All categories

3 years ago

Find the sum of the first 10 terms of the following series to the nearest Integra 3, 5, 25/3,...

Mathematics

1 answer:

Julli [10]3 years ago

3 0

Answer:

S(10)=3*(1-(5/3)^10)/(1-5/3)

Step-by-step explanation:

It is Geometric series with r:

r=5/3=(25/3)/5=5/3

a=3

Sum of G.S. in formula:

S(n)=a(1-r^n)/(1-r)

S(10)=3*(1-(5/3)^10)/(1-5/3)

=246.5725...

You might be interested in

Can somebody please please help me I'm desperate!!!!!!

FromTheMoon [43]

You would have to solve this by Lxwxh.

25.4*19.05*76.2= 36870.89 centimeters

1 in= 2.54 centimeters

36870.89 centimeters*1 in/2.54 centimeters= 14516.09843 inches.

I hope this is helpful :)

4 0

3 years ago

Read 2 more answers

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.