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Tpy6a [65]
2 years ago
12

Please answer this question

Mathematics
1 answer:
AfilCa [17]2 years ago
4 0

Answer:

it is C

Step-by-step explanation:

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Mila [183]

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I think $10 is the answer

5 0
3 years ago
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Please Help! Thanks!
svetlana [45]

Answer:

C, 20 units

Step-by-step explanation:

We see that both angles QRS and QTR are 90 degrees. In addition, angles SQR and RQT are equivalent (because they're both angle Q).

By AA Similarity, we know that triangle QTR is similar to triangle QRS.

With this similarity in mind, we can look at the ratios of corresponding lengths to set up a proportion. QR from triangle QTR is the hypotenuse, and it corresponds to hypotenuse QS from triangle QRS. So, we can write the ratio x/(9 + 16) = x/25.

Now, we see that long leg QT of triangle QTR corresponds to long leg QR of triangle QRS. So, another ratio we can write is: 16/x.

Finally, we set these two ratios equal to each other:

\frac{x}{25} =\frac{16}{x}

Cross-multiplying, we get: x^{2} =16*25=400.

Thus, x = \sqrt{400} =20. The answer is C, 20 units.

Hope this helps!

4 0
3 years ago
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A + c + 2b = a + 2b + c property
Hatshy [7]
<span>A + c + 2b = a + 2b + c 
Commutative property

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6 0
3 years ago
Please help me with this problem immediately
kari74 [83]

Answer:

Step-by-step explanation:

I don't know

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3 years ago
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Help me please. <br>Help me please. <br>Help me please.
Gnesinka [82]
A)
To be similar triangles have to have equal angles
triangle ZDB' 
1)angle Z=90 degrees
triangle B'CQ
1) angle C 90 degrees

angle A'B'Q=90
DB'Z+A'B'Q+CB'Q=180, straight angle
DB'Z+90+CB'Q=180
DB'Z+CB'Q=90

triangle ZDB'
DZB'+DB'Z=180-90=90

DB'Z+CB'Q=90
DZB'+DB'Z=90
DB'Z+CB'Q=DZB'+DB'Z
2)CB'Q=DZB' (these angles from two triangles ZDB' and B'CQ )
3)so,angles DB'Z and B'QC are going to be equal because of sum of three angles in triangles =180 degrees and 2 angles already equal.
so this triangles are similar by tree angles
b)
B'C:B'D=3:4
B'D:DZ=3:2
CQ-?
DC=AB=21
DC=B'C+B'D (3+4= 7 parts)
21/7=3
B'C=3*3=9
B'D=3*4=12
B'D:DZ=3:2
12:DZ=3:2
DZ=12*2/3=8
B'D:DZ=CQ:B'C
3:2=CQ:9
CQ=3*9/2=27/2

c)
BC=BQ+QC=B'Q+QC
BQ' can be found by pythagorean theorem



3 0
3 years ago
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