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http://mathhelpforum.com/statistics/218368-number-ways-climb-11-stairs-if-you-can-climb-1-2-stairs-every-time.html
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# Thread: Number of ways to climb 11 stairs if you can climb 1 or 2 stairs every time
Hi,
2. ## Re: Number of ways to climb 11 stairs if you can climb 1 or 2 stairs every time
Originally Posted by verasi
$\begin{array}{*{20}{c}}5&{twos}&{}&1&{one}\\4&{two s}&{}&3&{ones}\\3&{twos}&{}&5&{ones}\\2&{twos}&{}& 7&{ones}\\1&{two}&{}&9&{ones}\\0&{twos}&{}&{11}&{o nes}\end{array}$
For example, we can arrange 4 twos and 3 ones in $\frac{7!}{4!\cdot 3!}$ ways.
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Convert Pixel (px) (Font size (CSS))
## Convert Pixel (px)
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Pixel.php
# Convert Pixel (px):
1. Choose the right category from the selection list, in this case 'Font size (CSS)'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Pixel [px]'.
4. The value will then be converted into all units of measurement the calculator is familiar with.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '535 Pixel'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Pixel' or 'px'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Font size (CSS)'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(65 * 91) px'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '535 Pixel + 1605 Pixel' or '84mm x 8cm x 96dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 2.964 197 503 89×1027. For this form of presentation, the number will be segmented into an exponent, here 27, and the actual number, here 2.964 197 503 89. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 2.964 197 503 89E+27. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 2 964 197 503 890 000 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications.
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Sketching vector fields
-EquinoX-
1. Homework Statement
I am asked to sketch the following vector field in the xy-plane
(a) F(r) = 2r
(b) F(r) = -r/||r||3
(c) F(x,y) = 4xi + xj
2. Homework Equations
3. The Attempt at a Solution
Can someone please give me some hints on how to proceed
Related Calculus and Beyond Homework Help News on Phys.org
lanedance
Homework Helper
hi EquinoX
first I would look at the direction and magitudes of the vectors & see if you can spot any symmetries
a good way to start is to sketch the vector at several points as well
-EquinoX-
yes but for F(r) = 2r this is a 3D vector field, so if I am asked to sketch it in an xy plane, do I just look at the i + j direction? and is 2r basically just 2xi + 2yj + 2zk?
because r is xi + yj + zk?
Last edited:
lanedance
Homework Helper
Yeah i think that should do
r = sqrt(x^2 + y^2 + z^2) and in the xy plane z = 0, so this shouldn't affect your r anyway
-EquinoX-
so r = sqrt(x^2)i + sqrt(y^2)j ?
lanedance
Homework Helper
sorry bit of confusion over notation
|r| = sqrt(x^2 + y^2 + z^2)
r = xi + yj + zk
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### Elan Ding¶
Modified: Aug. 17, 2018 $\newcommand{\bs}{\boldsymbol}$ $\newcommand{\argmin}[1]{\underset{\bs{#1}}{\text{arg min}}\,}$ $\newcommand{\argmax}[1]{\underset{\bs{#1}}{\text{arg max}}\,}$ $\newcommand{\tr}{^{\top}}$ $\newcommand{\norm}[1]{\left|\left|\,#1\,\right|\right|}$ $\newcommand{\given}{\,|\,}$
Recall from my previous post, the Forward Stagewise Additive Modeling (Algorithm 1) requires the following "greedy" minimization scheme:
$$(\beta_m, \gamma_m) = \argmin{\beta, \gamma} \sum_{i=1}^N L(r_{m-1}^{(i)}, \beta b(\bs{x}^{(i)}; \gamma)),\tag{1}$$
where $r_{m-1}^{(i)} = y^{(i)} - f_{m-1} (\bs{x}^{(i)})$ is the residual from $f_{m-1}$.
For two-class classification, the function $b(\bs{x}^{(i)}; \gamma_m) = \gamma_{j,m}$ is assumed to be a classification tree whose leaves gave outputs $\gamma_{j,m} \in \{-1, 1\}$ for $j=1,..., J_m$ where $J_m$ is the number of leaves. Using the exponential loss, we derived the AdaBoost algorithm.
Now we are ready to generalize the results. Suppose that in general, we have a tree
$$T(\bs{x}; \Theta) = \sum_{j=1}^J \gamma_j I(\bs{x} \in R_j),$$
where the parameter $\Theta = \{R_j, \gamma_j\}_{j=1}^{J}$ defines the tree's structure and the leaf values. Note that the leaf values $\gamma$ is no longer restricted by the set $\{-1, 1\}$. Hence, in the optimization process, we can drop the parameter $\beta$ in (1). Hence, the Forward Stagewise Additive Modeling, yields the following optimization subproblem in each iteration:
$$\hat{\Theta}_m = \argmin{\Theta_m} \sum_{i=1}^N L(r_{m-1}^{(i)} + T(\bs{x}^{(i)}; \Theta_m)).\tag{2}$$
In general, simple fast algorithms do not exist for solving (2); instead, we can use gradient descent to approximate the solution.
In general, if we are trying to minimize the loss function,
$$L(f) = \sum_{i=1}^N L(y^{(i)}, f(\bs{x}^{(i)})),$$
where $f(\bs{x})$ is any model, such as the sum of trees.
At the $m$th iteration of gradient descent, for each training example $(\bs{x}^{(i)}, y^{(i)})$, we compute the partial derivative evaluated at the previous model $f_{m-1}$:
$$g_{m-1}^{(i)} = \left[\partial_{f(\bs{x}^{(i)})} L(y^{(i)}, f(\bs{x}^{(i)})) \right]_{f(\bs{x}^{(i)}) = f_{m-1}(\bs{x}^{(i)})}.\tag{3}$$
Now, let's go back to (2). Instead of fitting a tree to the residuals $r_{m-1}^{(i)}$, we fit trees to the negative gradient $-g_{m-1}^{(i)}$. Why is this a good idea? Recall that the gradient value (3) is the direction in which the loss function has the fastest rate of decrease. Hence, for the fastest convergence toward the minimum, the residuals $\bs{r}=(r^{(1)},..., r^{(N)})'$ should point in the same direction as the gradient $\bs{g} = (g^{(1)},.., g^{(m)})'$. This gave rise to the Gradient Tree Boosting Algorithm.
Algorithm 1 Gradient Tree Boosting Algorithm.
• Initialize $f_0(\bs{x}) = \argmin{\gamma}\sum_{i=1}^N L(y^{(i)}, \gamma)$.
• For $m=1$ to $M$:
• For $i=1,2,..., N$ compute $$g_{m-1}^{(i)} = \left[\partial_{f(\bs{x}^{(i)})} L(y^{(i)}, f(\bs{x}^{(i)})) \right]_{f(\bs{x}^{(i)}) = f_{m-1}(\bs{x}^{(i)})}.$$
• Fit a regression tree to the netative gradients $-g_{m-1}^{(i)}$, producing leaves $R_{j,m}, j=1,..., J_m$.
• For $j=1,2,..., J_m$, compute $$\gamma_{j, m} = \argmin{\gamma} \sum_{\bs{x}^{(i)} \in R_{j, m}} L(y^{(i)}, f_{m-1}(\bs{x}^{(i)})+ \gamma).$$
• Update $f_m(\bs{x}) = f_{m-1}(\bs{x}) + \sum_{j=1}^{J_m} \gamma_{j,m} I(\bs{x}\in R_{j, m}).$
• Output $f_M(\bs{x})$.
