All categories

2 years ago

Given h(x)=-3x-4h(x)=−3x−4, find h(0)

Mathematics

1 answer:

bogdanovich [222]2 years ago

7 0

Answer:

h(0) = -4

Step-by-step explanation:

Put the value where the variable is and do the arithmetic.

h(x) = -3x -4

h(0) = -3·0 -4 = 0 -4

h(0) = -4

_____

<em>Additional comment</em>

In general, a polynomial function evaluated when the variable is zero will have the value of any added constant. All of the variable terms will be zero. Since the constant in this function is -4, we know immediately that h(0) = -4.

You might be interested in

You receive X dollars an hour for babysitting you babysit three hours on Friday and five hours on Saturday you receive $40 for t

malfutka [58]

5 hours + 3 hours = 8 hours

8x = $40. Divide each side by 8.

x = $6.

You earn $6 / hour

6 0

3 years ago

Helppppppppppoppppp??

valentina_108 [34]

Answer:

1st option because those angles are a linear pair.

3 0

3 years ago

In the figure below, mZAHE = (6x + 7)º

Andru [333]

Answer:

Step-by-step explanation:

∠AHE + ∠AHK = 180 {linear pair}

6x + 7 + 35 = 180

6x + 42 = 180

6x = 180 - 42

6x = 138

x = 138/6

x = 23

8 0

3 years ago

How do you write 41,300 in scientific notation?

Nuetrik [128]

Answer: 4.13 x 10^4

Step-by-step explanation: You have to just count the places such as the tens all the way to the thousands which leaves you with the exponent of 4.

5 0

3 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.