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3 years ago

Can you please help ?

Mathematics

1 answer:

madam [21]3 years ago

7 0

Answer:

Step-by-step explanation:

if the question is 4 * -6 ... that's four times negative 6 then the question about susans' quiz question is good

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5. Which of the following shows the equation after the variables have been collected on the left

Anestetic [448]

The answer is B 5x=15

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3 years ago

Select the correct answer

Tasya [4]

The correct answer is the 5th option (one before the last one).

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3 years ago

Given f(x) = 4x^4 find f^-1(x) Then state whether f^-1(x) is a function.

sdas [7]

Answer:

$f^{-1}(x)=\pm\sqrt[4]{\dfrac{x}{4}}$
$f^{-1}(x) \quad\text{is not a function}$

Step-by-step explanation:

To find the inverse function, solve for y:

$x=f(y)\\\\x=4y^4\\\\\dfrac{x}{4}=y^4\\\\\pm\sqrt[4]{\dfrac{x}{4}}=y\\\\f^{-1}(x)=\pm\sqrt[4]{\dfrac{x}{4}}$

f(x) is an even function, so f(-x) = f(x). Then the inverse relation is double-valued: for any given y, there can be either of two x-values that will give that result.

___

A function is single-valued. That means any given domain value maps to exactly one range value. The test of this is the "vertical line test." If a vertical line intersects the graph in more than one point, then that x-value maps to more than one y-value.

The horizontal line test is similar. It is used to determine whether a function has an inverse function. If a horizontal line intersects the graph in more than one place, the inverse relation is not a function.

Since the inverse relation for the given f(x) maps every x to two y-values, it is not a function. You can also tell this by the fact that f(x) is an even function, so does not pass the horizontal line test. When f(x) doesn't pass the horizontal line test, f^-1(x) cannot pass the vertical line test.

_____

The attached graph shows the inverse relation (called f₁(x)). It also shows a vertical line intersecting that graph in more than one place.

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3 years ago

Determine the most precise name for ABCD (parallelogram, rhombus, rectangle, or square). Explain how you determined your answer.

Sonja [21]

<h3>Answer: Rhombus</h3>

======================================================

Reason:

Let's find the distance from A to B. This is equivalent to finding the length of segment AB. I'll use the distance formula.

$A = (x_1,y_1) = (3,5) \text{ and } B = (x_2, y_2) = (7,6)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-7)^2 + (5-6)^2}\\\\d = \sqrt{(-4)^2 + (-1)^2}\\\\d = \sqrt{16 + 1}\\\\d = \sqrt{17}\\\\d \approx 4.1231\\\\$

Segment AB is exactly $\sqrt{17}$ units long, which is approximately 4.1231 units.

If you were to repeat similar steps for the other sides (BC, CD and AD) you should find that all four sides are the same length. Because of this fact, we have a rhombus.

-------------------------

Let's see if this rhombus is a square or not. We'll need to see if the adjacent sides are perpendicular. For that we'll need the slope.

Let's find the slope of AB.

$A = (x_1,y_1) = (3,5) \text{ and } B = (x_2,y_2) = (7,6)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{6 - 5}{7 - 3}\\\\m = \frac{1}{4}\\\\$

Segment AB has a slope of 1/4.

Do the same for BC

$B = (x_1,y_1) = (7,6) \text{ and } C = (x_2,y_2) = (6,2)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{2 - 6}{6 - 7}\\\\m = \frac{-4}{-1}\\\\m = 4\\\\$

Unfortunately the two slopes of 1/4 and 4 are not negative reciprocals of one another. One slope has to be negative while the other is positive, if we wanted perpendicular lines. Also recall that perpendicular slopes must multiply to -1.

We don't have perpendicular lines, so the interior angles are not 90 degrees each.

Therefore, this figure is not a rectangle and by extension it's not a square either.

The best description for this figure is a <u>rhombus</u>.