Note that for a regression problem, if we use RSS as the loss function, then the negative gradient $-g_{m-1}^{(i)}$ is
$$-\partial_{f(\bs{x}^{(i)})} \frac{1}{2}(y^{(i)} - f(\bs{x}^{(i)}))^2 = y^{(i)} - f(\bs{x}^{(i)}).$$
This means that for squared loss, using the negative gradient is exactly the same as using the ordinary residual! For $K$-class classification, suppose that the response variable $Y$ takes values in $\mathcal{G} = \{\mathcal{G}_1,..., \mathcal{G}_K\}$. We fit $K$ regressions trees, each producing a score $f_k(\bs{x}^{(i)}), k=1,2,..., K$, and final model uses the softmax function,
\begin{aligned} f(\bs{x}^{(i)}) &= \argmax{k} p_k(\bs{x}^{(i)}) \\ &= \argmax{k} \frac{e^{f_k(\bs{x}^{(i)})}}{\sum_{l=1}^K e^{f_l(\bs{x}^{(i)})}}. \end{aligned}\tag{4}
Using the multinomial cross entropy (cross entropy between the empirical distribution and a multinomial distribution, $p_{\text{model}}$) as the the loss, we have,
\begin{aligned} L(y^{(i)}, f(\bs{x}^{(i)})) &= E_{y\sim \widehat{p}} -\log p_{\text{model}}(y) \\ &= - \sum_{j=1}^K I(y^{(i)}=\mathcal{G}_j) \log p_j(\bs{x}^{(i)}) \\ &= -\sum_{j=1}^K I(y^{(i)}=\mathcal{G}_j) f_k(\bs{x}^{(i)}) + \log\left(\sum_{l=1}^K e^{f_l(\bs{x}^{(i)})}\right). \end{aligned}\tag{5}
Each tree $f_k, k=1,2,..., K$, is fitted to its respective negative gradient given by
\begin{aligned} -g_k^{(i)} &= \partial_{f_k(\bs{x}^{(i)})} L(y^{(i)}, f(\bs{x}^{(i)})) \\ &= I(y^{(i)} = \mathcal{G}_k) - \frac{e^{f_k(\bs{x}^{(i)})}}{\sum_{l=1}^K e^{f_l(\bs{x}^{(i)})}}. \end{aligned}\tag{6}
We have the following algorithm for classification.
Algorithm 2 Gradient Boosting Classification Model.
• Fit $K$ constant models $f_{0,1}, f_{0,1}, ..., f_{0,K}$ by solving $$f_{0,k}(\bs{x}) = \argmin{\gamma}\sum_{i=1}^N L(y^{(i)}, \gamma).$$
• For $m=1$ to $M$:
• For $k=1,2,...,K$,
• For $i=1,2,...,N$ compute: $$-g_{m-1, k}^{(i)} = I(y^{(i)} = \mathcal{G}_k) - \frac{e^{f_{m-1,k}(\bs{x}^{(i)})}}{\sum_{l=1}^K e^{f_{m-1,l}(\bs{x}^{(i)})}}.$$
• Fit a regression tree against the values $-g_{m-1,k}^{(i)}$, producing leaves $R_{j, m, k}, j=1,..., J_{m,k}$.
• For $j=1,2,..., J_{m,k}$, compute $$\gamma_{j, m, k} = \argmin{\gamma} \sum_{\bs{x}^{(i)} \in R_{j, m}} L(y^{(i)}, f_{m-1}(\bs{x}^{(i)})+ \gamma).$$
• Update $f_{m,k}(\bs{x}) = f_{m-1,k}(\bs{x}) + \sum_{j=1}^{J_{m,k}} \gamma_{j,m,k} I(\bs{x}\in R_{j, m,k}).$
• Output the final prediction by $$f_M(\bs{x}) = \argmax{k} \frac{e^{f_{M,k}(\bs{x}^{(i)})}}{\sum_{l=1}^K e^{f_{M,l}(\bs{x}^{(i)})}}.$$
## Implementation¶
Using sklearn, we can easily implement the model on the same dataset we used on the previou post by simply calling
from sklearn.ensemble import GradientBoostingClassifier
gbm.fit(X,y)
The decision boundary is found to be,
In [5]:
plot_decision_boundary(gbm, X, y)
Finally, just out of curiosity, I tried a modern gradient boosting method called XGBoost. You can read the paper here.
from xgboost import XGBClassifier
xgb = XGBClassifier()
xgb.fit(X,y)
Let's visualize the decision boundary.
In [11]:
plot_decision_boundary(xgb, X, y)
Just by looking at the decision boundaries, we see that the XGBoost model captures the true shape of the distribution better than traditional boosting models, such as AdaBoost and Gradient Boosting Machine.
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# SOUND with MATLAB. SOUND INPUT [a, fa, na]= wavread(’mim-44100.wav') Sound data Sampling Frequency #bit representation.
## Presentation on theme: "SOUND with MATLAB. SOUND INPUT [a, fa, na]= wavread(’mim-44100.wav') Sound data Sampling Frequency #bit representation."— Presentation transcript:
SOUND with MATLAB
SOUND INPUT [a, fa, na]= wavread(’mim-44100.wav') Sound data Sampling Frequency #bit representation
a = ….. -0.0001 0.0001 ……. fa = 44100 na = 16 Teknik Multimedia Fakultas Ilmu Komputer UI 3
SAMPLING sound(a, 44100) sound(a, 16000) sound(a, 11025) sound(a, 8000) 4
WRITING SAMPLED SOUND wavwrite(a, 30000, ‘gbush-30000.wav’) PLOTTING plot(a) Teknik Multimedia Fakultas Ilmu Komputer UI 5
Working with Frequency Domain - use discrete cosine transform or discrete fourier transform to convert from wave spatial signal to frequency domain; in matlab we could use fft function transform = fft(a, fs) fs = the number of sample frequency misal transform = fft (a, 41500) Learning Introductory Signal Processing Using Multimedia 6
SOUND with AUDACITY Demonstrated in class
Download ppt "SOUND with MATLAB. SOUND INPUT [a, fa, na]= wavread(’mim-44100.wav') Sound data Sampling Frequency #bit representation."
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# Sum of lengths of a finite number of overlapping segments > length of their union.
1. Nov 20, 2009
### Esran
1. The problem statement, all variables and given/known data
I know this is probably fairly trivial, but for the life of me I cannot remember or reconstruct the proof for the proposition, "The sum of the lengths of a finite number of overlapping open intervals is greater than the length of their union."
2. Relevant equations
Not Applicable.
3. The attempt at a solution
Suppose $$G=\left\{\left(a_{i},b_{i}\right)\right\}^{n}_{i=1}$$ is a finite collection of overlapping open intervals (any two intervals in $$G$$ have nonempty intersection). We wish to show:
$$b_{n}-a_{1}\leq\sum^{n}_{i=1}\left(b_{i}-a_{i}\right)$$.
My current approach is the following:
$$\sum^{n}_{i=1}\left(b_{i}-a_{i}\right)=b_{n}-a_{1}+\left(b_{n-1}+b_{n-2}+\ldots+b_{1}\right)-\left(a_{n}+a_{n-1}+\ldots+a_{2}\right)$$
It remains to show that $$\left(b_{n-1}+b_{n-2}+\ldots+b_{1}\right)-\left(a_{n}+a_{n-1}+\ldots+a_{2}\right)$$ is a positive number.
