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exis [7]
1 year ago
15

Which is the completely factored form of the following equation? 4a4b2−21a2b+5=0

Mathematics
1 answer:
r-ruslan [8.4K]1 year ago
4 0

The completely factored equation is (a²b - 5)(4a²b - 1)

<h3>How to factor equation?</h3>

4a⁴b² - 21a²b + 5 = 0

let's expand the equation

Therefore,

4a⁴b² - 21a²b + 5 = 0

4a⁴b² - 20a²b  - a²b + 5 = 0

Rearrange the equation

4a⁴b² - a²b  - 20a²b + 5 = 0

Let's factorise the equation

4a⁴b² - a²b  - 20a²b + 5 = 0

a²b(4a²b - 1) - 5(4a²b - 1) = 0

Therefore, the completely factored equation is as follows:

a²b(4a²b - 1) - 5(4a²b - 1) = 0

(a²b - 5)(4a²b - 1)

learn more on equation here: brainly.com/question/8842252

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3 years ago
What is the value of x?
kipiarov [429]
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7 0
2 years ago
Find the second derivative of 2x^3-3y^2=8​
Umnica [9.8K]

Answer:

\frac{d^2y}{dx^2}=\frac{x(2y-x^3)}{y^3}

Step-by-step explanation:

<u>Find the first implicit derivative using implicit differentiation</u>

<u />2x^3-3y^2=8\\\\6x^2-6y\frac{dy}{dx}=0\\ \\-6y\frac{dy}{dx}=-6x^2\\ \\\frac{dy}{dx}=\frac{x^2}{y}

<u>Use the substitution of dy/dx to find the second derivative (d²y/dx²)</u>

<u />\frac{d^2y}{dx^2}=\frac{(y)(2x)-(x^2)(\frac{dy}{dx})}{y^2}\\ \\\frac{d^2y}{dx^2}=\frac{2xy-(x^2)(\frac{x^2}{y})}{y^2}\\\\\frac{d^2y}{dx^2}=\frac{2xy-\frac{x^4}{y}}{y^2}\\\\\frac{d^2y}{dx^2}=\frac{x(2y-x^3)}{y^3}

5 0
2 years ago
(1/3+9) x ( 8-3) tell me the steps help
mafiozo [28]
Use PEMDAS to solve
(Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction)

Solve parenthesis first
<span>(1/3+9) x ( 8-3)
</span>28/3 x 5

Use Multiplication 
28/3 x 5 ≈ 4.7
After rounding 4.7 should be your answer
3 0
3 years ago
PLEASE help me solve this problem. I really need help.
mote1985 [20]
When:
a = b = 0 we have point (0,0).
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a ≠ 0 and b ≠ 0 a = b circle
3 0
3 years ago
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