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3 years ago

Which expression is equivalent to picture attached

Mathematics

1 answer:

Sedbober [7]3 years ago

7 0

Answer:

$\sum_{n=3}^{20}(n(n+1))=\sum _{n=3}^{20} n^2+ \sum _{n=3}^{20} n$

Step-by-step explanation:

Given expression : $\sum_{n=3}^{20}(n(n+1))$

Solving further :

$\Rightarrow \sum_{n=3}^{20}(n^2+n)$

$\Rightarrow \sum _{n=3}^{20} n^2+ \sum _{n=3}^{20} n$

So, $\sum_{n=3}^{20}(n(n+1))=\sum _{n=3}^{20} n^2+ \sum _{n=3}^{20} n$

So, The given expression is equivalent to Option A

So, Option A is the answer

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Brianna and let kelci are raising money for a class field trip. Brianna has raised 3 times more money than kelci, but together t

ddd [48]

Let $x be the amount of money which Kelci has raised. Brianna has raised 3 times more money than kelci, then she has raised $3x. Totally both have aised $(x+3x)=$4x.

Since together they have raised more than $300, then 4x>300,

$x\ \textgreater \ \dfrac{300}{4}$ .
Answer: the inequality $x\ \textgreater \ \dfrac{300}{4}$ can be used to determine the amount of money kelci has raised

7 0

3 years ago

HELP NOW!!!!

VMariaS [17]

Using the concept of probability and the arrangements formula, there is a

0.002 = 0.2% probability that the first 8 people in line are teachers.

----------------------------------

A probability is the number of desired outcomes divided by the number of total outcomes.
The order in which they are positioned is important, and all people will be positioned, and thus, the arrangements formula is used to find the number of outcomes.

The number of possible arrangements from a set of n elements is given by:

$A_n = n!$

----------------------------------

The desired outcomes are:

First 8 people are teachers, in 8! possible ways.
Last 4 are students, in 4! possible ways.

Thus, $D = 8! \times 4!$

----------------------------------

For the total outcomes, number of arrangements of 12 people, thus:

$T = 12!$

The probability is:

$p = \frac{D}{T} = \frac{8! \times 4!}{12!} = 0.002$

0.002 = 0.2% probability that the first 8 people in line are teachers.

A similar problem is given at brainly.com/question/24650047

7 0

2 years ago

Use the Chain Rule (Calculus 2)

atroni [7]

1. By the chain rule,

$\dfrac{\mathrm dz}{\mathrm dt}=\dfrac{\partial z}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial z}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}$

I'm going to switch up the notation to save space, so for example, $z_x$ is shorthand for $\frac{\partial z}{\partial x}$ .

$z_t=z_xx_t+z_yy_t$

We have

$x=e^{-t}\implies x_t=-e^{-t}$

$y=e^t\implies y_t=e^t$

$z=\tan(xy)\implies\begin{cases}z_x=y\sec^2(xy)=e^t\sec^2(1)\\z_y=x\sec^2(xy)=e^{-t}\sec^2(1)\end{cases}$

$\implies z_t=e^t\sec^2(1)(-e^{-t})+e^{-t}\sec^2(1)e^t=0$

Similarly,

$w_t=w_xx_t+w_yy_t+w_zz_t$

where

$x=\cosh^2t\implies x_t=2\cosh t\sinh t$

$y=\sinh^2t\implies y_t=2\cosh t\sinh t$

$z=t\implies z_t=1$

To capture all the partial derivatives of $w$ , compute its gradient:

$\nabla w=\langle w_x,w_y,w_z\rangle=\dfrac{\langle1,-1,1\rangle}{\sqrt{1-(x-y+z)^2}}}=\dfrac{\langle1,-1,1\rangle}{\sqrt{-2t-t^2}}$

$\implies w_t=\dfrac1{\sqrt{-2t-t^2}}$

2. The problem is asking for $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ . But $z$ is already a function of $x,y$ , so the chain rule isn't needed here. I suspect it's supposed to say "find $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$ " instead.