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faltersainse [42]
2 years ago
5

PLS HELP IM FAILING MATH!

Mathematics
1 answer:
oksano4ka [1.4K]2 years ago
7 0

Answer:

1. Never true

2. Never true

3. Bad question!  Typo? (see discussion below)

4. Always true

5. Always false

Step-by-step explanation:

1. x - 13 = x + 1.  Subtract  x  from both sides of the equation, and you get -13 = 1.  False!  (Never true!)

2.  x + 1/2 = x - 1/2.  Subtract x from both sides:  1/2 = - 1/2.  False! (Never true!)

3. This one's weird, with two equal signs.  That's not unusual; multiple equal signs are often used when a chain of equalities is being displayed.  For instance, you can write  (3)(2x + 5) = (3)(2x) + (3)(5) = 6x + 15.  That's an example of the Distributive Property--the 3 is multiplied by both of the terms inside parentheses (shown after the first equal sign), then actually doing those multiplications produces the expression after the second equal sign.

OK!  This question says 2(x + 3) = 5x = 6 - 3x.  Take it in two parts.

a. 2(x + 3) = 5x   Distribute the 2.

2x + 6 = 5x   Now subtract 2x from both sides.

6 = 3x   Divide both sides by 3.

2 = x  So that first part is true only when x = 2.  Gaa!  That's not one of your answer choices.  Now, check out the second part

b.  5x = 6 - 3x  Add 3x to both sides.

8x = 6   Divide by 8

x = 6/8 = 3/4

It doesn't help to use the first and <u>third</u> expressions, either.

2(x + 3) = 6 - 3x   Distribute the 2

2x + 6 = 6 - 3x  Add 3x.

5x + 6 = 6   Subtract 6.

5x = 0   Divide by 5

x = 0

None of these is always false.  None of these is always true.

4.  x - 3 = 2x - 3 - x   Simplify the right side by combining the x terms.

x - 3 = x - 3  Aha!  Same expression on both sides!  Always true

5.  3(x - 5) = 2(x - 5) + x  Distribute the 3. Distribute the 2.

3x - 15 = 2x - 10 + x  Combine the x terms on the right.

3x - 15 = 3x - 10  Subtract 3x

-15 = -10  (Always false!)

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Answer:

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Step-by-step explanation:

By the double angle identity of sines:

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Rewrite the original equation with this identity:

2\, (2\, \sin x \cdot \cos x) - 2\, \sin x + 2\sqrt{3}\, \cos x - \sqrt{3} = 0.

Note, that 2\, (2\, \sin x \cdot \cos x) and (-2\, \sin x) share the common factor (2\, \sin x). On the other hand, 2\sqrt{3}\, \cos x and (-\sqrt{3}) share the common factor \sqrt[3}. Combine these terms pairwise using the two common factors:

(2\, \sin x) \cdot (2\, \cos x - 1) + \left(\sqrt{3}\right)\, (2\, \cos x - 1) = 0.

Note the new common factor (2\, \cos x - 1). Therefore:

\left(2\, \sin x + \sqrt{3}\right) \cdot (2\, \cos x - 1) = 0.

This equation holds as long as either \left(2\, \sin x + \sqrt{3}\right) or (2\, \cos x - 1) is zero. Let k be an integer. Accordingly:

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Any x that fits into at least one of these patterns will satisfy the equation. These pattern can be further combined:

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7 0
3 years ago
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podryga [215]

Answer:

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Step-by-step explanation:

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