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Citrus2011 [14]
2 years ago
6

What value of x makes the equation true? 3(4x-5)-4x+1=-6 Please show work

Mathematics
1 answer:
V125BC [204]2 years ago
3 0

Answer:

x = 1

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

Step-by-step explanation:

<u>Step 1: Define Equation</u>

3(4x - 5) - 4x + 1 = -6

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Distribute 3:                    12x - 15 - 4x + 1 = -6
  2. Combine like terms:       8x - 14 = -6
  3. Isolate <em>x</em> term:                 8x = 8
  4. Isolate <em>x</em>:                          x = 1

<u>Step 3: Check</u>

<em>Plug in x into the original equation to verify it's a solution.</em>

  1. Substitute in <em>x</em>:                    3(4(1) - 5) - 4(1) + 1 = -6
  2. Multiply:                               3(4 - 5) - 4 + 1 = -6
  3. Subtract:                              3(-1) - 4 + 1 = -6
  4. Multiply:                               -3 - 4 + 1 = -6
  5. Subtract:                              -7 + 1 = -6
  6. Add:                                     -6 = -6

Here we see that -6 does indeed equal -6.

∴ x = 1 is the solution to the equation.

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Determine whetherfis a function from the set of all bitstrings to the set of integers if
ipn [44]

Answer:

a) Not a function

b) It is a function

c) It is a function

Step-by-step explanation:

a) f(S) is not completly well defined, becuase if the bistring has multiple zeros, you need to pick one and there is not specified method in how you would do so. Another reason for which f is not well defined as a function is that for a bistring with only 1's, f is not defined because there is no zero.

b) This is a function. f(S) returns a concrete and unique integer number for each bistring S. If the bistring has only zeros, the f(S) = 0.

c) This is well defined as a function. It compensates for the failures of the first definition:

- If multiple zeros appears in S, you pick the first one, so there is one possible value for f(S)

-If no zeros appear, f is still well defined because it takes the value 0, as specified in the definition.

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3 years ago
What do 110°, 250°, and -470° angles have in common?
Nadusha1986 [10]
They are all larger than 90 degrees, and in a triangle they would all be obtuse angles.
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3 years ago
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4k – 3 = -2k + 3 help me please lol
hoa [83]

Answer: k=1

Step-by-step explanation:

4k-3=-2k+3

take -3 and 3 and add them up on one side and change the sign

4k=-2k+3

           +3

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take the -2k and change the sign and add to 4k

4k=6

2k

6k=6

now divide both sides with x

6k/6= 6/6

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PLEASE HELP QUICKLY 25 POINTS
Natalija [7]

Answer:

○ \displaystyle \pi

Step-by-step explanation:

\displaystyle \boxed{y = 3sin\:(2x + \frac{\pi}{2})} \\ y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{-\frac{\pi}{4}} \hookrightarrow \frac{-\frac{\pi}{2}}{2} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\pi} \hookrightarrow \frac{2}{2}\pi \\ Amplitude \hookrightarrow 3

<em>OR</em>

\displaystyle \boxed{y = 3cos\:2x} \\ y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\pi} \hookrightarrow \frac{2}{2}\pi \\ Amplitude \hookrightarrow 3

You will need the above information to help you interpret the graph. First off, keep in mind that although this looks EXACTLY like the cosine graph, if you plan on writing your equation as a function of <em>sine</em>, then there WILL be a horisontal shift, meaning that a C-term will be involved. As you can see, the photograph on the right displays the trigonometric graph of \displaystyle y = 3sin\:2x,in which you need to replase "cosine" with "sine", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the <em>sine</em> graph [photograph on the left], accourding to the horisontal shift formula above. Also keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY <em>REALLY</em> ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the <em>sine</em> graph [photograph on the right] is shifted \displaystyle \frac{\pi}{4}\:unitto the right, which means that in order to match the <em>cosine</em> graph [photograph on the left], we need to shift the graph BACKWARD \displaystyle \frac{\pi}{4}\:unit,which means the C-term will be negative, and by perfourming your calculations, you will arrive at \displaystyle \boxed{-\frac{\pi}{4}} = \frac{-\frac{\pi}{2}}{2}.So, the sine graph of the cosine graph, accourding to the horisontal shift, is \displaystyle y = 3sin\:(2x + \frac{\pi}{2}).Now, with all that being said, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph WILL hit \displaystyle [-1\frac{3}{4}\pi, 0],from there to \displaystyle [-\frac{3}{4}\pi, 0],they are obviously \displaystyle \pi\:unitsapart, telling you that the period of the graph is \displaystyle \pi.Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the <em>midline</em>. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at \displaystyle y = 0,in which each crest is extended <em>three units</em> beyond the midline, hence, your amplitude. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

I am delighted to assist you at any time.

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Is 4-heap Nim with heaps of sizes 22, 19, 14, and 11 balanced or unbalanced? Player I's first move is to remove 6 coins from the
nadya68 [22]

Answer:

Player II should remove 14 coins from the heap of size 22.

Step-by-step explanation:

To properly answer this this question, we need to understand the principle and what it is exactly is being asked.

This question revolves round a game of Nim

What is a game of Nim: This is a strategic mathematical game whereby,  two opposing sides or opponent take turns taking away objects from a load or piles. On each turn, a player remove at least an object and may remove any number of objects provided they all come from the same heap/pile.

Now, referring back to the question, we should first understand that:

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                         19₂= 1 0 0 1 1

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and also that the “bit sums” are all even, so this is a balanced game.

However, after Player I removes 6 coins from the heap of size 19, Player II should remove 14 coins from the heap of size 22.

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