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Alika [10]
3 years ago
13

Find the length of segment DE.

Mathematics
1 answer:
Dovator [93]3 years ago
6 0

Answer:

segments forming a square at the vertex

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Find the missing information. <br><br> PLEASE HELP ITS A QUIZ DUE SOON
xxTIMURxx [149]

Answer:

x = 13

∠F = 28°

∠S = 28°

y = 9

DE = 38 ft

Step-by-step explanation:

We are told that △DEF = △QRS

Looking at both triangles and by comparison, it is clear that;

∠R = ∠E = 123°

∠Q = ∠D = 29°

∠S = ∠F = 180 - (123 + 29) = 28°

Also,QR = DE

Thus; 5y - 7 = 38

5y = 38 + 7

5y = 45

y = 45/5

y = 9

Thus,y = 9 and DE = 38 ft

We see in the image that ∠S = (2x + 2)°

From earlier we deduced that ∠S = 28°

Thus; 2x + 2 = 28

2x = 28 - 2

2x = 26

x = 26/2

x = 13

7 0
3 years ago
In △BFD, m∠B = 34 ° and m∠FDC = 76°. What is m∠F and m∠FDB?
Paul [167]

Answer:

f=42

fdb=104

Step-by-step explanation:

7 0
3 years ago
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Solve for x (geometry)
vitfil [10]
I got 3.75. not sure if it’s right but hope i helped
6 0
3 years ago
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Please help me with this!
marishachu [46]
It is this photo if it has been added

5 0
3 years ago
Determine the volume of the parallelepiped with one vertex at the origin and the three vertices adjacent to it at (2, â1, â1), (
valentinak56 [21]

Answer:

23

Step-by-step explanation:

Here is the complete question

Find the volume of the parallelepiped with one vertex at the origin and adjacent vertices at (1, 0, -3), (1, 2, 4), and (5, 1, 0).

Solution

We find the volume of the parallelepiped by making a 3 × 3 column matrix whose columns are the corresponding coordinates of the vertices of the parallelepiped.

So, (1, 0, -3), (1, 2, 4)  and (5, 1, 0)

A = \left[\begin{array}{ccc}1&1&5\\0&2&1\\-3&4&0\end{array}\right]

The determinant of A is the volume of the parallelepiped. So,

detA = 1(2 × 0 - 4 × 1) - 1(0 × 0 - (-3) × 1) + 5(0 × 4 - (-3) × 2)

= 1(0 - 4) - 1(0 + 3) + 5(0 + 6)

= 1(-4) - 1(3) + 5(6)

= -4 - 3 + 30

= 23

So the volume of the parallelepiped is 23

4 0
3 years ago
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