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Zielflug [23.3K]
2 years ago
13

PLS HELPPPPP Find the perimeter of AXYZ. *

Mathematics
2 answers:
kifflom [539]2 years ago
7 0

Answer:

XY=2 ON mid point theorem

so XY=8

yz=2 OM

so yz= 6

since o is the mid point

xz=2 OZ

so xz=10

perimeter =8+6+10

=24 cm is the answer

hope it helps..

Sergeu [11.5K]2 years ago
5 0
Yo estoy en el hospital en el trabajo que me voy al baño a las tres para ir al baño a la casa y me hhhhhb. Yo estoy en la oficina y me voy al baño y
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Please find the exact length of the midsegment of trapezoid JKLM with vertices J(6, 10), K(10, 6), L(8, 2), and M(2, 2). Thank y
I am Lyosha [343]

Answer:

the exact length of the midsegment of trapezoid JKLM  = \mathbf{ = 3 \sqrt{5} } i.e 6.708 units on the graph

Step-by-step explanation:

From the diagram attached below; we can see a graphical representation showing the mid-segment of the trapezoid JKLM. The mid-segment is located at the line parallel to the sides of the trapezoid. However; these mid-segments are X and Y found on the line JK and LM respectively from the graph.

Using the expression for midpoints between two points to determine the exact length of the mid-segment ; we have:

\mathbf{ YX = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }

\mathbf{ YX = \sqrt{(8-5)^2+(8-2)^2} }

\mathbf{ YX = \sqrt{(3)^2+(6)^2} }

\mathbf{ YX = \sqrt{9+36} }

\mathbf{ YX = \sqrt{45} }

\mathbf{ YX = \sqrt{9*5} }

\mathbf{ YX = 3 \sqrt{5} }

Thus; the exact length of the midsegment of trapezoid JKLM  = \mathbf{ = 3 \sqrt{5} } i.e 6.708 units on the graph

8 0
2 years ago
Помогите математика !!!!!!
ohaa [14]

Answer:

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4 0
2 years ago
What is this. Please help!!
Mekhanik [1.2K]

Answer: 8x+24


Step-by-step explanation:


8 0
3 years ago
Bob needs to wash the windows on his house. He has a 25-foot ladder and places the base of the ladder 10 feet from the wall on t
Andrei [34K]

Answer:

22.9 ft

Step-by-step explanation:

the height of the wall that reached by the ladder = √ (25²-10²)

=√(625-100)

=√525

= 22.9 ft

7 0
2 years ago
Sergio works at a bakery and needs to cover
Elan Coil [88]

Answer:

The surface area that Sergio needs to frost on each cake is 3,238 square centimeters

Step-by-step explanation:

The picture of the question in the attached figure

we know that

The surface area of the cylindrical cake is equal to the lateral area plus the area of the top of cylinder (remember that the bottom of the cake does not need frosting)

so

SA=\pi r^{2}+2\pi r h

we have

r=40.2/2=20.1\ cm ---> the radius is half the diameter

h=15.6\ cm

\pi=3.14

substitute the given values

SA=(3.14)(20.1)^{2}+2(3.14)(20.1)(15.6)

SA=1,268.59+1,969.16\\SA=3,237.75\ cm^2

Round up to the  nearest whole number

SA=3,238\ cm^2

therefore

The surface area that Sergio needs to frost on each cake is 3,238 square centimeters

5 0
3 years ago
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