Well, we know $$\left(b_{n-1}+b_{n-2}+\ldots+b_{2}\right)-\left(a_{n-1}+a_{n-2}+\ldots+a_{2}\right)$$ is a positive number, so we just need to establish that $$b_{1}-a_{n}<\left(b_{n-1}+b_{n-2}+\ldots+b_{2}\right)-\left(a_{n-1}+a_{n-2}+\ldots+a_{2}\right)$$.
This is where I am having problems.
2. Nov 20, 2009
### clamtrox
Re: Sum of lengths of a finite number of overlapping segments > length of their union
$$b_n - a_n + b_{n-1} - a_{n-1} + ... = b_n +(b_{n-1}-a_n) + (b_{n-2} - a_{n-1}) +... -a_1$$
and since the intervals overlap, you can sort them in such a way that $$b_{k-1} > a_k$$ for all k.
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## is possible use some function to find derivatives of a vector?
### jonathan valle (view profile)
on 30 Nov 2012
``` by example:
NO2=( 1.1 2.4 3.3 4.7 5.9 6.0)' that corresponding to depth: Z=(4.5 6.2 8.4 10.3 12.5 14.8)'
I want find d(NO2)/dz and d^2(NO2)/dz^2```
Exist some function that calculate this?
## Products
### Azzi Abdelmalek (view profile)
on 30 Nov 2012
Edited by Azzi Abdelmalek
### Azzi Abdelmalek (view profile)
on 2 Dec 2012
Edit
```NO2=[1.1 2.4 3.3 4.7 5.9 6.0]
Z=[4.5 6.2 8.4 10.3 12.5 14.8]
d1=diff(NO2)./diff(Z)
d2=diff(NO2,2)./diff(Z(2:end)).^2
```
Azzi Abdelmalek
### Azzi Abdelmalek (view profile)
on 2 Dec 2012
1. diff(y)./diff(t) is an approximation of the first derivative g=dy/dt, In general diff(t) is a constant, then diff(y)./diff(t)=cst*diff(y) , with cst=unique(1/diff(t))
2. the second derivative f=d(dy/dt)/dt is approximated by diff(cst*diff(y))./diff(t)=cst*diff(cst*diff(y))=cst^2*diff(diff(y))
3. finally f=diff(diff(y))./diff(t).^2=diff(y,2)./diff(t).^2
Jan Simon
### Jan Simon (view profile)
on 2 Dec 2012
As far as I can see, your approximation is based on the assumption, that Z is equidistant. This is neither the general case, nor does it match the question. Therefore I think, that this approximation in unnecessarily rough, especially if the 2nd derivative is wanted.
```d2 = [-0.0826, 0.1385, -0.0413, -0.2079]
```
Suggest 2nd order method, one-sided differences at the edges:
```d2 = [-0.0912, -0.0563, 0.0126, -0.0555, -0.1354, -0.1116]
```
Azzi Abdelmalek
### Azzi Abdelmalek (view profile)
on 2 Dec 2012
No, even Z is not equidistant, there is no reason that diff(Z) will change at each point, we are not looking for the variation of Z, it's No2. if the approximation is bad, it's because the distance between Z's value is big. To improve the result, maybe we can interpolate.
on 30 Nov 2012
### Jan Simon (view profile)
on 1 Dec 2012
Matlab's GRADIENT is accurate in the fist order only for not equidistant input. See FEX: DGradient and FEX: central_difference.
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math
posted by .
How tall would a glass (in the shape of a right circular cylinder)with a diameter of 3 inches need to be to hold 12 ounces.
• math -
Hint:
1 fl. oz (US) = 29.57 ml
let h=height,
π4²h/4 = 12*29.57
Solve for h.
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8. NCERT Solutions class 12...
# NCERT Solutions class 12 Maths Exercise 11.3
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## NCERT Solutions class 12 Maths Three Dimensional Geometry
Formula for equation number 1 and 2
If is the length of perpendicular from the origin to a plane and is a unit normal vector to the plane, then equation of the plane is (where of course being length is > 0)
1. In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
(a)
(b)
(c)
(d)
Ans. (a) Given: Equation of the plane is
In vector form it is where
(here = 2 > 0)
Here, = (Position vector of point P )
And
Now let us reduce to
Now [Dividing both sides by ]
where
and
Therefore, direction cosines of normal to the plane are coefficients of in , i.e., 0, 0, 1 and length of perpendicular from the origin to the plane is .
(b) Given: Equation of the plane is (here )
i.e., where
Dividing both sides by = , we get
where
and
Therefore direction cosines of the normal to the plane are the coefficients of in , i.e., and length of perpendicular from the origin to the plane is .
(c) Equation of the plane is (here )
i.e., where
Dividing both sides by = , we get
where
and
Therefore direction cosines of the normal to the plane are the coefficients of in , i.e., and length of perpendicular from the origin to the plane is .
(d) Given: Equation of the plane is
(here )
i.e., where
Dividing both sides by = , we get
where
and
Therefore direction cosines of the normal to the plane are the coefficients of in , i.e., and length of perpendicular from the origin to the plane is .
#### 2. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector
Ans. Here
The unit vector perpendicular to the plane is
Also (given)
Therefore the equation of the required plane is
#### 3. Find the Cartesian equation of the following planes:
(a)
(b)
(c)
Ans. (a) Vector equation of the plane is ……….(i)
Putting in eq. (i) as in 3-D, Cartesian equation of the plane is
(b) Since, is the position vector of any arbitrary point P on the plane.
which is the required Cartesian equation.
(c) Vector equation of the plane is
Since, Since, is the position vector of any arbitrary point P on the plane.
which is the required Cartesian equation.
#### 4. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin:
(a)
(b)
(c)
(d)
Ans. (a) Given: Equation of the plane is …….(i) and point is O (0, 0, 0)
Let M be the foot of the perpendicular drawn from the origin (0, 0, 0) to the given plane.
Since, direction ratios of perpendicular OM to plane are coefficients of in , i.e., 2, 3, 4 = (say)
Equation of the perpendicular OM is (say)
Therefore, point M on this line OM is M ………..(ii)
But point M lies on plane (i)
Putting in eq. (i), we have
##### Hence, putting in equation (ii), the coordinates of foot of the perpendicular is
(b) Given: Equation of the plane is …….(i) and point is O (0, 0, 0)
Let M be the foot of the perpendicular drawn from the origin (0, 0, 0) to the given plane.
Since, direction ratios of perpendicular OM to plane are coefficients of in , i.e., 0, 3, 4 = (say)
Equation of the perpendicular OM is (say)
Therefore, point M on this line OM is M ………..(ii)
But point M lies on plane (i)
Putting in eq. (i), we have
Hence, putting in equation (ii), the coordinates of foot of the perpendicular is
(c) Given: Equation of the plane is …….(i) and point is O (0, 0, 0)
Let M be the foot of the perpendicular drawn from the origin (0, 0, 0) to the given plane.
Since, direction ratios of perpendicular OM to plane are coefficients of in , i.e., 1, 1, 1 = (say)
Equation of the perpendicular OM is (say)
Therefore, point M on this line OM is M ………..(ii)
But point M lies on plane (i)
Putting in eq. (i), we have
Hence, putting in equation (ii), the coordinates of foot of the perpendicular is
(d) Given: Equation of the plane is …….(i) and point is O (0, 0, 0)
Let M be the foot of the perpendicular drawn from the origin (0, 0, 0) to the given plane.
