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vichka [17]
3 years ago
7

Can some one plz help me?

Mathematics
1 answer:
Anna11 [10]3 years ago
4 0
You are multiplying the numbers by 8 so it would most likely be patrick
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What is the y-intercept of the line shown below?<br><br>A:3/4<br>B:2<br>C:3<br>D:4​
olya-2409 [2.1K]

The y-intercept is the y value where the blue line crosses the Y axis which is the vertical black line.

The line crosses at the number 4, so the y-intercept is 4

Answer: D. 4

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2 years ago
A triangle has a 90° angle. What type of triangle is it?
gavmur [86]
A right triangle has one angle equal to 90 degrees. A right triangle can also be an isosceles triangle--which means that it has two sides that are equal. A right isosceles triangle has a 90-degree angle and two 45-degree angles. This is the only right triangle that is an isosceles triangle. But the answer is right triangle
7 0
2 years ago
Read 2 more answers
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.

If that's the case, then

z_s=z_xx_s+z_yy_s

z_t=z_xx_t+z_yy_t

as the hint suggests. We have

z=\sin x\cos y\implies\begin{cases}z_x=\cos x\cos y=\cos(s+t)\cos(s^2t)\\z_y=-\sin x\sin y=-\sin(s+t)\sin(s^2t)\end{cases}

x=s+t\implies x_s=x_t=1

y=s^2t\implies\begin{cases}y_s=2st\\y_t=s^2\end{cases}

Putting everything together, we get

z_s=\cos(s+t)\cos(s^2t)-2st\sin(s+t)\sin(s^2t)

z_t=\cos(s+t)\cos(s^2t)-s^2\sin(s+t)\sin(s^2t)

8 0
3 years ago
A contradiction has no solutions. True or False?
Eddi Din [679]

Answer:

false

Step-by-step explanation:

6 0
3 years ago
For , find the value of x for which .
Mice21 [21]
Uh what did you mean to upload a picture or somthing
8 0
3 years ago
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