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Olegator [25]
3 years ago
13

An exponential function is graphed for all real numbers. This includes which of the following sets of numbers?

Mathematics
1 answer:
anygoal [31]3 years ago
8 0

<span>In addition to linear, quadratic, rational, and radical functions, there are exponential functions. Exponential functions have the form f(x) = <span>bx</span>, where b > 0 and b ≠ 1. Just as in any exponential expression, b is called the base and x is called the exponent.</span>

 

<span>An example of an exponential function is the growth of bacteria. Some bacteria double every hour. If you start with 1 bacterium and it doubles every hour, you will have 2x bacteria after x hours. This can be written as f(x) = 2x.</span>

 

<span>Before you start,   f(0) = 2<span>0 </span>= 1</span>

<span>After 1 hour           f(1) = 21 = 2</span>

<span>In 2 hours              f(2) = 22 = 4</span>

<span>In 3 hours              f(3) = 23 = 8</span>

and so on.

 

 

<span>With the definition f(x) = <span>bx</span> and the restrictions that b > 0 and that b ≠ 1, the domain of an exponential function is the set of all real numbers. The range is the set of all positive real numbers. The following graph shows f(x) = 2x.</span>

<span> </span>

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kirill115 [55]

Answer:

The triangle ABC is an isosceles right triangle

Step-by-step explanation:

we have

The coordinates of triangle ABC are

A (0, 2), B (2, 5), and C (−1, 7)

we know that

An isosceles triangle has two equal sides and two equal internal angles

The formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

step 1

Find the distance AB

substitute in the formula

d=\sqrt{(5-2)^{2}+(2-0)^{2}}

d=\sqrt{(3)^{2}+(2)^{2}}

dAB=\sqrt{13}\ units

step 2

Find the distance BC

substitute in the formula

d=\sqrt{(7-5)^{2}+(-1-2)^{2}}

d=\sqrt{(2)^{2}+(-3)^{2}}

dBC=\sqrt{13}\ units

step 3

Find the distance AC

substitute in the formula

d=\sqrt{(7-2)^{2}+(-1-0)^{2}}

d=\sqrt{(5)^{2}+(-1)^{2}}

dAC=\sqrt{26}\ units

step 4

Compare the length sides

dAB=\sqrt{13}\ units

dBC=\sqrt{13}\ units

dAC=\sqrt{26}\ units

dAB=dBC

therefore

Is an isosceles triangle

Applying the Pythagoras Theorem

(AC)^{2} =(AB)^{2}+(BC)^{2}

substitute

(\sqrt{26})^{2} =(\sqrt{13})^{2}+(\sqrt{13})^{2}

26=13+13

26=26 -----> is true

therefore

Is an isosceles right triangle

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