Since, direction ratios of perpendicular OM to plane are coefficients of in , i.e., 0, 5, 0 = (say)
Equation of the perpendicular OM is (say)
Therefore, point M on this line OM is M ………..(ii)
But point M lies on plane (i)
Putting in eq. (i), we have
Hence, putting in equation (ii), the coordinates of foot of the perpendicular is .
#### 5. Find the vector and Cartesian equations of the planes
(a) that passes through the point and the normal to the plane is
(b) that passes through the point (1, 4, 6) and the normal vector to the plane is
Ans. (a) Vector form: The given point on the plane is
The position vector of the given point is =
Also Normal vector to the plane is
Vector equation of the required line is
Putting the values of and ,
Cartesian form: The plane passes through the point =
Normal vector to the plane is
##### Direction ratios of normal to the plane are coefficients of in are
Cartesian form of equation of plane is
(b) Vector form: The given point on the plane is
The position vector of the given point is
Also Normal vector to the plane is
Vector equation of the required line is
Putting the values of and ,
Cartesian form: The plane passes through the point =
Normal vector to the plane is
Direction ratios of normal to the plane are coefficients of in are
Cartesian form of equation of plane is
#### 6. Find the equations of the planes that passes through three points:
(a)
(b)
Ans. We know that through three collinear points A, B, C i.e., through a straight line, we can pass an infinite number of planes.
(a) The three given points are A B and C
Now direction ratios of line AB are
=
Again direction ratios of line BC are
=
Now
Since,
Therefore, line AB and BC are parallel and B is their common point.
Points A, B and C are collinear and hence an infinite number of planes can be drawn through the three given collinear points.
(b) The three given points are A B and C
Now direction ratios of line AB are
=
Again direction ratios of line BC are
=
Now
Since,
##### Points A, B and C are not collinear and hence the unique plane can be drawn through the three given collinear points, i.e.,
Expanding along first row,
Hence the equation of required plane is .
#### 7. Find the intercepts cut off by the plane
Ans. Equation of the plane is
Comparing with intercept form , we have which are intercepts cut off by the plane on axis, axis and axis respectively.
#### 8. Find the equation of the plane with intercept 3 on the axis and parallel to ZOX plane.
Ans. Since equation of ZOX plane is
Equation of any plane parallel to ZOX plane is ……….(i)
[ Equation of any plane parallel to the plane is i.e., change only the constant term]
Now, Plane (i) makes an intercept 3 on the axis ( and ) i.e., plane (i) passes through (0, 3, 0).
Putting and in eq. (i),
Putting in eq. (i), equation of required plane is
#### 9. Find the equation of the plane through the intersection of the planes and and the point (2, 2, 1).
Ans. Equations of given planes are and
Since, equation of any plane through the intersection of these two planes is
L.H.S. of plane I + (L.H.S. of plane II) = 0
……….(i)
Now, required plane (i) passes through the point (2, 2, 1).
Putting in eq. (i),
Now putting in eq. (i) of required plane is
#### 10. Find the vector equation of the plane passing through the intersection of the planes and through the point (2, 1, 3).
Ans. Equation of first plane is
……….(i)
Again equation of the second plane is
……….(ii)
Since, equation of any plane passing through the line of intersection of two planes is
L.H.S. of plane I + (L.H.S. of plane II) = 0
……….(iii)
Now, the plane (iii) passes through the point (2, 1, 3) =
Putting this value of in eq. (iii),
Putting in eq. (iii) of required plane is
#### 11. Find the equation of the plane through the line of intersection of the planes and which is perpendicular to the plane
Ans. Equations of the given planes are and
and
Since, equation of any plane passing through the line of intersection of two planes is
L.H.S. of plane I + (L.H.S. of plane II) = 0
……….(i)
According to the question, this plane is perpendicular to the plane
Putting in eq. (i) of required plane is
#### 12. Find the angle between the planes whose vector equations are and
Ans. Equation of one plane is ……….(i)
Comparing this equation with , we have
Normal vector to plane (i) is
Again, equation of second plane is ……….(ii)
Comparing this equation with , we have
Normal vector to plane (i) is
Let be the acute angle between plane (i) and (ii).
angle between normals and to planes (i) and (ii) is also
=
=
#### 13. In the following cases, determine whether the given planes are parallel or perpendicular and in case they are neither, find the angle between them.
(a) and
(b) and
(c) and
(d) and
(e) and
Ans. (a) Equations of the given planes are and
Here,
Since
Therefore, the given two planes are not parallel.
Again = 21 – 5 – 60 = 21 – 65 = –44
Since
Therefore, the given two planes are not perpendicular.
Now let be the angle between the two planes.
=
=
= = =
(b) equations of the given planes are and i.e.,
Here,
Since
Therefore, the given two planes are not parallel.
Again = = 2 – 2 + 0 = 0
Since
Therefore, the given two planes are perpendicular.
(c) equations of the given planes are and
Here,
Since
Therefore, the given two planes are parallel.
##### (d) equations of the given planes are and
Here,
Since
Therefore, the given two planes are parallel.
(e) equations of the given planes are and i.e.,
Here,
Since
Therefore, the given two planes are not parallel.
Again = 4 x 0 + 8 x 1 + 1 x 1 = 0 + 8 + 1 = 9
Since
Therefore, the given two planes are not perpendicular.
Now let be the angle between the two planes.
=
= = = =
#### 14. In the following cases find the distances of each of the given points from the corresponding given plane:
(a) Point (0, 0, 0)
Plane
(b) Point
Plane
(c) Point
Plane
(d) Point
Plane
##### Ans. (a) Distance (of course perpendicular) of the point (0, 0, 0) from the plane
is
=
= =
(b) Length of perpendicular from the point on the plane is
=
= =
##### (c) Length of perpendicular from the point on the plane is
=
= =
(d) Length of perpendicular from the point on the plane is
=
= =
## NCERT Solutions class 12 Maths Exercise 11.3
NCERT Solutions Class 12 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 12 Maths includes text book solutions from both part 1 and part 2. NCERT Solutions for CBSE Class 12 Maths have total 13 chapters. 12 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 12 solutions PDF and Maths ncert class 12 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide
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### 3 thoughts on “NCERT Solutions class 12 Maths Exercise 11.3”
1. very nice it is very helpful for me.
2. thanks for it.
3. That was really helpful..thank you…..great help for students
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# Is quadrilateral ABCD is a parallelogram?
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Is quadrilateral ABCD is a parallelogram?
1) All four angles in quadrilateral ABCD are right angles
2) All four sides of quadrilateral ABCD are equal length
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24 Jan 2018, 16:10
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A parallelogram has opposite sides parallel to each other.
Statement 1: All angles are 90 so that means that the quadrilateral is either a square or a rectangle and both have parallel sides equal. Hence A is sufficient.
Statement 2: All sides are equal which means that the qudrilateral is a square, rhombus or a parallelogram and since all of them have two pairs of parallel verteces, they are all parallelograms. Hence B is sufficient.
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GMATPrepNow wrote:
Is quadrilateral ABCD is a parallelogram?
1) All four angles in quadrilateral ABCD are right angles
2) All four sides of quadrilateral ABCD are equal length
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Check Properties of Polygons Questions for similar problems.
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# Integers
Maths
Class 7
## Overview
Suppose you take a jar and add 5 chocolates into the jar. It is represented by a positive value +5. Now remove 5 chocolates from a jar so this is represented by a negative value -5. But can you remove half of or a third of chocolate? NO! So here the integer always represent the number of chocolates added or taken.
