With the information given, can you prove this quadrilateral is a parallelogram?

Solve each of these equations. Explain or show your reasoning.

defon

Answer: Answers are in the pictures

4 0

3 years ago

Does anyone know how to solve this?

erik [133]

Answer:

A point in polar coordinates is written as (R, θ)

If we want to transform this point to rectangular coordinates, we get:

x = R*cos(θ)

y = R*sin(θ)

Now we can remember that the sine and cosine functions have a period of 2*pi, then:

cos(θ) = cos(θ + 2*pi)

or:

cos(θ) = cos(θ + 2*pi + 2*pi)

and so on.

Then the point (R, θ) is the same as (R, θ + 2*pi) and (R, θ + k*(2*pi))

where k can be any integer number.

Then if we have a point in polar coordinates:

(-4, -5*π/3)

Then another two polar representations of this point are:

(-4, -5*π/3 + 2*π) = (-4, -5*π/3 + 6*π/3) = (-4, π/3)

Now we can not add 2*π (nor subtract) because we would have an angle outside the range [-2*π, 2*π]

For example, if we have:

(-4, π/3 + 2*π) = (-4, 7*π/3)

And we can not change the value of the radius and get the coordinates for the same point.

So another representation could be something like:

(-8/2, π/3)

Where i just wrote -4 in another way.

Now, a really important point.

When working with polar coordinates, we always use R as a positive number (here you can see that R is negative) so this is not the standard notation for the polar representation of a point.

8 0

3 years ago

3 2/4 pieces f cake after party eats 1 /2/3 next night how much does she have left

xz_007 [3.2K]

The person has 1 5/6 cake left because 3 2/4 - 1 2/3 = 1 5/6! Hope this helps ^0^

Explanation
Rewriting Expression in different parts.
= 3 + 2/4 + 1 + 2/3.

Solving The Whole Number Parts.
3 + 1 = 4

Solving The Fraction Parts.
2/4 + 2/3 = ?

Find the LCD of 2/3 and 2/4 and Rewrite to solve with equivalent Fractions.
LCD = 12
6/12 + 8/12 = 14/12

Simplify the Fraction Part.
14/12 = 7/6

Simplify The Fraction Part Again.
7/6 = 1 1/6

Combining The Whole Numbers With The Fractions.
4 + 1 + 1/6 = 5 1/6

Hope this helps ^0^! ;D

8 0

3 years ago

72% of ______ days is 18 days.

maks197457 [2]

Answer:

25 days

Step-by-step explanation:

lets the blank = x

Convert 72% to decimal: 0.72

0.72x = 18

x = 18/0.72

x = 25

-Chetan K

4 0

2 years ago

Could the inverse of a non-function be a function? Explain or give an example.

Kitty [74]

Answer:

The inverse of a non-function mapping is not necessarily a function.

For example, the inverse of the non-function mapping $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ is the same as itself (and thus isn't a function, either.)

Step-by-step explanation:

A mapping is a set of pairs of the form $(a,\, b)$ . The first entry of each pair is the value of the input. The second entry of the pair would be the value of the output.

A mapping is a function if and only if for each possible input value $x$ , at most one of the distinct pairs includes $x\!$ as the value of first entry.

For example, the mapping $\lbrace (0,\, 0),\, (1,\, 0) \rbrace$ is a function. However, the mapping $\lbrace (0,\, 0),\, (1,\, 0),\, (1,\, 1) \rbrace$ isn't a function since more than one of the distinct pairs in this mapping include $1$ as the value of the first entry.

The inverse of a mapping is obtained by interchanging the two entries of each of the pairs. For example, the inverse of the mapping $\lbrace (a_{1},\, b_{1}),\, (a_{2},\, b_{2})\rbrace$ is the mapping $\lbrace (b_{1},\, a_{1}),\, (b_{2},\, a_{2})\rbrace$ .

Consider mapping $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ . This mapping isn't a function since the input value $0$ is the first entry of more than one of the pairs.

Invert $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ as follows:

$(0,\, 0)$ becomes $(0,\, 0)$ .
$(0,\, 1)$ becomes $(1,\, 0)$ .
$(1,\, 0)$ becomes $(0,\, 1)$ .
$(1,\, 1)$ becomes $(1,\, 1)$ .

In other words, the inverse of the mapping $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ would be $\lbrace (0,\, 0),\, (1,\, 0),\, (0,\, 1),\, (1,\, 1) \rbrace\!$ , which is the same as the original mapping. (Mappings are sets. There is no order between entries within a mapping.)

Thus, $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ is an example of a non-function mapping that is still not a function.

More generally, the inverse of non-trivial ellipses (a class of continuous non-function $\mathbb{R} \to \mathbb{R}$ mappings, including circles) are also non-function mappings.

3 0

2 years ago