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Kisachek [45]
3 years ago
8

Factor 6 z^3/3 + 1 z^1/3

Mathematics
1 answer:
balandron [24]3 years ago
7 0

Use Rule of One: x^1 = x

6 × x^3 + 1 × z/3

Simplify 6 × z^3/3 to 2z^3

2z^3 + 1 × z/3

Simplify 1 × z/3 to z/3

<u>= 2z^3 + z/3</u>

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1 year ago
2. Consider the circle x^2−4x+y^2+10y+13=0. There are two lines tangent to this circle having a slope of 2/3.
likoan [24]

Answer:

a) The coordinates are (0.431, -2.646) and (3.568,-7.354)

b) The tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

Step-by-step explanation:

First, lets complete squares, by taking for each cordinate the square of a linear expression

0= x²−4x+y²+10y+13 = (x-2)²+ 4  + (y+5)²-25+13 = (x-2)²+(y+5)² - 8

Hence (x-2)² + (y+5)² = 8

Lets put y in function of x. We should obtain 2 functions f and g that represent the circle.

(y+5)² = 8 - (x-2)² = -x² + 4x + 4

y = ^+_- \sqrt{-x^2+4x+4} -5

Thus

f(x) = \sqrt{-x^2+4x+4} - 5

g(x) = -\sqrt{-x^2+4x+4} - 5

Lets find the derivate of each function and the points in which they reach the value 2/3. In those points the tangent line will have a slope of 2/3.

We may just find the values of x which the derivate of f is either 2/3 or -2/3.

\frac{-x+2}{\sqrt{-x^2+4x+4}} = ^+_-\frac{2}{3} \Leftrightarrow \frac{x^2-4x+4}{-x^2+4x+4} = \frac{4}{9} \Leftrightarrow 9(x^2-4x+4) = 4(-x^2+4x+4) \\\Leftrightarrow 13x^2-52x+20 = 0

The quadratic has roots

\frac{52 ^+_-\sqrt{1664}}{26}

one root is 3.568, which corresponds with g, and the other root is 0.431, corresponding to f.

Also g(3.568) = - 7.354 and f(0.431) = -2.646

This means that the coordinates are (0.431 , -2.646), (3.568 , -7.354) and the tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

5 0
3 years ago
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