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3 years ago

What is f(x)=x on a graphing table

Mathematics

1 answer:

labwork [276]3 years ago

4 0

Answer:

On a graphing table, f(x) is y.

Step-by-step explanation:

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pishuonlain [190]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

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lesantik [10]

Answer:

First answer.

Step-by-step explanation:

This is called the Arc length, here is the formula:

Arc length = (angle/360)*(2Pi*radius)

So,

(45/360)*(2*3.14*4)

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3 years ago

3. Would it be more efficient to use an inequality vo

myrzilka [38]

Answer:

Inequality

Step-by-step explanation:

There are an infinite number of quantities less than 6 (5, 4, 3, 2, 1, 0, -1, -2, -3...) so it is much easier to write an inequality

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3 years ago

MOON The expression gives the weight of an object on the Moon in pounds with a weight of w pounds on Earth. What is the weight o

dybincka [34]

Answer: 29.7 pounds

Step-by-step explanation:

The expression in question is w/6.

If the weight of an object on earth is w, the weight of the object on the moon is w/2.

The weight of the space suit is therefore:

= 178.2/6

= 29.7 pounds

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3 years ago

Graph and label the image of the figure below after a dilation by a factor of 1/2.

neonofarm [45]

Answer:

M' (1.5, -1), F' (2, -1), L' (0.5 -2.5), W' (2.5, -2.5)

see graph below

Explanation:

Given:

The image of a quadrilateral on a coordinate plane

To find:

The coordinates of the new image after dilation of 1/2 have been applied to the original image.

Then graph the coordinates

First, we need to state the coordinates of the original image:

M = (3, -2)

F = (4, -2)

L = (1, -5)

W = (5, -5)

Next, we will apply a scale factor of 1/2:

$\begin{gathered} Dilation\text{ rule:} \\ (x,\text{ y\rparen}\rightarrow(kx,\text{ ky\rparen} \\ where\text{ k = scale factor} \\ \\ scale\text{ factor = 1/2} \\ M^{\prime}\text{ = \lparen}\frac{1}{2}(3),\text{ }\frac{1}{2}(-2)) \\ M^{\prime}\text{ = \lparen}\frac{3}{2},\text{ -1\rparen} \\ \\ F\text{ = \lparen}\frac{1}{2}(4),\text{ }\frac{1}{2}(-2)) \\ F^{\prime}\text{ = \lparen2, -1\rparen} \end{gathered}$ $\begin{gathered} L\text{ = \lparen}\frac{1}{2}(1),\text{ }\frac{1}{2}(-5)) \\ L^{\prime}\text{ = \lparen}\frac{1}{2},\text{ }\frac{-5}{2}) \\ \\ W\text{ = \lparen}\frac{1}{2}(5),\text{ }\frac{1}{2}(-5)) \\ W^{\prime}\text{ = \lparen}\frac{5}{2},\text{ }\frac{-5}{2}) \end{gathered}$

The new coordinates:

M' (3/2, -1), F' (2, -1), L' (1/2, -5/2), W' (5/2, -5/2)

M' (1.5, -1), F' (2, -1), L' (0.5 -2.5), W' (2.5, -2.5)

Plotting the coordinates:

3 0

1 year ago