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1 year ago

You took one-third of an hour to walk home. How many minutes did you walk?

Mathematics

2 answers:

miskamm [114]1 year ago

6 0

Answer:

20 minutes

Step-by-step explanation:

60÷3=20

Step 1 you divide 60 minutes by 1/3 which is gonna give you 20 mins

MA_775_DIABLO [31]1 year ago

4 0

<h3>Answer:</h3>

20 min

<h3>Solution:</h3>

We know that an hour has 60 minutes.
In order to find one-third of a hour, we should divide 60 by 3:
60÷3=20 min.
So if you took one-third of an hour to walk home, you walked for 20 minutes.

Hope it helps.

Do comment if you have any query.

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2 years ago

Factor by grouping. 9c^3-12c^2+18c-24

xeze [42]

Answer:

$9c^3 - 12c^2 - 18c - 24= [3c^2 - 6][3c - 4]$

Step-by-step explanation:

Given

$9c^3 - 12c^2 - 18c - 24$

Required

Factor

Group into 2

$[9c^3 - 12c^2] - [18c + 24]$

Factorize each group

$3c^2[3c - 4] - 6[3c - 4]$

Factor out 3c - 4

$[3c^2 - 6][3c - 4]$

Hence:

$9c^3 - 12c^2 - 18c - 24= [3c^2 - 6][3c - 4]$

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2 years ago

Last year the backgammon club had 30 members. This year the club had 24 members. Find the percent of decrease in the number of m

Gnom [1K]

There is 20% decrease in the number of members.

Step-by-step explanation:

Given,

Members last year = Old value = 30

Members this year = New year = 24

Change = New value - Old value

Change = 24 - 30

Change = -6

The negative sign indicates the decrease.

Decrease percent = $\frac{Change}{Old\ value}*100$

Decrease percent = $\frac{6}{30}*100$

Decrease percent = $\frac{600}{30}$

Decrease percent = 20%

There is 20% decrease in the number of members.

Keywords: percentage, division

Learn more about division at:

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2 years ago

For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f

Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0$

We want to find $f$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)$

$\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C$

$\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)$

$\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C$

$\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0$

so $\vec F$ is not conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0$