Please help me! I'm having trouble solving this! In the △PQR, PQ = 39 in, PR = 17 in, and the height PN = 15 in. Find QR. Consid
er all possible cases.
1 answer:
Answer: 44
Step-by-step explanation:
we will find RN and NQ, then add together to give us RQ.
To find RN;
RP= 17 PN = 15 and RN =?
using pythagoras theorem,
adj^2 = hyp^2 - opp^2
RN^2 = RP^2 - PN^2
?^2 = 17^2 - 15^2
?^2 = 17^2 - 15^2
?^2 = 289 - 225
?^2 = 64
? = √64
? = 8
RN=8
To find NQ,
PN = 15 PQ=39 and NQ=?
using pythagoras theorem
NQ^2 = PQ^2 - PN^2
?^2 = 39^2 - 15^2
?^2 = 1521 - 225
?^2 = 1296
? = √1296
? = 36
NQ= 36
RQ = RN + NQ
RQ= 8 + 36
RQ=44
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Step-by-step explanation:
Thank you buddy
have a great day ahead
Slope = (y2-y1)/(x2-x1) = (8-2)/(3-1) = 6/2 = 3. So the answer is c.
Answer:
#markasbrainliest
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Hello,

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