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ivolga24 [154]
3 years ago
10

Find < SPT in degrees

Mathematics
2 answers:
Jobisdone [24]3 years ago
5 0

If angle QR is 180° and arc QR is 10z, set them equal to find the value of z

180 = 10z; divide both sides by 10

18 = z

Then use this value in 6z to find the arc SPT

6(18) = SPT

108 = SPT

Since this is a central angle, the arc length and angle measurement are the same. Therefore angle SPT = 108°

(Also, hey XD)

kupik [55]3 years ago
3 0

Answer: 108deg

From picture:

10z=pi and m<SPT =6z So we know 20z=2pi or 360deg

Let’s solve:

20z=2pi

z=pi/10

m<SPT=6z=6pi/10=3pi/5 now lets take the angle from radians to degrees by unit conversion:

3pi/5*(180deg/pi)=108deg

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Step-by-step explanation:

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Horizontal:

If we have a function like this

\frac{1}{x}

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Remember a constant can be represent by

a \times  {x}^{0}

For example,

1 = 1 \times  {x}^{0}

2 =  2 \times {x}^{0}

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and

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\frac{1 \times  {x}^{0} }{ {x}^{1} }

Look at the degrees, since the numerator has a smaller degree than the denominator, the denominator will grow larger than the numerator as x gets larger, so since the larger number is the denominator, our y values will approach 0.

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Consider the function

\frac{3 {x}^{2} }{ {x}^{2}  + 1}

As x get larger, the only thing that will matter will be the leading coefficient of the leading degree term. So as x approach infinity and negative infinity, the horizontal asymptote will the numerator of the leading coefficient/ the leading coefficient of the denominator

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x =  \frac{3}{1}

Finally, if the numerator has a greater degree than denominator, the value of horizontal asymptote will be larger and larger such there would be no horizontal asymptote instead of a oblique asymptote.

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2 years ago
What is 4.2 kilometers in meters
vovikov84 [41]
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4 0
3 years ago
X+y+z =4
cestrela7 [59]
<span><span> <span>Akar akar persamaan kuadrat 2x² - 3x -1 = 0 adalah x1 dan x2. Persamaan kuadrat baru yang akar akarnya satu lebih kecil dari dua kali akar akar persamaan kuadrat di atas adalah ........</span></span><span><span><span>A.x² - x - 4 = 0</span><span>B.x² + 5x - 4 = 0</span><span>C.x² - x + 4 = 0</span></span><span><span>D.x² + x + 4 = 0</span><span>E.x² - 5x - 4 = 0</span></span></span><span>Jawaban : A
Penyelesaian : 
Akar-akar persamaan lama :  x1 dan x2
                                             
Akar-akar persamaan baru :  xA dan xB
                                             xA = 2x1 - 1
                                             xB = 2x2 - 1
                                             xA + xB  = (2x1 - 1) + (2x2 - 1)
                                                           = 2 (x1 + x2) - 2
                                                           = 2 () - 2
                                                           = 3 - 2
                                              xA + xB = 1
    
                                              xA . xB = (2x1 - 1) (2x2 - 1)
                                                          = 4 x1.x2 - 2(x1 + x2) + 1
                                                          = 4.(-) - 2() + 1
                                                          = -2 - 3 + 1
                                              xA . xB = -4
    
Jadi persamaan kuadrat baru : x² - (xA + xB)x + xA . xB = 0
                                              x² - x - 4 = 0

</span></span>
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