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Slav-nsk [51]
3 years ago
9

Use the formula to evaluate the series 1+2+4+8+...a10

Mathematics
1 answer:
valina [46]3 years ago
8 0

Answer:

The sum of the given series is 1023

Step-by-step explanation:

Geometric series states that a series in which a constant ratio is obtained by multiplying the previous term.

Sum of the geometric series is given by:

S_n = \frac{a(1-r^n)}{1-r}

where a is the first term and n is the number of term.

Given the series: 1+2+4+8+.................+a_{10}

This is a geometric series with common ratio(r) = 2

We have to find the sum of the series for 10th term.

⇒ n = 10 and a = 1

then;

S_n = \frac{1(1-2^{10})}{1-2} =\frac{1-1024}{-1} =\frac{-1023}{-1}=1023

Therefore, the sum of the given series is 1023



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Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1
Varvara68 [4.7K]
I assume there are some plus signs that aren't rendering for some reason, so that the plane should be x+y+z=1.

You're minimizing d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2} subject to the constraint f(x,y,z)=x+y+z=1. Note that d(x,y,z) and d(x,y,z)^2 attain their extrema at the same values of x,y,z, so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

The Lagrangian is

L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)

Take your partial derivatives and set them equal to 0:

\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}

Adding the first three equations together yields

2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43

and plugging this into the first three equations, you find a critical point at (x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right).

The squared distance is then d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43, which means the shortest distance must be \sqrt{\dfrac43}=\dfrac2{\sqrt3}.
7 0
3 years ago
180 minus what = 45<br><br> i really need this answer thank you
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ABCD is a rectangle. Find the length of each diagonal.ac=2(5a+1) bd=2(a+1).
Igoryamba
   
Rectangle diagonals are equal.

2(5a+1) = 2(a+1)
10a + 2 = 2a + 2
10a -2a = 2 - 2
8a = 0
⇒ a = 0

AC = 2(5a+1)  = 2(5 × 0 +1)= 2(0 + 1) = 2 × 1 = 2 <span> 
</span>BD = 2(a+1) = 2(0 + 1) = 2 <span>× 1 = 2
</span>
Ansver:   AC = BD = 2 



7 0
3 years ago
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