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e-lub [12.9K]
3 years ago
7

Find the equation of the exponential function represented by the table below:

Mathematics
1 answer:
shepuryov [24]3 years ago
3 0

Given:

The table of values.

To find:

The equation of the exponential function represented by the table.

Solution:

The general form of an exponential function is

y=ab^x          ...(i)

where, a is initial value and b is growth factor.

From the given table it is clear that the function passes through (0,3) and (1,1.5). So, the must be satisfy by these points.

Putting x=0 and y=3 in (i), we get

3=ab^0

3=a(1)

3=a

Putting a=3, x=1 and y=1.5 in (i), we get

1.5=3b^1

\dfrac{1.5}{3}=b

\dfrac{1}{2}=b

Putting a=3 and b=\dfrac{1}{2} in (i), we get

y=3\left(\dfrac{1}{2}\right)^x

Therefore, the required equation for exponential function is y=3\left(\dfrac{1}{2}\right)^x.

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The area of a rectangle is 506 square metre. If its length is 23 metre, then find its breadth?
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Answer:

The area of a triangle can be determined with the two side lengths, the length and breadth/width multiplied.

x = unknown value of breadth/width

23 = length

506 = the area of the two lengths multiplied

23(x) = 506

/23        /23

x = <u>22</u>, 22 metres is it's breadth.

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Answer:

1780

Step-by-step explanation:

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A disk with radius 3 units is inscribed in a regular hexagon. Find the approximate area of the inscribed disk using the regular
Oksi-84 [34.3K]

Answer:

The approximate are of the inscribed disk using the regular hexagon is A=18\sqrt{3}\ units^2

Step-by-step explanation:

we know that

we can divide the regular hexagon into 6 identical equilateral triangles

see the attached figure to better understand the problem

The approximate area of the circle is approximately the area of the six equilateral triangles

Remember that

In an equilateral triangle the interior measurement of each angle is 60 degrees

We take one triangle OAB, with O as the centre of the hexagon or circle, and AB as one side of the regular hexagon

Let

M  ----> the mid-point of AB

OM ----> the perpendicular bisector of AB

x ----> the measure of angle AOM

m\angle AOM =30^o

In the right triangle OAM

tan(30^o)=\frac{(a/2)}{r}=\frac{a}{2r}\\\\tan(30^o)=\frac{\sqrt{3}}{3}

so

\frac{a}{2r}=\frac{\sqrt{3}}{3}

we have

r=3\ units

substitute

\frac{a}{2(3)}=\frac{\sqrt{3}}{3}\\\\a=2\sqrt{3}\ units

Find the area of six equilateral triangles

A=6[\frac{1}{2}(r)(a)]

simplify

A=3(r)(a)

we have

r=3\ units\\a=2\sqrt{3}\ units

substitute

A=3(3)(2\sqrt{3})\\A=18\sqrt{3}\ units^2

Therefore

The approximate are of the inscribed disk using the regular hexagon is A=18\sqrt{3}\ units^2

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WHAT IS X???<br> 1.2/0.96 = 3.0/x
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Answer:

x= 2.4

Step-by-step explanation:

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