How long does it take a kangaroo to travel 5 kilometer

Mathematics

1 answer:

vagabundo [1.1K]3 years ago

8 0

1 Second is how long

You might be interested in

Which equation represents a proportional relationship?

storchak [24]

Answer:

y = 4x + 1

Step-by-step explanation:

I'm sorry if i am wrong

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=15x%5E%7B2%7D" id="TexFormula1" title="15x^{2}" alt="15x^{2}" align="absmiddle" class="latex-f

Genrish500 [490]

Answer:

not sure of what you were asking so it's simplified below.

Step-by-step explanation:

$15x^2-10x-6x+4\\\\15x^2-16x+4$

Note: best i could do with what was given.

8 0

3 years ago

-3(n-2)=-8(n-2) im solving equations

olga2289 [7]

Here my work! Hope this help!

5 0

3 years ago

Read 2 more answers

For the given functions, (a) express dw/dt as a function of t, both by using the chain rule and by expressing w in terms of t an

ch4aika [34]

By the chain rule,

$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\partial w}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial w}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac{\partial w}{\partial z}\dfrac{\mathrm dz}{\mathrm dt}$

We have

$w=7ye^x-\ln z\implies\begin{cases}\dfrac{\partial w}{\partial x}=7ye^x\\\\\dfrac{\partial w}{\partial y}=7e^x\\\\\dfrac{\partial w}{\partial z}=-\dfrac1z\end{cases}$

and

$\begin{cases}x=\ln(t^2+1)\\y=\tan^{-1}t\\z=e^t\end{cases}\implies\begin{cases}\dfrac{\mathrm dx}{\mathrm dt}=\dfrac{2t}{t^2+1}\\\\\dfrac{\mathrm dy}{\mathrm dt}=\dfrac1{t^2+1}\\\\\dfrac{\mathrm dz}{\mathrm dt}=e^t\end{cases}$