>

\displaystyle b^2 + a^2 - 2ba\:cos∠C = c^2 \\ c^2 + a^2 - 2ca\:cos∠B = b^2 \\ c^2 + b^2 - 2cb\:cos∠A = a^2

You would use this law under <em>two</em> conditions:

  • One angle and two edges defined, while trying to solve for the <em>third edge</em>
  • ALL three edges defined

* Just make sure to use the <em>inverse</em> function towards the end, or elce you will throw your answer off!

_____________________________________________

Now, JUST IN CASE, you would use the Law of Sines under <em>three</em> conditions:

  • Two angles and one edge defined, while trying to solve for the <em>second edge</em>
  • One angle and two edges defined, while trying to solve for the <em>second angle</em>
  • ALL three angles defined [<em>of which does not occur very often, but it all refers back to the first bullet</em>]

* I HIGHLY suggest you keep note of all of this significant information. You will need it going into the future.

I am delighted to assist you at any time.

7 0
2 years ago
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asambeis [7]

Answer:

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Step-by-step explanation:

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7 0
3 years ago
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kaheart [24]

Answer: The answer is A

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7 0
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