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Deffense [45]
3 years ago
13

Find the value of x. If necessary, round your answer to the nearest tenth. The figure is not drawn to scale. (1 point) 9.7 9.2 9

.2 13.9 85

Mathematics
1 answer:
Vilka [71]3 years ago
4 0
X= 9.2

From the figure, the chord (12) is drawn with a line (from the center of the circle) perpendicular to it. It can be said that the chord 12 is bisected by the 7 long line from the center which gives us with 6 each of the two parts of the chord. Since they are forming (the 6 and 7) a right angled triangle, we can then use Pythagorean theorem to find X, which can be seen as the radius at the same time hypotenuse of the right angled triangle.
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Hope this isn’t too difficult, my teacher said to show you're work... so I hope you don’t mind explaining how you found it.
SVEN [57.7K]

Answer:

3/4 / 1/6

keep the first fraction the same

3/4

flip the second fraction

6/1

change the division to a multiplication it should then read

3/4 X 6/1

now multiply the fractions

3 X 6 = 18

4 X 1 = 4

18/4 this can then be turned into a mixed number.

there are four 4's in 18 with two left over. so it becomes

4 2/4 and simplified 4 1/2.

8 0
3 years ago
Look at the way that Shira answered the question above.
guapka [62]

Answer:

x = 8

Step-by-step explanation:

Her mistake was subtracting 2 on both sides. Instead she should've added 2 on both sides.Getting a result of 2x=16. Then you divide 2 on both sides to get x = 8.

4 0
4 years ago
Read 2 more answers
Find the roots of the equation<br> x ^ 2 + 3x-8 ^ -14 = 0 with three precision digits
scoray [572]

Answer:

Step-by-step explanation:

Given quadratic equation:

x^{2} + 3x - 8^{- 14} = 0

The solution of the given quadratic eqn is given by using Sri Dharacharya formula:

x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}

The above solution is for the quadratic equation of the form:

ax^{2} + bx + c = 0  

x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}

From the given eqn

a = 1

b = 3

c = - 8^{- 14}

Now, using the above values in the formula mentioned above:

x_{1, 1'} = \frac{- 3 \pm \sqrt{3^{2} - 4(1)(- 8^{- 14})}}{2(1)}

x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})})

x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})} - 3)

Now, Rationalizing the above eqn:

x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(- 8^{- 14})} - 3)\times (\frac{\sqrt{9 - 4(- 8^{- 14})} + 3}{\sqrt{9 - 4(- 8^{- 14})} + 3}

x_{1, 1'} = \frac{1}{2}.\frac{(\pm {9 - 4(- 8^{- 14})^{2}} - 3^{2})}{\sqrt{9 - 4(- 8^{- 14})} + 3}

Solving the above eqn:

x_{1, 1'} = \frac{2\times 8^{- 14}}{\sqrt{9 + 4\times 8^{-14}} + 3}

Solving with the help of caculator:

x_{1, 1'} = \frac{2\times 2.27\times 10^{- 14}}{\sqrt{9 + 42.27\times 10^{- 14}} + 3}

The precise value upto three decimal places comes out to be:

x_{1, 1'} = 0.758\times 10^{- 14}

5 0
3 years ago
-5x&lt;25 what is te answer to this equation
poizon [28]

Answer:

x > -5

Step-by-step explanation:

Sign change when you divide it with negative number

x > 25/-5

x > -5

3 0
3 years ago
Someone help me please
natali 33 [55]
Part A:
1) 16
2) -25
3) 9
4) -7

I'm too lazy to do the rest, just pay attention in class next time
5 0
3 years ago
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