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Ludmilka [50]
3 years ago
5

Triangle ABC has vertices at A(3, 8) , B(11, 8) , and C(9, 12) . Triangle FGH has vertices at F(1, 3) , G(9, 3) , and H(7, 7) .

Which sequence of transformations shows that triangle ABC and triangle FGH are congruent? Select each correct answer. Translate ​ triangle ABC ​ up 5 units then translate ​ triangle ABC ​ right 2 units. Translate ​ triangle ABC ​ down 5 units then translate ​ triangle ABC ​ left 2 units. Translate ​ triangle FGH ​ up 5 units then translate ​ triangle FGH ​ right 2 units. ​ Translate ​ triangle FGH ​ down 5 units then translate ​ triangle FGH ​ left 2 units. ​
Mathematics
2 answers:
Alex17521 [72]3 years ago
6 0
Hmm, (x,y)
x is horizontal or left right
y is vertical or up down


so

we see that A to F is move left 2 and down 5
B to G is left 2 and down 5
C to H is left 2 and down 5


so translatete ABC left 2 and down 5 and
translate FGH right 2 and up 5

not 1st option
2nd works
3rd works
4th is false


2nd and 3rd
Andreas93 [3]3 years ago
4 0
Translate triangle ABC down 5 units then left 2 units 
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Segment wx is shown Explain how you would construct a perpendicular bisector of wx using a compass and a straightedge
n200080 [17]

Answer:

To perpendicular bisector of line segment WX. There are following steps:

1) Draw arcs or circles from points A and B on the both sides of WX.

2) Name the intersection points as W and X.

3) Use the straightedge to draw a line through points W and X.

4) Name the point as O

hence we have construct perpendicular bisector AB of WX which bisects at O.

4 0
3 years ago
The sum of the diagonals of a rhombus is 5√2.
Alex
Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC
= \frac{1}{2} BD (AO + OC)
\frac{1}{2} BD \times AC

So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}
BD \times AC = 8
Now, AC + \: BD = 5 \sqrt{2}

Squaring both sides, we get
AC {}^{2} + BD {}^{2} + 2 AC.BD =50
AC {}^{2} + BD {}^{2} + 2 \times 8 = 50
AC {}^{2} + BD {}^{2} = 50 - 16
AC {}^{2} + BD {}^{2} = 34

In △AOB, we have
OA {}^{2} + OB {}^{2} =AB {}^{2}
( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}
\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}
AC {}^{2} + BD {}^{2} = 4 AB {}^{2}
34 = 4 AB {}^{2}

Square rooting both sides
\sqrt{34} = 2 AB
Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units.

Hope it helps!

7 0
3 years ago
May I please have help with 2-14 even on this ? I'm horrible at math.
Vilka [71]
2. x = 1/4 ( 0.25)
4. y = 122, x = 42  
6. x = 63, y = 8
8. is missing
10. 13.5 cm
12. x = 10.6
14. x = 5.45
6 0
3 years ago
a fish is 12 m below the surface of the ocean what is the elevation of sea bird is 28 m above the surface of the ocean what is t
ExtremeBDS [4]

Answer:

40m???

Step-by-step explanation:

7 0
3 years ago
What is the perimeter of the actual garden?
Lilit [14]

Answer: 576 inches or 48 feet

Step-by-step explanation:

set up a proportion where \frac{drawing AB}{actual AB} =\frac{drawing perimeter}{actual perimeter}

so you get drawingAB = 24 in

actualAB = 18 feet = 18 * 12 in = 216 in

drawing perimeter = 64 in

so plug these in to get\frac{24}{216} =\frac{64}{actual perimeter}

rearrange to get actual perimeter = \frac{216 * 64}{24}

and solve to get 576 inches or 576/12 = 48 feet

4 0
3 years ago
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