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Sergio039 [100]
3 years ago
7

Select ALL the correct answers. Given that line AB is parallel to line DE, determine how triangle ABC and triangle EDC can be sh

own to be similar. Since CD ≅ CE and ∠BAC ≅ ∠DEC, the triangles are similar by the angle-side criterion. Since ∠ABC ≅ ∠EDC and AC ≅ BC, the triangles are similar by the angle-side criterion. Since ∠ABC ≅ ∠EDC and ∠BAC ≅ ∠DEC, the triangles are similar by the angle-angle criterion. Since ∠ABC ≅ ∠EDC and ∠ACB ≅ ∠ECD, the triangles are similar by the angle-angle criterion. Since ∠ACB ≅ ∠ECD and ∠BAC ≅ ∠DEC, the triangles are similar by the angle-angle criterion. Since ∠ABC ≅ ∠DEC and ∠BAC ≅ ∠EDC, the triangles are similar by the angle-angle criterion.
Mathematics
1 answer:
Evgen [1.6K]3 years ago
8 0

cheeseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

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Find the area for the following figure
Shtirlitz [24]

Area of triangle = 1/2 (base) (height)

base = 20.3 cm and height = 6.2 cm

Plug in

A = 1/2 (20.3)(6.2)

A = 62.93

Answer

62.93 cm^2

4 0
4 years ago
Which of the following statements about photocopier A1 is correct compared to the other photocopiers?
Sati [7]

Answer:

Option 2

Step-by-step explanation:

Number of breakdowns of the photocopiers in year 2011,

A1 = 15

A2 = 21

B1 = 34

B2 = 32

B3 = 23

Option (1)

A1 had just under a half the number of breakdowns compared to photocopier B3.

Half of B3 = 12

And A1 = 15

Therefore, A1 > B3

So the statement is False.

Option (2)

A1 had just under 50% of the breakdowns compared to photocopier B2

50% of the breakdowns of B2 = 16

A1 = 15

So, A1 < 50% of B2

Statement (2) is True.

Option (3)

A1 had more than 25 extra breakdowns compared to B1.

A1= 15

B1 = 34

A1 < B1

Therefore, statement is False.

Option (4)

A1 had 6% less breakdowns than photocopier A2.

A1 = 15

A2 = 21

A2 - A1 = 21 - 15 = 6

Percentage less breakdowns = \frac{6}{21}\times 100 = 28.6%

So this Option is False.

Therefore, Option (2) is the correct option.

3 0
3 years ago
Plz Help (Not really tho). Maath stuff.
Pani-rosa [81]

Answer:

I don't know the answer

7 0
4 years ago
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An ellipse has a co-vertex at (–8, 9) and a foci at (4, 4). If the center of the ellipse is located below the given co-vertex, t
vovikov84 [41]

Answer:

Step-by-step explanation:

“the center of the ellipse is located below the given co-vertex”

Co-vertex and center are vertically aligned, so the ellipse is horizontal.

Equation for horizontal ellipse:

(x-h)²/a² + (y-k)²/b² = 1

with

a² ≥ b²

center (h,k)

vertices (h±a, k)

co-vertices (h, k±b)

foci (h±c,k), c² = a² -b²

One co-vertex is (-8,9), so h = -8.

One focus is (4,4), so k = 4.

Center (h,k) = (-8,4)

c = distance between center and focus = |-8 - 4| = 12

b = |9-k| = 5

a² = c² + b² = 169

(x+8)²/169 + (y-4)²/25 = 1

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3 years ago
How do you simplify 4(5+x)
alexira [117]

Answer:

20+4x

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4*5+4*x=

20+4x.

*Distributive law upon addition.*

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