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daser333 [38]
3 years ago
6

What is the approximate length of arc QR? Round to the nearest tenth of a centimeter.

Mathematics
2 answers:
NeX [460]3 years ago
8 0

Step-by-step explanation:

theta = 180 - 123 = 57 degree

arc length = 2 * pi * R * ( theta / 360)

= 2 * pi * R * ( 57 / 360)

= 0.994R cm

aleksandrvk [35]3 years ago
3 0

Answer: 19.9

Step-by-step explanation:

just took the test e2020 and it said this one:)

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42 ft

Step-by-step explanation:

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Solve for x: 2(x 3)2 − 4 = 0 Round your answer to the nearest hundredth. X = 4. 41, 1. 59 x = 1. 34, 5. 24 x = −1. 34, −5. 24 x
gladu [14]

The two values when the provided quadratic equation is solved for the x are -4.41 and -1.59 to the nearest hundredth.

<h3>What is a quadratic equation?</h3>

A quadratic equation is the equation in which the unknown variable is one and the highest power of the unknown variable is two.

The standard form of the quadratic equation is,

ax^2+bx+c=0

Here, (<em>a,b,c</em>) are the real numbers and <em>x </em>is the variable.

To find the value of x, the following formula is used,

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

The given equation is,

2(x+ 3)^2 - 4 = 0

To solve this equation, we need to apply some mathematical operations over it. Let's start with opening the brackets.

2(x+ 3)^2 - 4 = 0\\2(x^2+6x+9)-4=0\\2x^2+12x+14-4\\2x^2+12x+12=0\\x^2+6x+7=0

On comparing with standard equation we get,

a=1, b=6, c=7

Put this values in the above formula,

x=\dfrac{-(6)\pm\sqrt{(6)^2-4(1)(7)}}{2(1)}\\x=\dfrac{-(6)\pm\sqrt{(6)^2-4(1)(7)}}{2(1)}\\x=-4.41,-1.59

Hence, the two values when the provided quadratic equation is solved for the x are -4.41 and -1.59 to the nearest hundredth.

Learn more about the quadratic equation here;

brainly.com/question/1214333

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3 years ago
Please help me with number #3 and check if the others are correct please I really need help
Sveta_85 [38]
All I can say is 1 and 2 are correct.
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Step-by-step explanation:

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shutvik [7]

Answer:

????

Step-by-step explanation:

??????????????????

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