Multiply the exponents.
So 10 ×4/5=
40/5
=8
Thus (243k^10)^4/5 is 243k^8
Range is eleven median is 27
If there are two hydrogen atom molecules in one atom of water you'd do 51 times 2, getting you 102 hydrogen atoms in 51 water molecules.
The solutions to the given quadratic equation {49n² - 301n + 42 = 0 } are 1/7 and 6.
<h3>What are the solutions to the given quadratic equation?</h3>
Quadratic equation is simply an algebraic expression of the second degree in x. Quadratic equation in its standard form is expressed as;
ax² + bx + c = 0
Where x is the unknown
To solve for x, we use the quadratic formula
x = (-b±√(b² - 4ac)) / (2a)
Given the equation in the question;
49n² - 301n + 42 = 0
Compared to the standard form of quadratic equation { ax² + bx + c = 0 }
We plug in these values into the quadratic formula.
x = (-b±√(b² - 4ac)) / (2a)
x = (-(-301) ±√((-301)² - 4 × 49 × 42 )) / (2 × 49)
x = ( 301 ±√( 90601 - 8232 )) / 98
x = ( 301 ±√( 82369 )) / 98
x = ( 301 ± 287) / 98
x = (301 - 287)/98, (301 + 287)/98
x = 14/98, 588/98
x = 1/7, 6
Therefore, the solutions to the given quadratic equation {49n² - 301n + 42 = 0 } are 1/7 and 6.
Learn more about quadratic equations here: brainly.com/question/1863222
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