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Terri and Mike each bought 70 donuts from a bakery. The total number of donuts they bought can be found using the expression shown below.
$$2 × 70$$
What is the total number of donuts Terri and Mike bought from the bakery?
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They bought 140
by Wooden (186 points)
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weight of one cubic meter of indian coal
#### Z K Morvay, D D Gvozdenac Part III FUNDAMENTALS FOR ,
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FUNDAMENTALS FOR ANALYSIS AND CALCULATION OF ENERGY AND ENVIRONMENTAL PERFORMANCE 1 , 1 1 m3 normal cubic meter n is the weight of gas whose volume is 1 m3 at normal condition 0 oC = 27315 K and 10325 bar 1 m3 s tan dard cubic meter s is frequently used...
#### What Is 1 Ton in Cubic Meters? Reference
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A ton is a measure of weight, whereas a cubic meter is a measure of volume, so the material must be known to be able to convert the units One ton of pure water is equal to 1 cubic meter...
#### Convert volume to weight Coal, Anthracite solid
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Cubic astronomical unit au³ is a volume measurement unit used in astronomy, with sides equal to one astronomical unit 1 au , or with sides equal to 149 597 871 kilometers or 92 955 8072 mil The area measurement was introduced to measure surface of a two-dimensional object...
#### Convert volume to weight Coal, Anthracite broken
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Coal, Anthracite broken weighs 1105 gram per cubic centimeter or 1 105 kilogram per cubic meter, ie it s density is equal to 1 105 kg/m³ In Imperial or US customary measurement system, the Coal, Anthracite broken density is equal to 68982896463 pound per cubic foot lb/ft³ , or 0638730523 ounce per cubic inch oz/inch³...
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How much does one cubic meter of concrete weigh One cubic meter of concrete that will be used for foundation weighs approximately 176 tons General purpose concrete weighs a bit less at 16 tons per cubic , 2007 Average weight of coal is 55 lb per cubic foot 1 tonne of coal would be 075 cubic metres so a square metre of coal with a ....
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Glass For Windowscubic meter volume ,Traditional oven density of broken coal in kilogram per meter cubic The glass for windows calculator for exchange of conversion factorcubic meter mequals =,kilograms kg , broken glass fragments , density per one ....
#### how to convert meter cube to tonnes for crusher run
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1 m3 Crusher Run = 245 Tonnes without wastage , of crusher run per cubic metre? One crusher run per cubic meter is typically the weight of two tons However, this conversion may vary in ,...
#### Wood Charcoal 1 kilogram weight to cubic meters converter
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The answer is The change of 1 kg - kilo kilogram weight unit of wood charcoal measure equals = to volume 00033 m3 cubic meter as the equivalent measure within the ,...
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#### cubic meter to ton conversion of iron ore
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weight 1 cubic meter iron ore ajssjaipur , cubic meter to ton conversion of iron ore 6132 USD Glossary Ministry of Steel Government of India Iron making is the process of Reduction of iron ore , 1,900 tonnes of coal for injection, and 12 million cubic metres of Live Chat REGULATIONS OF PORT DUES AND CHARGES No Kinds Of Cargo Unit ....
#### density of broken coal in kilogram per meter cubic
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What is the density of coal in cubic meter - Answers Density is irrespective of quantityDifferent types of coal have different densities Anthracite has a density of about 1,5 1500 kg/m3 Bituminous has a density of about 1,35 1350 kg/m3 These densities are for the solid oreWhen broken the density is about 3/4 of the solid value...
#### How is the bulk density of coal determined?
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Then average figure kilogram/cubic meter is arrived that is the bulk density of the particular coal seam The bulk density of coal is one of the determining factors to calculate grade of coal For a coal seam the bulk density is determined for every 6 months...
#### How much does a cubic foot of coal weight?
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A cubic foot of coal pieces weighs between 40 and 60 pounds A solid cubic foot of coal can weight up to 84 pouinds A coal hopper rail car contains 100 tons of coal...
#### What Is the Bulk Density of Coal? Reference
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The bulk density of coal depends on the specific type of coal It can range from 40 to 58 pounds per cubic feet, or 641 to 929 kilograms per cubic meter The three types of coal are anthracite coal, bituminous coal and lignite coal...
#### Convert Cubic Yards Of Crusher Run To Tons
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convert cubic yards of crusher run to tons , what does 1 cubic meter of crusher run weight-crusher and mill the weight of 446 cubic meters of , Read more convert tonne crusher run - NO1...
#### Specific Gravity Of General Materials Table
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As 1000kg of pure water 4 176 C = 1 cubic meter, those materials under 1000kg per cubic meter will float more dense materials will obviously sink Those materials have a specific gravity more than 1 Pure water at 4 176 C the maximum density was chosen as the accepted standard for specific gravity and given the value of 1...
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weight of 1 cubic meter of hardstone gravel The weight of gravel per cubic yard is approximately,pounds, ortons, if the gravel is dry If the gravel is out of water, the weight per cubic yard is , Get Price gt gt...
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What does 1 cubic yard of coal weight? Find answers now No 1 Questions Answers Place Weknowtheanswer ABOUT FIND THE ANSWERS What does 1 cubic yard of coal weight? Free e-mail watchdog Tweet Answer this question What does 1 cubic yard of coal weight? Answer for question Your name , Who has written Coal India Limited corporate song?...
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Coal mining processing plant in Nigeria This coal mining project is an open pit mine located in Nigeria, announced by mining company - Western Goldfields - that it has discovered 62,400,000 tonnes of proven reserves of coal deposits worth US 12 billion which could be used for the generation of electric power....
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As 1000kg of pure water = 1 cubic metre, those materials under 1000kg/cum will float more dense will sink ie those materials with a specific gravity more than 1 Pure water was chosen as the base line for specific gravity and given the value of 1...
#### cost of 1 cubic meter of m20 concrete in india
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May 05, 2013 0183 32 Cost of 1 cubic meter of M30 concrete in India How much does a cubic meter of concrete cost? grade concrete will cost about Rs 2900 per meter cubic \n Weight of 1 cubic meter of concrete? , Fly ash is one of the types of coal combustion by- products...
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indonesian coal weight per metre cubic indonesian coal weight per metre cubic We are the manufacturer of coal mining machine,roadheader,coal loader,tunnel mucking loader,backfilling machine,concerte pumping machine and so on indonesian coal weight per metre cubic...
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m25 concrete per m3 cost india - BINQ Mining 48/5 concrete rate per cubic meter in india This will give us the weight of reinforcement steel per cubic meter of concrete ....
#### How do I Convert M3 into Kilograms? Sciencing
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Since 1 kg = 2204 lb, a portion of gold one meter by one meter by one meter about the size of a smallish table would have a mass of 42,537 pounds, in excess of 21 tons Applications Since a modern seafaring ship is made primarily of metal, how does it float?...
#### Cubic Meter To Ton Conversion Of Iron Ore
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Rip Rap Rock Convert Tonne To Meter Cube how many cubic metres of iron ore in metric tonne how many meter cube in 1 tonne of 04 46 AM 1 metric tonne how many cubic meters iron ore rip rap rock rip rap rock - convert tonne to meter cubeHow to...
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weight of one cubic meter of indian coal indonesian coal weight per metre cubic weight of concrete per cubic meter in wikki - Gold Ore Crusher , grams per cubic centimetre cubic , 1 cubic meter weight Know More Convert volume to weight Coal, anthracite broken - AquaCalc...
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This page contains information to help content creators create particular commodities for use with Trainz, by providing basic information as to the density, weight, or volume of particular commodities, so that those commodities can be used realistically in game...
