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3 years ago

12

Is hunter license a form of a tax

1 answer:

puteri [66]3 years ago

8 0

A hunting license is not a from of tax it is a form of license that allows you to hunt game and only costs each time for renewal of the right
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What is the domain of this function in the graph below? <br><br> What is the interval?

Maurinko [17]

Domain in interval: [6,infinity)

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2 years ago

Solve the equation: 5y+2=22 Please help me!! :(

AysviL [449]

Answer: 4

Step-by-step explanation:

1. Subtract 2 from both sides.

5y=22 -2

This equals 5y=20

2.Divide by 5 on each side. That gets 4.

Hope it helped :)

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3 years ago

Read 2 more answers

Which triangle’s unknown side length measures 7 units?

kow [346]

A (A right triangle with a side length of 5 and hypotenuse with length \sqrt{74})

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3 years ago

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What is the approximate volume of a cylinder with a height of 2 ft and a radius of 6 ft? Use 3.14 to approximate pi. Express you

marissa [1.9K]

The volume of a cylinder is Pi times the radius^2 times the height

Volume = Pi x 6^2 x 2
Volume = 226.2

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3 years ago

What two rational expressions sum to 2x+3/x^2-5x+4

Anni [7]

Answer:

$\frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)}$

Step-by-step explanation:

Given the rational expression: $\frac{2x + 3}{x^2 - 5x + 4}$ , to express this in simplified form, we would need to apply the concept of partial fraction.

Step 1: factorise the denominator

$x^2 - 5x + 4$

$x^2 - 4x - x + 4$

$(x^2 - 4x) - (x + 4)$

$x(x - 4) - 1(x - 4)$

$(x- 1)(x - 4)$

Thus, we now have: $\frac{2x + 3}{(x- 1)(x - 4)}$

Step 2: Apply the concept of Partial Fraction

Let,

$\frac{2x + 3}{(x- 1)(x - 4)}$ = $\frac{A}{x- 1} + \frac{B}{x - 4}$

Multiply both sides by (x - 1)(x - 4)

$\frac{2x + 3}{(x- 1)(x - 4)} * (x - 1)(x - 4)$ = $(\frac{A}{x- 1} + \frac{B}{x - 4}) * (x - 1)(x - 4)$

$2x + 3 = A(x - 4) + B(x - 1)$

Step 3:

Substituting x = 4 in $2x + 3 = A(x - 4) + B(x - 1)$

$2(4) + 3 = A(4 - 4) + B(4 - 1)$

$8 + 3 = A(0) + B(3)$

$11 = 3B$

$\frac{11}{3} = B$

$B = \frac{11}{3}$

Substituting x = 1 in $2x + 3 = A(x - 4) + B(x - 1)$

$2(1) + 3 = A(1 - 4) + B(1 - 1)$

$2 + 3 = A(-3) + B(0)$

$5 = -3A$

$\frac{5}{-3} = \frac{-3A}{-3}$

$A = -\frac{5}{3}$

Step 4: Plug in the values of A and B into the original equation in step 2

$\frac{2x + 3}{(x- 1)(x - 4)} = \frac{A}{x- 1} + \frac{B}{x - 4}$

$\frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)}$

7 0

3 years ago