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4 years ago

Do you round up on kilowatts or down

2 answers:

8 0

If the last number is 5 or larger than 5, then you round up, but if it less than 5, than you round down.
$Ex: 42.591$ , round to nearest hundredth:
Because 1 is less than 5, then we keep the 9 there so the answer is $42.59$

$Ex: 31.5727$ , round the number to the nearest thousand:
Because 7 is greater than 5, then we round the 2 to 3 so the answer will be $31.573$ .
Hoped i helped:)

Stels [109]4 years ago

6 0

It depends. If it's .5 or above, round up.
If it's .4, round down

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Solve the equation<br> -6.8w = 3.4

Westkost [7]

Answer:

-6.8w = 3.4

divide both side by -6.8

the left side become w

and the right side become - 0.5

w = -0.5

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3 years ago

A company collects data on its employees. It finds that

alexdok [17]

Answer:

9 Part-time employees

36 full-time employee

Step-by-step explanation:

20% multiply by 45 = 9

80% multioly by 45 =36

therefore to check if your answers are correct add 9 + 36 which gives you 45, the total number of employees in each company's four department.

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2 years ago

Read 2 more answers

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$