#### what weight is cubic meter of coal
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OIL INDUSTRY CONVERSIONS Reference Barrel Oil Ton Oil Ton Coal,Cubic =xkilocalorie per normal cubic meter what weight is cubic meter of coal ,CR4ThreadCoal to Ash Calculation Volume This does not, however, , weight of one cubic meter of indian coal...
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A cubic foot of coal can weight from under 40 to over 60 pounds, so a cubic yard of coal 27 cubic feet would run from roughly 1,080 to 1,620 pounds How much does a cubic meter of lead weight? According to the Wikipedia, the density of lead is 1134 g/cm 3...
#### Weight Of 1 Cubic Meter Of Crusher Run Stone
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Weight Of 1 Cubic Meter Of Crusher Run Stone , what does 1 cubic meter of crusher run weight i would like to convert the crusher run from tons to cubic meter , Read more Weight Of 1 Yard Of Crusher Run - Protable Plant...
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Class 11
MATHS
Conic Sections
# The range of values of lambda,lambda>0 such that the angle theta between the pair of tangents drawn from (lambda,0) to the circle x^2+y^2=4 lies in (pi/2,(2pi)/3) is (a) (4/(sqrt(3)),2/(sqrt(2))) (b) (0,sqrt(2)) (c) (1,2) (d) none of these
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# Search by Topic
#### Resources tagged with Cubes & cuboids similar to Triple Cubes:
Filter by: Content type:
Age range:
Challenge level:
### There are 34 results
Broad Topics > 3D Geometry, Shape and Space > Cubes & cuboids
### Triple Cubes
##### Age 5 to 11 Challenge Level:
This challenge involves eight three-cube models made from interlocking cubes. Investigate different ways of putting the models together then compare your constructions.
### Holes
##### Age 5 to 11 Challenge Level:
I've made some cubes and some cubes with holes in. This challenge invites you to explore the difference in the number of small cubes I've used. Can you see any patterns?
### Little Boxes
##### Age 7 to 11 Challenge Level:
How many different cuboids can you make when you use four CDs or DVDs? How about using five, then six?
### Four Layers
##### Age 5 to 11 Challenge Level:
Can you create more models that follow these rules?
### Green Cube, Yellow Cube
##### Age 7 to 11 Challenge Level:
How can you paint the faces of these eight cubes so they can be put together to make a 2 x 2 cube that is green all over AND a 2 x 2 cube that is yellow all over?
##### Age 7 to 11 Challenge Level:
Make a cube with three strips of paper. Colour three faces or use the numbers 1 to 6 to make a die.
### Dicey
##### Age 7 to 11 Challenge Level:
A game has a special dice with a colour spot on each face. These three pictures show different views of the same dice. What colour is opposite blue?
### A Puzzling Cube
##### Age 7 to 11 Challenge Level:
Here are the six faces of a cube - in no particular order. Here are three views of the cube. Can you deduce where the faces are in relation to each other and record them on the net of this cube?
##### Age 7 to 11 Challenge Level:
We went to the cinema and decided to buy some bags of popcorn so we asked about the prices. Investigate how much popcorn each bag holds so find out which we might have bought.
### Construct-o-straws
##### Age 7 to 11 Challenge Level:
Make a cube out of straws and have a go at this practical challenge.
### Cube Drilling
##### Age 7 to 11 Challenge Level:
Imagine a 4 by 4 by 4 cube. If you and a friend drill holes in some of the small cubes in the ways described, how many will not have holes drilled through them?
### Three Sets of Cubes, Two Surfaces
##### Age 7 to 11 Challenge Level:
How many models can you find which obey these rules?
### Painted Faces
##### Age 7 to 11 Challenge Level:
Imagine a 3 by 3 by 3 cube made of 9 small cubes. Each face of the large cube is painted a different colour. How many small cubes will have two painted faces? Where are they?
### Start Cube Drilling
##### Age 5 to 7 Challenge Level:
Imagine a 3 by 3 by 3 cube. If you and a friend drill holes in some of the small cubes in the ways described, how many will have holes drilled through them?
### Next Size Up
##### Age 7 to 11 Challenge Level:
The challenge for you is to make a string of six (or more!) graded cubes.
### Three Cubed
##### Age 7 to 11 Challenge Level:
Can you make a 3x3 cube with these shapes made from small cubes?
### Making Cuboids
##### Age 7 to 11 Challenge Level:
Let's say you can only use two different lengths - 2 units and 4 units. Using just these 2 lengths as the edges how many different cuboids can you make?
### Cubes Cut Into Four Pieces
##### Age 5 to 7 Challenge Level:
Eight children each had a cube made from modelling clay. They cut them into four pieces which were all exactly the same shape and size. Whose pieces are the same? Can you decide who made each set?
### Cereal Packets
##### Age 7 to 11 Challenge Level:
How can you put five cereal packets together to make different shapes if you must put them face-to-face?
### Cuboid-in-a-box
##### Age 7 to 11 Challenge Level:
What is the smallest cuboid that you can put in this box so that you cannot fit another that's the same into it?
##### Age 7 to 11 Challenge Level:
If you had 36 cubes, what different cuboids could you make?
### Christmas Presents
##### Age 7 to 11 Challenge Level:
We need to wrap up this cube-shaped present, remembering that we can have no overlaps. What shapes can you find to use?
### More Christmas Boxes
##### Age 7 to 11 Challenge Level:
What size square should you cut out of each corner of a 10 x 10 grid to make the box that would hold the greatest number of cubes?
### The Third Dimension
##### Age 5 to 11 Challenge Level:
Here are four cubes joined together. How many other arrangements of four cubes can you find? Can you draw them on dotty paper?
### Cubes
##### Age 7 to 11 Challenge Level:
How many faces can you see when you arrange these three cubes in different ways?
### Drilling Many Cubes
##### Age 7 to 14 Challenge Level:
A useful visualising exercise which offers opportunities for discussion and generalising, and which could be used for thinking about the formulae needed for generating the results on a spreadsheet.
### Dice, Routes and Pathways
##### Age 5 to 14
This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . .
### All Wrapped Up
##### Age 7 to 11 Challenge Level:
What is the largest cuboid you can wrap in an A3 sheet of paper?
### Always, Sometimes or Never? Shape
##### Age 7 to 11 Challenge Level:
Are these statements always true, sometimes true or never true?
### Thinking 3D
##### Age 7 to 14
How can we as teachers begin to introduce 3D ideas to young children? Where do they start? How can we lay the foundations for a later enthusiasm for working in three dimensions?
### Paper Folding - Models of the Platonic Solids
##### Age 7 to 16
A description of how to make the five Platonic solids out of paper.
### Cubic Conundrum
##### Age 7 to 16 Challenge Level:
Which of the following cubes can be made from these nets?
### Castles in the Middle
##### Age 7 to 14 Challenge Level:
This task depends on groups working collaboratively, discussing and reasoning to agree a final product.
### Which Solid?
##### Age 7 to 16 Challenge Level:
This task develops spatial reasoning skills. By framing and asking questions a member of the team has to find out what mathematical object they have chosen.
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Groups, An Introduction
Introduction
A group is a set of elements S and an operator * possessing the following properties:
1. The binary operator * is a well defined function mapping S cross S into S.
2. The operator * is associative.
3. There is a unique identity element 1, such that x*1 = 1*x = x for every x in S. This identity element is sometimes denoted e. When the group operator is commutative the identity element is often denoted 0, and * is replaced with +.
