Find the area of a circle with radius of 6<br> inches. (Use 3.14 for 1.)

Find the length of the hypotenuse of a right triangle whose legs are 5 and √2.

Blizzard [7]

Answer:

$\sqrt{27}$

Step-by-step explanation:

Using Pythagoras' identity in the right triangle

The square on the hypotenuse is equal to the sum of the squares on the other 2 sides.

let h represent the hypotenuse, then

h² = 5² + ( $\sqrt{2}$ )² = 25 + 2 = 27 ( take the square root of both sides )

h = $\sqrt{27}$ ← exact value

6 0

3 years ago

Read 2 more answers

What value of x satisfies the equation below?<br> 12 ( x - 2 ) + 3 x = 1 2 ( x + 6 ) + 2

Sergeu [11.5K]

Answer:

x = 32.6666667

Step-by-step explanation:

12(x - 2) + 3x = 12(x + 6) + 2

Distribute;

12x - 24 + 3x = 12x + 72 + 2

Collect like terms;

15x - 24 = 12x + 74

Subtract 12x from both sides;

3x - 24 = 74

Add 24 to both sides;

3x = 98

Divide both sides by 3;

x = 32.6666667

4 0

3 years ago

You are given the following equation.

saul85 [17]

Answer:

Step-by-step explanation:

Given the equation 4x²+ 49y² = 196

a) Differentiating implicitly with respect to y, we have;

$8x + 98y\frac{dy}{dx} = 0\\98y\frac{dy}{dx} = -8x\\49y\frac{dy}{dx} = -4x\\\frac{dy}{dx} = \frac{-4x}{49y}$

b) To solve the equation explicitly for y and differentiate to get dy/dx in terms of x,

First let is make y the subject of the formula from the equation;

If 4x²+ 49y² = 196

49y² = 196 - 4x²

$y^{2} = \frac{196}{49} - \frac{4x^{2} }{49} \\y = \sqrt{\frac{196}{49} - \frac{4x^{2} }{49} \\} \\$

Differentiating y with respect to x using the chain rule;

Let $u= \frac{196}{49} - \frac{4x^{2} }{49}$

$y = \sqrt{u} \\y =u^{1/2} \\$

$\frac{dy}{dx} = \frac{dy}{du} * \frac{du}{dx}$

$\frac{dy}{du} = \frac{1}{2}u^{-1/2} \\$

$\frac{du}{dx} = 0 - \frac{8x}{49} \\\frac{du}{dx} =\frac{-8x}{49} \\\frac{dy}{dx} = \frac{1}{2} ( \frac{196}{49} - \frac{4x^{2} }{49})^{-1/2} * \frac{-8x}{49}\\\frac{dy}{dx} = \frac{1}{2} ( \frac{196-4x^{2} }{49})^{-1/2} * \frac{-8x}{49}\\\frac{dy}{dx} = \frac{1}{2} ( \sqrt{ \frac{49}{196-4x^{2} })} * \frac{-8x}{49}\\\frac{dy}{dx} = \frac{1}{2} *{ \frac{7}\sqrt {196-4x^{2} }} * \frac{-8x}{49}\\$

$\frac{dy}{dx} = \frac{-4x}{7\sqrt{196-4x^{2} } }$

c) From the solution of the implicit differentiation in (a)

$\frac{dy}{dx} = \frac{-4x}{49y}$

Substituting $y = \sqrt{\frac{196}{49} - \frac{4x^{2} }{49} \\$ into the equation to confirm the answer of (b) can be shown as follows

$\frac{dy}{dx} = \frac{-4x}{49\sqrt{\frac{196-4x^{2} }{49} } }\\\frac{dy}{dx} = \frac{-4x}{49\sqrt{196-4x^{2}}/7} }\\\\\frac{dy}{dx} = \frac{-4x}{7\sqrt{196-4x^{2}}}$