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ICE Princess25 [194]
3 years ago
12

Find the distance of a chord 8cm long from the centre of a circle radius 5cm​

Mathematics
1 answer:
KiRa [710]3 years ago
4 0

Answer:

3 cm

Step-by-step explanation:

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The slope is -10, I used a graph for this.
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Rewrite the equation 2x + 3y = 6 in slope-intercept form.
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(-6time + 3) + (5time -4)
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18+-20

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3 years ago
An object is launched from a launching pad 208 ft. above the ground at a velocity of 192ft/sec. what is the maximum height reach
Alexeev081 [22]

Answer:

h(x) = -16x² + 192x + 208

784ft

6 sec

13 sec

Step-by-step explanation:

a)

h(x) = -16x² +vx + h_{o}

here v represent velocity

         h_{o} represent initial height of launch

       

h(x) = -16x² + 192x + 208

b)

h(x) = -16x² + 192x + 208

here a = -16

        b = 192

        c = 208

x = -b/2a

  = -192/2(-16)

  = 6

plug this value in the equation

h(x) = -16(6)² + 192(6) + 208

      = 784ft

e)

Plug h(x)=0 in the equation

0 = -16x² + 192x + 208

divide equation by -16

x² - 12x - 13 = 0

Factors

1x * -13x = -13

1x - 13x = -12

Factorised form

x² - 12x - 13 = 0

x² + x - 13x - 13 = 0

x(x+1) -13(x+1) = 0

(x+1)(x-13) = 0

x = -1

x = 13

Since time can not be negative so we will reject x = -1  

3 0
3 years ago
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