4. every x has a unique inverse y such that x*y = y*x = 1.
The group is abelian if * is commutative. Why don't they just call it a commutative group? It is called abelian in honor of Niels Abel. (biography)
As with regular multiplication, the star is often omited. Thus x*y is simply written xy, and x*x*x*x is written x4. Realize that x4yx is not x5y, since x and y may not commute.
If the group is abelian, we often use + instead of * and 0 instead of 1. This reminds us that the elements can be rearranged as we wish. For instance, x+x+y+x+y = 3x+2y.
The integers form an abelian group under addition, with 0 as the identity element. The positive rationals form an abelian group under multiplication, with 1 as the identity element.
The integers mod n, Zn, form a group under addition, and the units mod n, denoted Zn*, form a group under multiplication.
Use the clock to illustrate addition mod 12, written Z12, which is an abelian group. Add 4 hours + 3 hours + 7 hours, in any order, and wind up at 2:00. And the inverse of 4 is 8, since 4+8 takes you back to the top. The set of 3 by 3 real orthonormal matrices form a nonabelian group under matrix multiplication. Thinking geometrically, this is the rigid rotations and reflections about the origin in 3 space. A rotation about the x axis and a rotation about the y axis do not commute, but a sequence of 3 rotations is always associative, and every rotation has an inverse. You can always spin the object back to start. Try it with a standard die. Spin it around x, then around y, and compare that with the rotation about y, then about x. The operations do not commute. A semigroup has no identity or inverses, such as the positive reals under addition. A monoid is a semigroup with an identity, such as the non-negative reals. Here 0 is the identity, but without negative numbers we have no inverses. The simplest monoid, that is not a group, consists of 0 and x, where x+x = x.
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# Best answer: How many homes can a nuclear power plant support?
Contents
A single nuclear power reactor generates enough electricity on average to power over 760,000 homes without emitting any greenhouse gases—that’s more than enough to power a city the size of Philadelphia.
## How many houses can nuclear energy power?
HOURS OF ELECTRICITY IN 2017
The United States is the world’s largest producer of nuclear power. It generated just under 805 billion kilowatt hours of electricity in 2017— enough to power 73 million homes. Commercial nuclear power plants have supplied around 20% of the nation’s electricity each year since 1990.
## How many homes can a small nuclear reactor power?
Small modular reactors (SMRs) are nuclear reactor units with an output of up to 300 megawatts of electricity. Since 2010, at least nine states introduced legislation supporting SMR development. A 300-megawatt SMR could generate enough electricity to power approximately 230,000 homes a year.
## How many homes can one power plant support?
For conventional generators, such as a coal plant, a megawatt of capacity will produce electricity that equates to about the same amount of electricity consumed by 400 to 900 homes in a year.
## How much area does a nuclear power plant cover?
A nuclear energy facility has a small area footprint, requiring about 1.3 square miles per 1,000 megawatts of installed capacity.
## How does nuclear energy get to homes?
The same commercial power grid carries electricity from nuclear and fossil-fuel plants as well as renewable sources. A series of power distribution lines carries the electricity from the sources to the end users, including homes, commercial customers, government and industry.
## Can I power my house with nuclear power?
If the question is whether you can generate nuclear energy in your home, scientifically the answer is yes, but legally, the answer is no. Nuclear energy has many risks associated with it and therefore is only produced at centralized large nuclear power plants.
## How many homes can 1 gigawatt power?
One gigawatt is roughly the size of two coal-fired power plants and is enough energy to power 750,000 homes.
## How big is the smallest nuclear reactor?
Bilibino Nuclear power plant (NPP) in the Chukotka Autonomous Okrug, Russia, houses the world’s smallest commercial nuclear reactor. The plant, owned and operated by state-owned Rosenergoatom, is equipped with four EGP-6 light water graphite reactors (LWGR) with gross power capacity of 12MWe each.
## How many homes can a nuclear power plant power in a year?
Supplies electricity each year to serve 60 million homes. Nuclear energy has one of the lowest environmental impacts of any electricity source. A wind farm would need 235 square miles to produce the same amount of electricity as a 1,000-megawatt nuclear power plant.
THIS IS UNIQUE: Which of these have a negative electrical charge?
## How many homes can 1.5 megawatts power?
A 1.5 MW turbine produces about 360,000 kWh per month. That’s enough to power 415 American homes and over 1,100 households in Europe.
## How many homes can one megawatt solar?
To put that number in perspective, the Solar Energy Industries Association (a U.S. trade association) calculates that on average 1 megawatt of solar power generates enough electricity to meet the needs of 164 U.S. homes. 100 megawatts of solar power is thus enough, on average, to power 16,400 U.S. homes.
## Is it safe to live near a nuclear plant?
Yes, is safe to live near Nuclear Power Plant.. The fact is, cancer rates and risks in general are lower around NPP. That has nothing to do with the plant itself, but instead with the higher standard of living of the people who live and work there.
## How many acres of land does a nuclear power plant need?
Thus, we will never see a group of 2077 2-MW (4154 MW name-plate capacity) wind turbines. The 1154-MW nuclear power plant can typically occupy about 50 acres of land, often with a buffer space of land area of at least 1 square mile.
## Is it bad to live near a power plant?
A review of studies over the past 30 years provides a body of evidence that people living near coal-fired power plants have higher death rates and at earlier ages, along with increased risks of respiratory disease, lung cancer, cardiovascular disease and other health problems.
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# First-order logic and second-order logic confusion
First-order logic uses only variables that range over individuals (elements of the domain of discourse); second-order logic has these variables as well as additional variables that range over sets of individuals. For example, the second-order sentence $\forall P\,\forall x (x \in P \lor x \notin P)$ says that for every set $P$ of individuals and every individual $x$, either $x$ is in $P$ or it is not (this is the principle of bivalence). ("Second-order logic", Wikipedia)
And
In the formal language of set theory, the axiom schema is: $\forall w_1,\ldots,w_n \, \forall A \, \exists B \, \forall x \, ( x \in B \Leftrightarrow [ x \in A \land \phi(x, w_1, \ldots, w_n, A) ] )$
Comparing and contrasting these two, I find that in the latter case, set $A$ can be considered as set of individuals that was quanitifed by universal quantifier. So, how is this different from second-order logic (in the first case, I would be referring to set $P$.)?
• So you mean that even in first-order logic, we can allow first-order logic to quantify over sets of individuals - set $A$ in the quoted axiom of specification (when the thing quantified is not a non-set class)? Jun 3, 2012 at 7:13
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longest
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http://www.girlsaskguys.com/other/q1580653-what-would-be-the-height-of-freeboard-on-this-barge
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What would be the height of freeboard on this barge?
If a barge is 3m in height, 50m in length, 10m wide and weighs 15,000 kN
when immersed, what would be the height of the freeboard (waterline to upper deck)?
0|0
12
• how deep is the water?
0|1
0|0
• Assume it's deep enough since rivers and sea coast are pretty deep.
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CC-MAIN-2017-26
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latest
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en
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https://scripts.top4download.com/free-optimum-route/xml/
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## Optimum Route XML scripts
Currently the optimum route - XML keyword combination you have chosen does not match any script on our site. Please search again using a different combination of optimum route - XML; or remove optimum route from search box and browse a scripts category instead.
#### Fixed Start Open Multiple Traveling Salesmen Problem Genetic Algorithm
... up a GA to search for the shortest route (least distance needed for each salesman to travel from the start location to unique individual cities without returning to the starting location).Each salesman starts at the first point, but travels to a unique set of cities after that (and none of ...
