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Slav-nsk [51]
3 years ago
6

X^2-8x+16/5x^2-19x-4

Mathematics
1 answer:
GREYUIT [131]3 years ago
4 0

Step-by-step explanation:

(x² - 8x + 16) / (5x² - 19x - 4)

= (x - 4)² / [(5x + 1)(x - 4)]

= (x - 4)/(5x + 1).

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The calculation of property tax is based on the
Tresset [83]
The assessment rate is a uniform percentage and varies by tax jurisdiction, and could be any percentage below 100%. After getting the assessed value, it is multiplied by the mill levy to determine your taxes due. For example, suppose the assessor determines your property value is $500,000 and the assessment rate is 8%.
5 0
3 years ago
Solve for x
ludmilkaskok [199]

Answer:

x = 3/4

Step-by-step explanation:

Step 1: Write equation

5x - 2(x + 1) = 1/4

Step 2: Solve for <em>x</em>

  1. <u>Distribute -2:</u> 5x - 2x - 2 = 1/4
  2. <u>Combine like terms:</u> 3x - 2 = 1/4
  3. <u>Add 2 to both sides:</u> 3x = 9/4
  4. <u>Divide both sides by 3:</u> x = 3/4

Step 3: Check

<em>Plug in x to verify it's a solution.</em>

5(3/4) - 2(3/4 + 1) = 1/4

15/4 - 2(7/4) = 1/4

15/4 - 14/4 = 1/4

1/4 = 1/4

3 0
3 years ago
3.6(-2y)(-5) simplify the equation
jonny [76]

Answer:   36y

---------------------------------------------------------------------------------------

4 0
3 years ago
Read 2 more answers
I need help on my math
bekas [8.4K]
The answer is the second choice I believe
8 0
3 years ago
Read 2 more answers
PLEASE HELP
lisabon 2012 [21]

Step-by-step explanation:

The figure below shows a portion of the graph of the function j\left(x\right) \ = \ 4^{x-2}, hence the average rate of change (slope of the blue line) between the x and x+h is

                     \text{Average rate of change} \ = \ \displaystyle\frac{\Delta y}{\Delta x} \\ \\ \rule{3.7cm}{0cm} = \dsiplaystyle\frac{f\left(x+h\right) \ - \ f\left(x\right)}{\left(x \ + \ h \right) \ - \ x} \\ \\ \\  \rule{3.7cm}{0cm} = \displaystyle\frac{f\left(x + h\right) \ - \ f\left(x\right)}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x+h-2} \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x-2+h} \ - \ 4^{x-2}}{h}

                                                            \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h}\right) \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h} \ - \ 1 \right)}{h}

7 0
2 years ago
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