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3 years ago

An angle turns through 40 one-degree angles. What is its angle measure?

Mathematics

2 answers:

Aleksandr-060686 [28]3 years ago

8 0

1 is C and 2 is D please say thanks

Nostrana [21]3 years ago

6 0

Answer:1)c 2)d

Step-by-step explanation:their you go :)

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100 POINTS + BRAINLYEST Evaluate 7⋅5+42−23÷4 ' 49 41 34 9

alexandr402 [8]

7.5+42−23÷4= 43.75

I hope its help you

3 0

2 years ago

Read 2 more answers

Solve for x. 2x+5+6x-1=120

svetlana [45]

Answer:

x=14.5

Step-by-step explanation:

2x+5+6x-1=120

Combine like terms (combine all of the variables and then combine all of the constants)

8x+4=120

subtract 4 on both sides

8x=116

Divide 8

x=14.5

Hope this helps!

8 0

3 years ago

Use synthetic division to solve(X^3+1) divide by. X-1

frozen [14]

The answer is =x^2 +x +1 +2/x-1

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.