Hi,

Sorry i don't understand you "determine the sigma notation of the sum for term 4 through term 15"
$a_1=4\\

a_2=-12=4*(-3)\\

a_3=36=(-12)*(-3)=4*(-3)^2
$

$a_4=4*(-3)^3\\

a_5=4*(-3)^4\\

...

a_{15}=4*(-3)^{14}\\


\sum_{i=4}^{15}a_i\\

=a_4+a_5+...+a_{15}\\
=4*(-3)^3 + 4*(-3)^4+...+4*(-3)^{14}\\
=4*(-3)^3*(1+(-3)+(-3)^2+(-3)^3+...+(-3)^{11}\\

=4*(-3)^3* \dfrac{(-3)^{12}-1}{(-3-1)} \\

=4*(-27)* \dfrac{3^{12}-1}{(-4)}\\

=27*(531441-1)\\

=14348880
$

5 0

3 years ago

Read 2 more answers

3. The angle between two equal sides of an isosceles triangle is 52º.

Naily [24]

Answer:

a) 64° each

b) 15.78 cm

c) 51.78 cm

Step-by-step explanation:

a) (180 - 52)/2 = 64°

b) cos 64 = x/18, x = 7.89, 2x = 15.78 cm

c) 18 + 18 + 15.78 = 51.78 cm

7 0

3 years ago

*help please thanks .*

help please thanks .