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Given an equation x3 - 8x2 - 9x + 72 = 0, find the zeros.

sveta [45]

Answer:

= 28/3

Step-by-step explanation:

Let's solve your equation step-by-step.

x(3)−(8)(2)−9x+72=0

Step 1: Simplify both sides of the equation.

x(3)−(8)(2)−9x+72=0

3x+−16+−9x+72=0

(3x+−9x)+(−16+72)=0(Combine Like Terms)

−6x+56=0

Step 2: Subtract 56 from both sides.

−6x+56−56=0−56

−6x=−56

Step 3: Divide both sides by -6.

−6x

−6

=

−56

−6

x=

28

3

Answer:

x=

28

3

7 0

3 years ago

Y = x + 2y = -2x + 2y = -3x + 2y = -5x + 2y = -

djverab [1.8K]

Answer:

152

Step-by-step explanation:

jtg djkgkjh uy eo y3oi4yio23 jdoie oi4h io

6 0

3 years ago

A = 9 m, b = 4 m and c = 13 m for the triangle shown below.

maw [93]

The answer is 9 shown below a b c triangle shown below ⬇️

7 0

2 years ago

Linear or non-linear?

Gnoma [55]

Answer:

the answer for this question is non-linear

Step-by-step explanation:

linear means in a straight line. the question you wrote is NOT in a strait line.

hoped the answer helped

7 0

3 years ago

2 tan 30° II 1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

3 0

3 years ago