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ivanzaharov [21]
2 years ago
10

Which could be the function graphed below?

Mathematics
2 answers:
stiv31 [10]2 years ago
3 0

The parent equation of the given graph is y=\sqrt[n]{x}

where, n is any even number.

When this function is vertically shifted ( vertical dilation) down in the y- axis,

we get a graph similar to the one given in the question.

For example: y=\sqrt[2]{x}-2

The graph of the above function is attached below.


Fed [463]2 years ago
3 0
Try this way:
f(x)= \sqrt[n]{x} -a,
where n - even number, a - const.
P.S. one example is in the attachment.

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2 years ago
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Which terms could be used as the first term of the expression below to create a polynomial written in standard form? Select five
kondaur [170]

Answer:

The last two terms of the expression are

  ----+8r^2s^4-3r^3s^3

Both the last terms has variable of degree equal to (2+4=6) and (3+3=6).So, the first term must have degree greater than 6.

Correct Options are

 1.\rightarrow 3r^4s^5\\\\2.\rightarrow -r^4s^6

3 0
2 years ago
Which statement is true regarding the functions on the graph?
Mrrafil [7]
Answer: f(3)=g(3)
I believe that is correct.
Tell me if I’m wrong
3 0
2 years ago
A region R in the xy-plane is given. Find equations for a transformation T that maps a rectangular region S in the uv-plane onto
Gre4nikov [31]
\begin{cases}y=2x-2\\y=2x+2\end{cases}\implies\begin{cases}-2x+y=-2\\-2x+y=2\end{cases}

For these lines, let u=-2x+y.

\begin{cases}y=2-x\\y=4-x\end{cases}\implies\begin{cases}x+y=2\\x+y=4\end{cases}

And for these, let v=x+y.

Now,

\begin{cases}u=-2x+y\\v=x+y\end{cases}\implies \begin{bmatrix}u\\v\end{bmatrix}=\underbrace{\begin{bmatrix}-2&1\\1&1\end{bmatrix}}_{\mathbf T}\begin{bmatrix}x\\y\end{bmatrix}

The vertices of S in the x-y plane are (0, 2), (2/3, 10/3), (2, 2), and (4/3, 2/3). Applying \mathbf T to each of these yields, respectively, (2, 2), (2, 4), (-2, 4), and (-2, 2), which are the vertices of a rectangle whose sides are parallel to the u-v plane.
6 0
2 years ago
Find the real numbers x and y if -3+ix^2y and x^2+y+4i are conjugate of each other. Pls solve with the steps
Firdavs [7]
ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work

EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same

For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y

For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4

Therefore, for the two expressions to be conjugates, we must satisfy the two conditions. 

Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the 

   x²y = -4 ... (I)

Condition 2: Real parts are the same

   x² + y = -3 ... (II)

We have a system of equations since both conditions must be satisfied

   x²y = -4 ... (I)
   x² + y = -3 ... (II)

We can rearrange equation (II) so that we have

   y = -3 - x² ... (II)

Substituting into equation (I)

   x²y = -4 ... (I)
   x²(-3 - x²) = -4
   -3x² - x⁴ = -4
   x⁴ + 3x² - 4 = 0
   (x² + 4)(x² - 1) = 0
   (x² + 4)(x-1)(x+1) = 0

Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.

Solve for y:

   y = -3 - x² ... (II)
   y = -3 - (±1)²
   y = -3 - 1
   y = -4

So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:

   -3 + ix²y 
   = -3 + i(±1)²(-4)
   = -3 - 4i

   x² + y + 4i
   = (±1)² - 4 + 4i
   = 1 - 4 + 4i
   = -3 + 4i

They result in conjugates
4 0
2 years ago
Read 2 more answers
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