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mrs_skeptik [129]
3 years ago
9

A puppy weighs 9 newtons on Earth and about 2 newtons on the planet Namar. A girl weighs 400 newtons on Earth. About how many ne

wtons does the girl weigh on Namar (to the nearest newton)?
Mathematics
1 answer:
serious [3.7K]3 years ago
4 0

Hi there!

In order to answer your question, you'll need to use the cross product method (let's say that Newtons = N):

9N on Earth = 2N on Namar

400N on Earth = xN on Namar

(2 × 400) ÷ 9 = xN on Namar

800 ÷ 9 = xN on Namar

88.888.. = xN on Namar


Since your answer must be to the nearest Newton, I'm guessing that you need to round your answer to the nearest whole number. This means that since the number in the tenths column is more than 5 (could also be equal to 5), you need to round the number up.


88.888... rounded to the nearest whole number = 89


Your answer is: The girl weigh 89 newtons on Namar.


There you go! I really hope this helped, if there's anything just let me know! :)

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At what angle does a diffraction grating produce a second-order maximum for light having a first-order maximum at 20.0 degrees?
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Answer:

At 43.2°.

Step-by-step explanation:

To find the angle we need to use the following equation:

d*sin(\theta) = m\lambda

Where:

d: is the separation of the grating

m: is the order of the maximum

λ: is the wavelength

θ: is the angle              

At the first-order maximum (m=1) at 20.0 degrees we have:

\frac{\lambda}{d} = \frac{sin(\theta)}{m} = \frac{sin(20.0)}{1} = 0.342

Now, to produce a second-order maximum (m=2) the angle must be:

sin(\theta) = \frac{\lambda}{d}*m

\theta = arcsin(\frac{\lambda}{d}*m) = arcsin(0.342*2) = 43.2 ^{\circ}

Therefore, the diffraction grating will produce a second-order maximum for the light at 43.2°.    

I hope it helps you!                                                        

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Express the function graphed on the axes below as a piecewise function.
adoni [48]

Answer:

  f(x) = {x-3 for x ≤ -1; -3x+14 for x > 5}

Step-by-step explanation:

To write the piecewise function, we can consider the pieces one at a time. For each, we need to define the domain, and the functional relation.

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<h3>Left Piece</h3>

The domain is the horizontal extent. It is shown as -∞ to -1, with -1 included.

The relation has a slope (rise/run) of +1, and would intersect the y-axis at -3 if it were extended.

The first piece can be written ...

  f(x) = x-3 for x ≤ -1

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<h3>Right Piece</h3>

The domain is shown as 5 to ∞, with 5<em> not included</em>.

The relation is shown as having a slope (rise/run) of (-3)/(1) = -3. If extended, it would intersect the point (5, -1), so we can write the point-slope equation as ...

  y -(-1) = -3(x -5)

  y = -3x +15 -1 = -3x +14

The second piece can be written ...

  f(x) = -3x +14 for x > 5

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<h3>Whole function</h3>

Putting these pieces together, we have ...

  \boxed{f(x)=\begin{cases}x-3&\text{for }x\le-1\\-3x+14&\text{for }5 < x\end{cases}}

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<em>Additional comment</em>

Sometimes it is convenient to write inequalities in number-line order (using < or ≤ symbols). This gives a visual indication of where the variable stands in relation to the limit(s). Perhaps a more conventional way to write the domain for the second piece is, <em>x > 5</em>.

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Answer:

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