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3 years ago

Can somebody please double-check these for me.

Mathematics

1 answer:

Sav [38]3 years ago

5 0

1. is a central angle, therefore, the arc will have the same measurement as the angle. Set the equation:

Arc = Central Angle

2x - 7 = 47

Isolate the variable, x. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.

First, add 7 to both sides of the equation:

2x - 7 (+7) = 47 (+7)

2x = 47 + 7

2x = 54

Next, divide 2 from both sides of the equation:

(2x)/2 = (54)/2

x = 54/2 = 27

27 is your answer.

2. is a inscribed angle, meaning that the angle measurement will be half of the arc. Set the equation:

212 = 2(13x - 24)

Isolate the variable, x. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.

First, divide 2 from both sides of the equation:

(212)/2 = (2(13x - 24))/2

106 = 13x - 24

Next, add 24 to both sides of the equation:

106 (+24) = 13x - 24 (+24)

106 + 24 = 13x

130 = 13x

Finally, divide 13 from both sides of the equation:

(130)/13 = (13x)/13

x = 130/13

x = 10

10 is your answer.

3. is a central angle. The arc and the angle will, therefore, have the same measurement:

137 = 3x + 5

First, subtract 5 from both sides of the equation:

137 (-5) = 3x + 5 (-5)

137 - 5 = 3x

132 = 3x

Next, divide 3 from both sides of the equation to isolate the variable, x:

(132)/3 = (3x)/3

x = 132/3 = 44

44 is your answer.

4. is a inscribed angle. The arc will be twice the measurement of the angle.

86 = 2(2x + 3)

First, isolate the variable x by dividing 2 from both sides of the equation.

(86)/2 = (2(2x + 3)/2

43 = 2x + 3

Next, subtract 3 from both sides of the equation:

43 (-3) = 2x + 3 (-3)

40 = 2x

Finally, divide 2 from both sides of the equation:

(40)/2 = (2x)/2

x = 40/2 = 20

20 is your answer.

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3 years ago

Y=sqrt(x)(8x-5) find the derivative

garri49 [273]

Answer:

$\displaystyle y' = \frac{24x - 5}{2\sqrt{x}}$

General Formulas and Concepts:

Algebra I

Exponentials [Fractions] - Are radicals
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Product Rule: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \sqrt{x}(8x - 5)$

Step 2: Differentiate

$\displaystyle f(x) = \sqrt{x}, \ g(x) = (8x - 5)$

Product Rule: $\displaystyle y' = \frac{d}{dx}[\sqrt{x}] \cdot (8x - 5) + \sqrt{x} \cdot \frac{d}{dx}[(8x - 5)]$
Rewrite: $\displaystyle y' = \frac{d}{dx}[x^{\frac{1}{2}}] \cdot (8x - 5) + \sqrt{x} \cdot \frac{d}{dx}[(8x - 5)]$
Basic Power Rule: $\displaystyle y' = \frac{1}{2}x^{\frac{1}{2} - 1} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8x^{1 - 1}$
Simplify: $\displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8x^{0}$
Rewrite: $\displaystyle y' = \frac{1}{2x^{\frac{1}{2}}} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8$
Multiply: $\displaystyle y' = \frac{8x + 5}{2x^{\frac{1}{2}}} + 8\sqrt{x}$
Rewrite: $\displaystyle y' = \frac{8x + 5}{2\sqrt{x}} + 8\sqrt{x}$
Add/Rewrite: $\displaystyle y' = \frac{24x - 5}{2\sqrt{x}}$