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2 years ago

Rachel paid $46 in tax for the money she earned

Mathematics

2 answers:

Harman [31]2 years ago

8 0

Answer:

not enough information to give accurate answer

Step-by-step explanation:

sorry :(

IceJOKER [234]2 years ago

7 0

no sense it has no problem

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What are the types of roots of the equation below?<br> - 81=0

Tju [1.3M]

Option B, that is Two Complex and Two Real which are x + 3, x - 3, x + 3i and x - 3i, are the types of roots of the equation x⁴ - 81 = 0. This can be obtained by finding root of the equation using algebraic identity.

<h3>What are the types of roots of the equation below?</h3>

Here in the question it is given that,

the equation x⁴ - 81 = 0

By using algebraic identity, (a + b)(a - b) = a² - b², we get,

⇒ x⁴ - 81 = 0

⇒ (x² + 9)(x² - 9) = 0

(x² - 9) = (x² - 3²) = (x - 3)(x + 3) [using algebraic identity, (a + b)(a - b) = a² - b²]
x² + 9 = 0 ⇒ x² = -9 ⇒ x = √-9 ⇒ x= √-1√9 ⇒x = ± 3i

⇒ (x² + 9) = (x - 3i)(x + 3i)

Now the equation becomes,

[(x - 3)(x + 3)][(x - 3i)(x + 3i)] = 0

Therefore x + 3, x - 3, x + 3i and x - 3i are the roots of the equation

To check whether the roots are correct multiply the roots with each other,

⇒ [(x - 3)(x + 3)][(x - 3i)(x + 3i)] = 0

⇒ [x² - 3x + 3x - 9][x² - 3xi + 3xi - 9i²] = 0

⇒ (x² +0x - 9)(x² +0xi - 9(- 1)) = 0

⇒ (x² - 9)(x² + 9) = 0

⇒ x⁴ - 9x² + 9x² - 81 = 0

⇒ x⁴ - 81 = 0

Hence Option B, that is Two Complex and Two Real which are x + 3, x - 3, x + 3i and x - 3i, are the types of roots of the equation x⁴ - 81 = 0.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: What are the types of roots of the equation below?

x⁴ - 81 = 0

A) Four Complex

B) Two Complex and Two Real

C) Four Real

Learn more about roots of equation here:

brainly.com/question/26926523

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5 0

1 year ago

Let L be a tangent line to the hyperbola x y = 2 at x = 9 . Find the area of the triangle bounded by L and the coordinate axes.

mafiozo [28]

Answer:

$A = 4$

Step-by-step explanation:

The equation of the slope of the tangent line L is obtained by deriving the equation of the hyperbola:

$y = \frac{2}{x}$

$y'=-2\cdot x^{-2}$

The numerical value of the slope is:

$y' = -2 \cdot (9)^{-2}\\y' = -\frac{2}{81}$

The component of the y-axis is:

$y = \frac{2}{9}$

Now, the tangent line has the following mathematical model:

$y = m \cdot x + b$

The value of the intercept is found by isolating it within the equation and replacing all known variables:

$b = y - m \cdot x$

$b = \frac{2}{9}-(-\frac{2}{81} )\cdot (9)\\b = \frac{4}{9}$

Thus, the tangent line is:

$y = -\frac{2}{81}\cdot x + \frac{4}{9}$

The vertical distance between a point of the tangent line and the origin is given by the intercept.

$d_{y} = \frac{4}{9}$

In order to find horizontal distance between a point of the tangent line and the origin, let equalize y to zero and clear x:

$-\frac{2}{81}\cdot x + \frac{4}{9}=0$

$-\frac{2}{9}\cdot x + 4 = 0$

$x = 18$

$d_{x} = 18$

The area of the triangle is computed by this formula:

$A = \frac{1}{2}\cdot d_{x}\cdot d_{y}$

$A = \frac{1}{2}\cdot (18)\cdot (\frac{4}{9} )$

$A = 4$

4 0

3 years ago

What does 1/4+1/3 equal

grandymaker [24]

Answer:

7/12

Step-by-step explanation:

1/4+1/3

the LCM is 12

4+3/7=7/12

=7/12

8 0

3 years ago

The question says 5k+6k+4k=90. What does k equal?

alina1380 [7]

$5k+6k+4k=90 \\ \\ 15k = 90 \ / \ simplify \\ \\ k = \frac{90}{15} \ / \ divide \ each \ side \ by \ 15 \\ \\ k = 6 \ / \ simplify \\ \\$

The final result is, k = 6.

5 0

3 years ago

A line with a slope of -2 passes thru the point (4,7)

Tpy6a [65]

Answer:

y - 7 = -2(x - 4)

Step-by-step explanation:

We are asked to write the equation of a line in point slope form

Step 1 : find slope

We are given the slope to be -2

Slope m = -2

Step 2: substitute m into point slope form

y - y_1 = m( x - x_1)

y - y_1 = -2 ( x - x _1)

Step 3: substitute the point into the equation

y - y_1 = -2( x - x _1)

( 4 , 7)

x_1 = 4

y_1 = 7

y - 7 = -2( x - 4)

We don't need to open the bracket because we are asked to write the equation in a point slope form

6 0

2 years ago