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Arada [10]
4 years ago
15

S the expression x3•x3•x3 equivalent to x3•3•3

Mathematics
1 answer:
Artyom0805 [142]4 years ago
3 0
Remember
(x^m)(x^n)=x^(m+n)
and difference of 2 perfect squres
(a²-b²)=(a-b)(a+b)
and sum or difference of 2 perfect cubes

so
(x^3)(x^3)(x^3)=x^(3+3+3)=x^9
so

x^9=3*3*x^3
x^9=9x^3
minus 9x^3 both sides
0=x^9-9x^3
factor
0=(x^3)(x^6-9)
factor difference of 2 perfect squraes
0=(x^3)(x^3-3)(x^3+3)
factor differnce or sum of 2 perfect cubes (force 3 into (∛3)³)
0=(x³)(x-∛3)(x²+x∛3+∛9)(x+∛3)(x²-x∛3+∛9)
set each to zero

x³=0
x=0

x-∛3=0
x=∛3

x²+x∛3+∛9=0 has no solution

x+∛3=0
x=-∛3

x²-x∛3+∛9=0 has no solution



so the solutions are
x=-∛3, 0, ∛3
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Answer:

It would take approximately 6.50 second for the cannonball to strike the ground.

Step-by-step explanation:

Consider the provided function.

h(t)=-4.9t^2+30.5t+8.8

We need to find the time takes for the cannonball to strike the ground.

Substitute h(t) = 0 in above function.

-4.9t^2+30.5t+8.8=0

Multiply both sides by 10.

-49t^2+305t+88=0

For a quadratic equation of the form ax^2+bx+c=0 the solutions are: x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

Substitute a = -49, b = 305 and c=88

t=\frac{-305+\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}=-\frac{-305+\sqrt{110273}}{98}\\t = \frac{-305-\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}= \frac{305+\sqrt{110273}}{98}

Ignore the negative value of t as time can't be a negative number.

Thus,

t=\frac{305+\sqrt{110273}}{98}\approx6.50

Hence, it would take approximately 6.50 second for the cannonball to strike the ground.

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Answer:

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