0Claudia0
Answered

At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

~please only answer if you know for sure~
-Thank you-
(Specify which one you are answering btw)

Please Only Answer If You Know For Sure Thank You Specify Which One You Are Answering Btw class=
Please Only Answer If You Know For Sure Thank You Specify Which One You Are Answering Btw class=

Sagot :

I can only see the similar triangle problem, with sides 6cm, 9cm, and 15cm. Since the triangles are similar, the ratio for side lengths is the same. 6:9 simplifies to 2:3. If we try to solve ?:15 we get 10 as 10:15 is equivalent to 2:3, so the length is 10 cm
TSO
JKL is similar to PQR

so they want PR 

JKL is similar to PQR

So PR is similar to JL

And they also give us the sides KL and QR

JKL is similar to PQR

Both of those sides are similar, so we can form a proportion.

[tex]\sf { \frac{PR}{JL} = \frac{QR}{KL} }[/tex]

Plug in the numbers:

[tex]\sf { \frac{PR}{6} = \frac{15}{9} }[/tex]

Cross multiply:

[tex]\sf{ PR \times 9 = 6 \times 15}[/tex]

Isolate PR

[tex]\sf{ PR = \frac{6 \times 15}{9}}[/tex]

Simplify it

[tex]\sf{ PR = 10}[/tex]

So your final answer is

[tex]\boxed{\bf{10 centimeters}}[/tex]
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.