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My name is Ann [436]
3 years ago
5

Triangle J K L is shown. Angle J K L is a right angle. An altitude is drawn from point K to point M on side J L, forming a right

angle.
Which similarity statements are true? Check all that apply.


△JKL ~ △KML

△JKM ~ △JKL

△JKM ~ △KML

△JMK ~ △JKL

△JMK ~ △KML

Mathematics
2 answers:
Murljashka [212]3 years ago
5 0

Answer:

yah its 1, 4, and 5. Best of luck on the assignment.

Step-by-step explanation:

<h3 />
vredina [299]3 years ago
4 0

Answer:

1, 4, 5

Step-by-step explanation:

Consider right triangle JKL with right angle JKL and height KM. This height divides triangle into two right triangles KML and KMJ.

Note that

m\angle MKL=90^{\circ}-m\angle MLK=m\angle KJM\\ \\m\angle JKM=90^{\circ}-m\angle KJM=m\angle MLK

So, by AA theorem, we have three similar triangles

\triangle JKL\sim \triangle JMK\sim \triangle KML

So, options 1, 4 and 5 are true

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A school has 63% percent girls and 37%, percent boys. If 23%, percent of the girls wears contacts and 42%, percent of the boys w
Degger [83]

Answer:

30% students

Step-by-step explanation:

Let x be the number of total students of a school.

63% are girls then total number of girls = (0.63)x

37% are boys then total number of boys = ( 0.37)x

Now 23% of girls wear contacts then total number of girls

= (0.63x) × 0.23

=  (0.1449x)

And 42% of boys wear contacts then total number of boys wearing contacts

= (0.37x) ( 0.42)

= (0.1554)x

Total students wears contacts = ( 0.1449x) + (0.1554x)

                                                  = (0.3003x)

Percentage of students wear contacts = \frac{\text{total students wear contacts}}{\text{total students in school}} × 100

= \frac{0.03003x}{x} × 100

= 30%

All 30% students wear contacts.

4 0
3 years ago
Find m\angle Cm∠Cm, angle, C.<br> Round to the nearest degree.<br><br> Help Course Challenge
astra-53 [7]
Hello,

1. Since Angle C has the longest side for this triangle, it will have the largest degree value.

2. Use the Law of Cosines and inverse properties of “theta” to solve for Angle C. (Ensure that the calculator used is in “degree mode”, not “radian mode”.

c^2 = a^2 + b^2 - 2(a)(b)(cos (C))
15^2 = 11^2 + 14^2 - 2(11)(14)(cos(C))
225 - 317 = -2(11)(14)(cos(C))
-92 / -2(11)(14) = cos(C)
cos(C) becomes ->> cos^-1[92 /-2(11)(14)] = 72.62° ->> to the nearest degree is 73°

The answer for angle C, 73°, is logical because the triangle in the picture represents a 60-60-60 triangle, known as an equilateral triangle.

Good luck to you!
5 0
3 years ago
Read 2 more answers
① 5J+2S =12.20 ② 5J+ 10S =15.80 elimination method
hram777 [196]

Answer:

<h2>S = 9/20</h2><h2>J = 113/50</h2>

Step-by-step explanation:

5J +2S =12.20 ........(1)\\5J+10S=15.80......(2)\\Multiply \:equation \:(1)\:by\:the\:coefficient\:of\:x\:in equation\:(2)\\\\Multiply\:equation(2)\: by\: the\: coefficient\:of \:x\: in \:equation (1)\\\\5J +2S =12.20 ........(1) \times 5\\5J+10S=15.80......(2)\times 5\\\\25J+10S=61......(3)\\25J+50S = 79......(4)\\Subtract\:equation\:(4)\:from\: equation\: (3)\\-40S =-18\\\frac{-40S}{-40}=\frac{-18}{-40}\\  S = 9/20\\

Substitute \:9/20\:for \:x \:in\:equation\:(1)\:or \:equation\:(2)\\5J+2S = 12.20\\5J +2(9/20) =12.20\\5J +9/10=12.20\\5J=12.20-9/10\\5J=113/10\\Cross-Multiply\\50J = 113\\J = 113/50

4 0
3 years ago
Need help with 2 math questions pls help me for 20 points.
s344n2d4d5 [400]
The answers are A and B
7 0
3 years ago
Find 3 consecutive odd integers with a sum of 63. PLEASE HELP!!!​
mamaluj [8]

Answer:

the numbers are 20, 21 and 22

Step-by-step explanation:

x+(x+1) + (x+2) = 63

3x +3 = 63

3x = 60

x = 20

doing this for ponits

4 0
3 years ago
Read 2 more answers
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