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3 years ago

Which expression is equivalent to 15a^8b^4/5a^4b? 3a^2b^4 3a^4b^3 10a^4b^3 10a^4b^4

Mathematics

1 answer:

EleoNora [17]3 years ago

5 0

Answer:

3 a^12 b^5

Step-by-step explanation:

Simplify the following:

(15 a^8 b^4 a^4 b)/5

15/5 = (5×3)/5 = 3:

3 a^8 b^4 a^4 b

3 a^8 b^4 a^4 b = 3 a^(8 + 4) b^(4 + 1):

3 a^(8 + 4) b^(4 + 1)

4 + 1 = 5:

3 a^(8 + 4) b^5

8 + 4 = 12:

Answer: 3 a^12 b^5

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For what value of k is there one solution to the given quadratic equation (k+1)x²+4kx+2=0.

andriy [413]

Answer:

If $k = 1$ or $k = -\frac{1}{2}$ , there is only one solution to the given quadratic equation.

Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

The signal of $\bigtriangleup$ determines how many real roots an equation has:

$\bigtriangleup > 0$ : Two real and different solutions

$\bigtriangleup = 0$ : One real solution

$\bigtriangleup < 0$ : No real solutions

In this problem, we have the following second order polynomial:

$(k+1)x^{2} + 4kx + 2 = 0$ .

This means that $a = k+1; b = 4k; c = 2$

It has one solution if

$\bigtriangleup = 0$

$b^{2} - 4ac = 0$

$16k^{2} -8(k+1) = 0$

$16k^{2} - 8k - 8 = 0$

We can simplify by 8

$2k^{2} - k - 1 = 0$

The solution is:

$k = 1$ or $k = -\frac{1}{2}$

So, if $k = 1$ or $k = -\frac{1}{2}$ , there is only one solution to the given quadratic equation.

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3 years ago

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Step-by-step explanation:

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So he would need 96 inches of wood!

Hope this helps!!! Good luck!!! ;)

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Answer:

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