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3 years ago

Anything to the power of 0 is always 1 right??

Mathematics

2 answers:

lana66690 [7]3 years ago

6 0

Yes, that is correct.

Shtirlitz [24]3 years ago

4 0

Yes anything to the power of 0 is one in the contrary anything to the power of 1 its it self

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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

A doctor wants to know if her patients are happy with her. She decides to survey a sample of her clients. What method would be b

Leokris [45]

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valkas [14]

6y=-36
y=-6
-4x+12=-12
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x=-6

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Solve for the variable x+9=12

labwork [276]

Answer:

Step-by-step explanation:

subtract 9 from both sides

x=3

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Help me assapppp!!! Pls

mafiozo [28]

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have you tries adding them together?

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