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3 years ago

Explain why the initial value of any function of the form f(x) = a(br) is equal to a.

1 answer:

7 0

Answer:When you substitute 0 for the exponent x, the expression simplifies to a times 1, which is just a. This is because any number to the 0 power equals 1. Since the initial value is the value of the function for an input of 0, the initial value for any function of this form will always be the value of a

Step-by-step explanation:

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Answer:

Step-by-step explanation:

26^2-24^2=100

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232/25.

Step-by-step explanation:

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Factor this expression completely and then place the factors in the proper location on the grid. 36y2 - 1

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3 years ago

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Which two values of x are roots of the polynomial below? x2-11x + 15

Alex787 [66]

The two values of roots of the polynomial $x^{2}-11 x+15$ are $\frac{11+\sqrt{61}}{2} \text { or } \frac{11-\sqrt{61}}{2}$

Solution:

Given, polynomial expression is $x^{2}-11 x+15$

We have to find the roots of the given expression.

In order to find roots, now let us use quadratic formula.

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Given that $x^{2}-11 x+15$

Here a = 1, b = -11 and c = 15

On substituting the values we get,

$x=\frac{-(-11) \pm \sqrt{(-11)^{2}-4 \times 1 \times 15}}{2 \times 1}$

$\begin{array}{l}{x=\frac{11 \pm \sqrt{121-60}}{2}} \\\\ {x=\frac{11 \pm \sqrt{61}}{2}} \\\\ {x=\frac{11+\sqrt{61}}{2} \text { or } \frac{11-\sqrt{61}}{2}}\end{array}$

Hence, the roots of given polynomial are $\frac{11+\sqrt{61}}{2} \text { or } \frac{11-\sqrt{61}}{2}$

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3 years ago