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xz_007 [3.2K]
3 years ago
13

If RX=4 and XS=9, then XT= And how do you get it?

Mathematics
1 answer:
konstantin123 [22]3 years ago
7 0

Answer:

XT=6 units

Step-by-step explanation:

The picture of the question is the attached figure

step 1

In the right triangle RST

Applying the Pythagorean theorem

RS^2=RT^2+TS^2

we have

RS=RX+XS=4+9=13\ units ---> by segment addition postulate

substitute

RT^2+TS^2=169  ----> equation A

step 2

In the right triangle RTX

Applying the Pythagorean theorem

RT^2=RX^2+XT^2

we have

RX=4\ units

substitute

RT^2=4^2+XT^2

RT^2=16+XT^2

XT^2=RT^2-16 ----> equation B

step 3

In the right triangle XTS

Applying the Pythagorean theorem

TS^2=XS^2+XT^2

we have

XS=9\ units

substitute

TS^2=9^2+XT^2

TS^2=81+XT^2

XT^2=TS^2-81 ----> equation C

step 4

equate equation B and equation C

TS^2-81=RT^2-16

TS^2-RT^2=81-16

TS^2-RT^2=65 ----> equation D

step 5

Solve the system

RT^2+TS^2=169 ----> equation A

TS^2-RT^2=65 ----> equation D

Solve by elimination

Adds equation A and equation D

RT^2+TS^2=169\\TS^2-RT^2=65\\---------\\TS^2+TS^2=169+65\\2TS^2=234\\TS^2=117

Find the value of  RT^2

RT^2+117=169\\RT^2=52

step 6

Find the value of XT

equation C

XT^2=117-81\\XT^2=36\\XT=6\ units

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2 years ago
Find the roots of h(t) = (139kt)^2 − 69t + 80
Sonbull [250]

Answer:

The positive value of k will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80, roots are those values of t so that h(t) = 0. That is:

19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0 (1)

Roots are determined analytically by the Quadratic Formula:

t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}

t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }

The smaller root is t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }, and the larger root is t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }.

h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80 has one real root when \frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0. Then, we solve the discriminant for k:

\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}

k \approx \pm 0.028

The positive value of k will result in exactly one real root is approximately 0.028.

7 0
2 years ago
Please help!!!! i will give max points i can
mixer [17]

Answer:

$9

Step-by-step explanation:

These are given:

Total = 22

Cost for shoes = 4

Total games played = 2

Cost per game (g)= ?

So,

22 = 4 + 2(g)

-4      -4

18 = 2g

divide by 2

9 = g

$9 per game

5 0
3 years ago
Read 2 more answers
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