Enter the number that belongs in the green box PLEASE HELP FAST

A number from 18 to 24 is drawn at random. What is the probability that the answer is a number less than 24?

icang [17]

So there are 6 numbers between 18 and 24. The probability of choosing a number less than 24 would be 5/24 or 20.83%

5 0

3 years ago

Read 2 more answers

Please help me. If △NMK ≅ △TRP, then answer the following questions:

Natalka [10]

Answer:

Answers are below.

Step-by-step explanation:

a. △MNK ≅ △RTP

b. TR ≅ NM

c. x = 7

hope this helps and is right!! p.s. i really need brainliest :)

3 0

3 years ago

Mutleys journey began on a boat, 2 feet above sea level. He dove down to twelve feet below sea level and then returned to the su

Lubov Fominskaja [6]

Answer:

Mutley would start at 2 feet above sea level or +2, he would then travel 12 feet below sea level (-12) to then return to the surface of the ocean (0). The integers from least to greatest: -12, 0, 2. Visual:

-12(dive level) * * * * * * * * * * * 0(sea level) * 2(boat)

Step-by-step explanation:

Using positive and negative integers, we can determine Mutleys journey above, below and at the surface of the ocean. The surface of the ocean represents his 'origin' or starting point, which is 0. The boat is above the surface or +2. When Mutley dives below the surface, he is at a negative level of the ocean. Think of it in terms of a number line - negative numbers are to the left of 0 and positive numbers are to the right of 0. The further we go to the left on the number line, the lower our number. In this case, -12 would be furthest to the left, then 0, followed by 2.

6 0

3 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.