Please help me with this

Mathematics

1 answer:

Gelneren [198K]3 years ago

4 0

9514 1404 393

Answer:

MN = 68°
LM = 25°
KLN = 180°
JLN = 248°
JN = 112°
NJL = 267°

Step-by-step explanation:

The relevant relationships are ...

the sum of arcs around a circle is 360°
a diameter divides the circle into two 180° arcs
the whole is the sum of the parts
the measure of an arc is the measure of the corresponding central angle

The arc measures can be found to be ...

MN = 68°

LM = 180° -68° -87° = 25° . . . . semicircle KLN less the marked angles

KLN = 180° . . . . a semicircle

JLN = JM +MN = 180° +68° = 248° . . . . semicircle JLM + MN

JN = 180° -68° = 112° . . . . semicircle JNM less MN

NJL = 180° +87° = 267° . . . . semicircle NJK + KL

You might be interested in

BRAINLIESTT ASAP! PLEASE HELP ME :)

grandymaker [24]

Answer:

yes the triangles are congruent

Step-by-step explanation:

hypotenuse-angle

3 0

3 years ago

A friend asks you to think of a number, double it, and then add 4. Write an algebraic expression using the variable x to describ

stich3 [128]

Answer:

Step-by-step explanation:

x= the number

2x+4

3= x

2(3)+4=

6+4= 10

3 0

3 years ago

Read 2 more answers

. (06.02)

lora16 [44]

Answers:

1.) C - Final exams: midterm = Q3 92 - Q1 80 = 12 and finals = Q3 85 - Q1 78 = 7, so finals exam has the smaller IQR.

2.) B - exam median is much higher than the class median: refer to the attached image of class and exam box and whisker graph

3.) A - IQR is a better measure of spread for movies than it is for basketball games: because the data for movies are quite wide or apart from each other than the data in the basketball games

4.) A - mean for April is higher than October's mean: mean in April is 67 and mean in October is 60

5.) A - Neither data set has suspected outliers: meaning there's no data that is further apart from the group or set

6.) B - There is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical

7.) C - college spread is best described by the IQR. The high school spread is best described by the standard deviation: this is because of the wide range of data in the college than the data in the high school.