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Juliette [100K]
3 years ago
13

Triangles R S T and V U T are connected at point T. Angles R S T and V U T are right angles. The length of side R S is 12 and th

e length of side S T is 16. The length of side T U is 8 and the length of U V is 6. Which statements can be concluded from the diagram and used to prove that the triangles are similar by the SAS similarity theorem? StartFraction R S Over V U EndFraction = StartFraction S T Over U T EndFraction and angle S is-congruent-to angle U StartFraction R S Over V U EndFraction = StartFraction S T Over U T EndFraction = StartFraction R T Over V T EndFraction StartFraction R S Over V U EndFraction = StartFraction T U Over T S EndFraction and angle S is-congruent-to angle U StartFraction R S Over V U EndFraction = StartFraction T U Over T S EndFraction = StartFraction R T Over V T EndFraction

Mathematics
2 answers:
nydimaria [60]3 years ago
8 0

Answer:

A

Step-by-step explanation:

I took the test

il63 [147K]3 years ago
6 0

Answer:

StartFraction R S Over V U EndFraction = StartFraction S T Over U T EndFraction and angle S is-congruent-to angle U

Step-by-step explanation:

The expression below means RS/VU = ST/UT

See the attachment for better explanation.

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S
Novay_Z [31]

Step-by-step explanation:

s is inversely proportional to t

If s= 0.6 , t= 4

s=k/t

0.6= k/4

k=2.4

If s=12, then

t=k/s

t=2.4/12

t=0.05

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2 years ago
How do you find slope
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Step-by-step explanation:

Hi there!

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Formula:

(Y1-Y2)/(X1-X2)

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3 years ago
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Liula [17]

Answer:

$35.76

Step-by-step explanation:

12.60

23.10

--------

35.76

4 0
3 years ago
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The function f(x)=-(x-3)^2 +9 can be used to represent the area of a rectangle with the perimeter of 12 units, as a function of
Ivenika [448]

Maximum area of the rectangle is 9cm^{2}

<u>Explanation:</u>

<u></u>

Considering the dimensions to be in cm

f(x) = -(x-3)^{2} +9\\f(x) = -(x^{2} +9 - 6x)+9\\f(x) = -x^{2} +6x\\f'(x) = -2x+6\\-2x+6 = 0\\2x=6\\x=3cm\\\\

Putting the value of x = 3

Perimeter = 2(x+b)\\12 = 2(3+b)\\6 = 3+b\\b= 3cm

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Therefore, maximum area of the rectangle is 9cm^{2}

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