Help (pic attached!!)

Mathematics

1 answer:

mash [69]3 years ago

7 0

$( \sqrt[5]{ 3^{2} } )^{ \frac{1}{3} } = (3^{ \frac{2}{5} } ) ^{ \frac{1}{3} } =$ $3^{ \frac{2}{15} }$

Answer: C.

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5. If your friend's doctor wants her target heart rate range to be between 143 and 170, and her

nikdorinn [45]

Answer:

a person can work out 50% of the maximum (50% of 194 is 97) and 85% of the maximum (85% of 194 is 164.9). This means that the target heart rate for a 20-year-old is approximately 97–165 bpm. so around 21.

Step-by-step explanation:

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3 years ago

Easy points for those who are good at math!

zalisa [80]

Answer:

The answer is A) y = -7x - 7.

Step-by-step explanation:

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2 years ago

11. Given: TQ bisects ZRTP, QS || PT RT = 30. Find QP, RS and QS

yaroslaw [1]

QP=24 cm

RS=11.25 cm

QS=18.75 cm

Explanation:

Given that TQ bisects <RTP

$therefore$ (1)

consider ΔRQS and ΔRPT

QS||PT,RP and RT are transversals

(alternate angles)(2)

comparing (1) and (2)

and triangle SQT is isocelus

Therefore SQ=ST(sides opposite to equal angles in an isocelus triangle)

Therefore <RQS=<RPT(corresponding angles)

<RSQ=<RTP(corresponding angles)

therefore by AA criterion for similarity

ΔRQS~ΔRPT

According to the property of similar triangles

$RQ/RP=RS/RT=QS/PT$ $9/RP=X/30=QS/50\\9/RP=X/30=(30-X)/50\\X/30=(30-X)/50\\50X=30(30-X)\\50X=900-30X\\50X+30X=900\\80X=900\\X=900/80=11.25\\RS=11.25 cm\\QS=30-X=30-11.25\\QS=18.75 cm\\9/RP=X/30\\9/RP=11.25/30\\9*30/11.25=RP\\RP=24 cm$