1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
soldier1979 [14.2K]
3 years ago
11

In a rhombus MPKN with an obtuse angle K the diagonals intersect each other at point E. The measure of one of the angles of a ∆P

KE is equal 16°. Find the measures of all angles of ∆PKE and ΔPMN.
Mathematics
2 answers:
allsm [11]3 years ago
5 0
The rhombus has a couple of very interesting properties. The first is that the diagonals meet at 90o angles.

The second is that all the sides are congruent. That's actually the key to the problem (or one of them.

The third is that the diagonals are line segments that bisect the angles where the vertex of the angle is. 

So just to make sure you understand what that last statement means <PKM  =  <NKM
K is the vertex of angle PKN 

Now the really heavy duty stuff about your question.
Given
K is obtuse. Therefore <PKM can't be 16o because MKN would also have to be 16 degrees and together they don't add up to anything over 90o.
So the 16o angle is at <EPK

Remember that the diagonals meet at right angles. <PEK = 90o

Finally all triangles have 180o
< PKE = 180 - 16 - 90 = 74. So to review
<PKE = 74o
<EPK = 16o
<PEK = 90o That's one half the problem.

Moving on to triangle PMN
By the properties of parallel lines and a rhombus  and isosceles triangles, that since PKN is bisected (rhombus property) Then PKN = 2* PKM =2*74 = 148o
The angle opposite <PKN is equal to <PKN so <PMN = <PKN
Since PKN = 148 then PMN = 148
Since KPN = 16o then PMN = 16o
Since triangle <PMN is isosceles <PNM = 16o

Summing up 
PMN = 148o
MPN = 16o
MNP = 16o

That's both triangles solved. This is a really nice little problem. If you google properties of a rhombus, you will find all the properties I have used proven. 
 
lana66690 [7]3 years ago
4 0

Answer:

m∠KEP = 90°

m∠EKP = 74°

m∠EPK = 16°

m∠N PM= 16°

m∠MNP = 16

m∠M = 148°

Step-by-step explanation:

If you do RSM those are the answers

You might be interested in
Liv hizo un ponche al mezclar 2,8 litros de jugo de naranja, 0,75 litros de jugo de piña y 1,2 litros de ginger ale. Liv dividir
Juli2301 [7.4K]
Answer is 24............
5 0
3 years ago
How do i write log 4^2=16 in logarithmic form
ohaa [14]
Log16 4 = 2      
log base 16 then the number four then set that equal to 2        

7 0
3 years ago
13. f(x) = x + 5<br> a. f(4)<br> b. f(7)<br> c. f(-3)<br> d. f(0)<br> e. f(2.4)<br> f. f(2/3)
SashulF [63]
A.9
B.12
C.2
D.5
E.7.4
F.5 2/3 or 5.6666
4 0
3 years ago
I need help solving this I just can’t get the right answer
hodyreva [135]

Answer:

-1.99, 1.99

Step-by-step explanation:

Use a table to find the z-scores.

For the first z-score:

P(x < z) = 0.0233

z = -1.99

For the second z-score:

P(x > z) = 0.0233

P(x < z) = 1 − 0.0233 = 0.9767

z = 1.99

7 0
3 years ago
Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean. See Attached Excel for Da
alexgriva [62]

Answer:

The 99% confidence interval for the mean germination time is (12.3, 19.3).

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>Recorded here are the germination times (in days) for ten randomly  chosen seeds of a new type of bean: 18, 12, 20, 17, 14, 15, 13, 11, 21, 17. Assume that the population germination time is normally distributed. Find the 99% confidence interval for the mean germination time.</em>

<em />

We start calculating the sample mean M and standard deviation s:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{10}(18+12+20+17+14+15+13+11+21+17)\\\\\\M=\dfrac{158}{10}\\\\\\M=15.8\\\\\\

s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{9}((18-15.8)^2+(12-15.8)^2+(20-15.8)^2+. . . +(17-15.8)^2)}\\\\\\s=\sqrt{\dfrac{101.6}{9}}\\\\\\s=\sqrt{11.3}=3.4\\\\\\

We have to calculate a 99% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=15.8.

The sample size is N=10.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{3.4}{\sqrt{10}}=\dfrac{3.4}{3.162}=1.075

The degrees of freedom for this sample size are:

df=n-1=10-1=9

The t-value for a 99% confidence interval and 9 degrees of freedom is t=3.25.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=3.25 \cdot 1.075=3.49

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 15.8-3.49=12.3\\\\UL=M+t \cdot s_M = 15.8+3.49=19.3

The 99% confidence interval for the mean germination time is (12.3, 19.3).

8 0
3 years ago
Other questions:
  • What is number 5???!!
    10·1 answer
  • What are the solutions to the equation 2h^2+8h=64
    11·2 answers
  • Is y=2000e^-0.273x a Growth or Decay ?
    6·1 answer
  • Which of these units should be used to measure the mass of a slice of bread? A. milligram B. centigram C. gram D. kilogram
    8·2 answers
  • Students are buying tickets for rides at the fall carnival. The student council keeps track of how many tickets they sell in one
    6·1 answer
  • What is the length of a side of the cube shown with given volume?
    6·1 answer
  • When 8 is added to 3times a certain number, the result is the same as adding 12 to twice the number. find the number​
    9·1 answer
  • CAN I HAVE HELP PLEASE
    6·1 answer
  • Explain, How hypothesis test is perform using statistics procedure?
    6·1 answer
  • PLEASE HELP<br> 5.<br> If t= -3, then 3t^2 +5t +6 equals<br> A. -36<br> B. -6<br> c. 6 <br> D. 18
    13·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!