Help me with this please

Determine whether each of the following pairs of angles have equal measures.

Hatshy [7]

The pairs of angles that have equal measures are:

LJP
NJR
MJP
PJR

The pairs of angles that do not have equal measures are:

MJK
PJR
KJR
MJP
KJL
LJM

<h3>What is an angle?</h3>

When two lines or rays converge at the same point, the measurement between them is called a "Angle."

It is given that:

Both MJK and PJR can be seen as (Unequal angles)

The angle MJK is ( 90 degrees), then angle PJR will be ( 48 + 46 )= 94 degrees

Both LJP and NJR can be seen as ( Equal angles)

The measure of angle LJP = 48 + 46 + 48 = 142 degrees

The measure of angle NJR = 48 + 48 + 46 = 142 degrees

KJR & MJP = Unequal

The angle KJR will now be [ 360 - (46+48+48+46+90) ]

= 82 degrees

The angle MJP will be ( 94 degrees)

Both KJL & LJM are Unequal, then

The angle KJL will be 42 degrees, then angle LJM will be ( 90 - 42 )

= 48 degrees

Read more about pairs of angles here:

brainly.com/question/17643033

#SPJ1

CHECK THE COMPLETE QUESTION BELOW:

Determine whether each of the following pairs of angles have equal measures. Drag and drop each pair

of angles into the correct catagory to show which pairs have equal measures.

MO

N

46°

42°

48°

46°

R

ZMJK and LPUR

ZLJP and ZNJR

ZKJR and ZMJP

MJP and ZPJR

ZKIL and ZUM

Equal Measures

Unequal Measures

4 0

2 years ago

A dvd case is 9/16 thick. how many of these cases will fit on a shelf that is 1 1/2 ft wide.

SVETLANKA909090 [29]

11/2 * 16/9

11/1 * 8/9

88/9

5 0

3 years ago

Help me it is below

Fofino [41]

I hope this helps you

7 0

3 years ago

Read 2 more answers

Lim (n/3n-1)^(n-1) n → ∞

n200080 [17]

Looks like the given limit is

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}$

With some simple algebra, we can rewrite

$\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)$

then distribute the limit over the product,

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}$

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, e :

$\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n$

To make our limit resemble this one more closely, make a substitution; replace 9/(n - 9) with 1/m, so that

$\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1$

From the relation 9m = n - 9, we see that m also approaches infinity as n approaches infinity. So, the second limit is rewritten as

$\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}$

Now we apply some more properties of multiplication and limits:

$\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9$

So, the overall limit is indeed 0:

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}$

7 0

3 years ago

Given that pentagon ABCDE ≅ pentagon FGHIJ, find the value of m.

I am Lyosha [343]

Thank you for posting your question here at brainly. Feel free to ask more questions. The best and most correct answer among the choices provided by the question is A. 2 . 

Hope my answer would be a great help for you.

7 0

3 years ago