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Elena-2011 [213]
3 years ago
10

HELP PLEASEEEE!What is the domain of the function shown in the graph below? THE DOMAIN

Mathematics
1 answer:
vredina [299]3 years ago
7 0

Answer:

It looks like (-1, ∞)

Step-by-step explanation:

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In parallelogram ABCD , diagonals AC¯¯¯¯¯ and BD¯¯¯¯¯ intersect at point E, AE=2x2−3x , and CE=x2+4 .
spin [16.1K]
If the diagonals of the parallelogram intersect with each other, this means that the diagonals bisect each other. Thus,

    AE = CE

Substituting the equations,

  2x² - 3x = x² + 4

The values of x from the equation are equal to 4 and -1. 

  AE = 2(4)² - 3(4) = 20

Hence, AC is equal to 40. 

Answer: 40 units
4 0
3 years ago
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0.038 x 10 with a power of negative 2
Ghella [55]
The correct answer will be 0.00038

The easiest way to solve is if the exponent is positive you move the decimal to the right and if the exponent is negative the decimal moves to the left.

3 0
3 years ago
Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal plac
Fed [463]

Answer:

\theta=k\pi, \hspace{3}k\in Z\\\\ or\\\theta =2\pi k \pm arccos(\frac{1 }{8} ), \hspace{3}k\in Z\\

Step-by-step explanation:

Factor constant terms:

-2(-4sin(2\theta)+sin(\theta))=0

Divide both sides by -2:

sin(\theta)-4sin(2 \theta)=0

Expand trigonometric functions using the fact:

sin(2 \theta) =2 sin(\theta) cos(\theta)

So:

sin(\theta) -8sin(\theta)cos(\theta)=0

Factor sin(x) and constant terms and multiply both sides by -1:

sin(\theta) (8cos(\theta)-1)=0

Split into two equations:

(1)=8cos(\theta)-1=0\\\\(2)=sin(\theta)=0

For (1)

Add 1 to both sides and divide both sides by 8:

cos(\theta)=\frac{1}{8}

Take the inverse cosine of both sides:

\theta =2\pi k \pm arccos(\frac{1 }{8} ), \hspace{3}k\in Z

For (2)

Simply take the inverse sine of both sides

\theta = k \pi, \hspace{3}k\in Z

Therefore, the solutions are given by:

\theta=k\pi, \hspace{3}k\in Z\\\\ or\\ \theta =2\pi k \pm arccos(\frac{1 }{8} ), \hspace{3}k\in Z\\

8 0
3 years ago
How would you describe the relationship between the real zeros and x-intercepts of the function y=log4(x-2)
notsponge [240]

Answer:

Step-by-step explanation:

First, look at y = log x.  The domain is (0, infinity).  The graph never touches the vertical axis, but is always to the right of it.  A real zero occurs at x = 1, as log 1 = 0 => (1, 0).  This point is also the x-intercept of y = log x.

Then look at y = log to the base 4 of x.  The domain is (0, infinity).  The graph never touches the vertical axis, but is always to the right of it.  Again, a real zero occurs at x = 1, as log to the base 4 of 1 = 0 => (1, 0).

Finally, look at y=log to the base 4 of (x-2).  The graph is the same as that of y = log to the base 4 of x, EXCEPT that the whole graph is translated 2 units to the right.  Thus, the graph crosses the x-axis at (3, 0), which is also the x-intercept.

5 0
3 years ago
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The product of 86 and the depth of the river
aniked [119]

Answer:

Step-by-step explanation:

Are you trying to find a variable expression? the product of 86 means multiplication so 86*n or 86n. Other than that I dont understand the question.

4 0
3 years ago
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