Look at the picture please

Mathematics

2 answers:

Svet_ta [14]2 years ago

6 0

Warmer in Columbus, Ohio

Sliva [168]2 years ago

4 0

Warmer in Columbus, Ohio

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What is the answer to this question 4(5x+30)=100?

Over [174]

Answer: x = -1

Step-by-step explanation: To solve this equation, we start by distributing the 4 through the parentheses on the left side.

So we have 4 times 5x which is 20x and 4 times 30 which is 120.

So our equation now reads 20x + 120 = 100.

Solving from here, we subtract 120 from both sides of the equation and we have 20x = -20 and dividing both sides by 20, we find that <em>x = -1</em>.

4 0

2 years ago

The graph of an arithmetic sequence is shown.

GalinKa [24]

Answer:

The fifth term is 7

Step-by-step explanation:

Looking at the graph

we have the ordered pairs

(1,5),(2,5.5),(3,6),(4,6.5),(5,7)

Let

$a_1=5\\a_2=5.5\\a_3=6\\a_4=6.6\\a_5=7$

The common difference in this arithmetic sequence is 0.5

The value of the fifth term is a_5

therefore

The fifth term is 7

7 0

3 years ago

Read 2 more answers

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 $\sqrt{\dfrac43}=\dfrac2{\sqrt3}$ .