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#### Fixed Endpoints Open Multiple Traveling Salesmen Problem Genetic Algorithm
... up a GA to search for the shortest route (least distance needed for each salesman to travel from the start location to unique individual cities and finally to the end location).Each salesman starts at the first point, and ends at the last point, but travels to a unique set of ...
Matlab
#### Fixed Start End Point Multiple Traveling Salesmen Problem Genetic Algorithm
... up a GA to search for the shortest route (least distance needed for each salesman to travel from the start location to individual cities and back to the original starting place).Each salesman starts at the first point, and ends at the first point, but travels to a unique set of ...
Matlab
#### Traveling Salesman Problem Nearest Neighbor
... the starting point.This function determines the Nearest Neighbor routes for multiple starting points and returns the best of those routes. Requirements: · MATLAB 7.6 or higher ...
Matlab
#### Fixed Start Open Traveling Salesman Problem Genetic Algorithm
The algorithm computes the optimal route from a fixed predetermined starting point to all other cities without returning to the starting point. Requirements: · MATLAB 7.6 or higher ...
Matlab
#### GODLIKE
GODLIKE (Global Optimum Determination by Linking and Interchanging Kindred Evaluators) compiles various population-based global optimization schemes.It handles both single- and multi-objective optimization, simply by adding additional objective functions.It solves optimization problems using relatively basic implementations of a genetic algorithm, differential evolution, particle swarm optimization and adaptive simulated annealing algorithms.The script ...
Matlab
#### Gravity Board X
... also click the IP address to trace its route. - Support for an unlimited amount of Categories, Boards, Threads and Posts - Customizable ranking system - Administrators may add ranks and specify the amount of posts needed to receive that rank, as well as the color of the ...
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#### Krumo
... going through your code and removing every dumping route you have ... right ? A lot of times you need to dump the contents of the superglobals (\$_GET, \$_POST, \$_SERVER, \$_SESSION, etc), or see the list of included files, or the declared classes and interfaces, or the defined constants, etc.Krumo ...
#### SMS Me
... mobile phones as well as the usual email route. ...
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https://gemseo.readthedocs.io/en/4.0.1/_modules/gemseo.problems.sellar.sellar.html
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gemseo / problems / sellar
# sellar module¶
The disciplines for the MDO problem proposed by Sellar et al. in
Sellar, R., Batill, S., & Renaud, J. (1996). Response surface based, concurrent subspace optimization for multidisciplinary system design. In 34th aerospace sciences meeting and exhibit (p. 714).
The MDO problem is written as follows:
\begin{split}\begin{aligned} \text{minimize the objective function }&obj=x_{local}^2 + x_{shared,2} +y_1^2+e^{-y_2} \\ \text{with respect to the design variables }&x_{shared},\,x_{local} \\ \text{subject to the general constraints } & c_1 \leq 0\\ & c_2 \leq 0\\ \text{subject to the bound constraints } & -10 \leq x_{shared,1} \leq 10\\ & 0 \leq x_{shared,2} \leq 10\\ & 0 \leq x_{local} \leq 10. \end{aligned}\end{split}
where the coupling variables are
$\text{Discipline 1: } y_1 = \sqrt{x_{shared,1}^2 + x_{shared,2} + x_{local} - 0.2\,y_2},$
and
$\text{Discipline 2: }y_2 = |y_1| + x_{shared,1} + x_{shared,2}.$
and where the general constraints are
\begin{align}\begin{aligned}c_1 = 3.16 - y_1^2\\c_2 = y_2 - 24\end{aligned}\end{align}
This module implements three disciplines to compute the different coupling variables, constraints and objective:
• Sellar1: this MDODiscipline computes $$y_1$$ from $$y_2$$, $$x_{shared,1}$$, $$x_{shared,2}$$ and $$x_{local}$$.
• Sellar2: this MDODiscipline computes $$y_2$$ from $$y_1$$, $$x_{shared,1}$$ and $$x_{shared,2}$$.
• SellarSystem: this MDODiscipline computes both objective and constraints from $$y_1$$, $$y_2$$, $$x_{local}$$ and $$x_{shared,2}$$.
class gemseo.problems.sellar.sellar.Sellar1[source]
The discipline to compute the coupling variable $$y_1$$.
Parameters
• name – The name of the discipline. If None, use the class name.
• input_grammar_file – The input grammar file path. If None and auto_detect_grammar_files=True, look for "ClassName_input.json" in the GRAMMAR_DIRECTORY if any or in the directory of the discipline class module. If None and auto_detect_grammar_files=False, do not initialize the input grammar from a schema file.
• output_grammar_file – The output grammar file path. If None and auto_detect_grammar_files=True, look for "ClassName_output.json" in the GRAMMAR_DIRECTORY if any or in the directory of the discipline class module. If None and auto_detect_grammar_files=False, do not initialize the output grammar from a schema file.
• auto_detect_grammar_files – Whether to look for "ClassName_{input,output}.json" in the GRAMMAR_DIRECTORY if any or in the directory of the discipline class module when {input,output}_grammar_file is None.
• grammar_type – The type of grammar to define the input and output variables, e.g. MDODiscipline.JSON_GRAMMAR_TYPE or MDODiscipline.SIMPLE_GRAMMAR_TYPE.
• cache_type – The type of policy to cache the discipline evaluations, e.g. MDODiscipline.SIMPLE_CACHE to cache the last one, MDODiscipline.HDF5_CACHE to cache them in a HDF file, or MDODiscipline.MEMORY_FULL_CACHE to cache them in memory. If None or if MDODiscipline.activate_cache is True, do not cache the discipline evaluations.
• cache_file_path – The HDF file path when grammar_type is MDODiscipline.HDF5_CACHE.
Return type
None
classmethod activate_time_stamps()
Activate the time stamps.
For storing start and end times of execution and linearizations.
Return type
None
Add inputs against which to differentiate the outputs.
This method updates MDODiscipline._differentiated_inputs with inputs.
Parameters
inputs (Iterable[str] | None) –
The input variables against which to differentiate the outputs. If None, all the inputs of the discipline are used.
By default it is set to None.
Raises
ValueError – When the inputs wrt which differentiate the discipline are not inputs of the latter.
Return type
None
This method updates MDODiscipline._differentiated_outputs with outputs.
Parameters
outputs (Iterable[str] | None) –
The output variables to be differentiated. If None, all the outputs of the discipline are used.
By default it is set to None.
Raises
ValueError – When the outputs to differentiate are not discipline outputs.
Return type
None
Add a namespace prefix to an existing input grammar element.
The updated input grammar element name will be namespace+:data:~gemseo.core.namespaces.namespace_separator+name.
Parameters
• name (str) – The element name to rename.
• namespace (str) – The name of the namespace.
Add a namespace prefix to an existing output grammar element.
The updated output grammar element name will be namespace+:data:~gemseo.core.namespaces.namespace_separator+name.
Parameters
• name (str) – The element name to rename.
• namespace (str) – The name of the namespace.
Add an observer for the status.
Add an observer for the status to be notified when self changes of status.
Parameters
obs (Any) – The observer to add.
Return type
None
auto_get_grammar_file(is_input=True, name=None, comp_dir=None)
Use a naming convention to associate a grammar file to the discipline.
Search in the directory comp_dir for either an input grammar file named name + "_input.json" or an output grammar file named name + "_output.json".
Parameters
• is_input (bool) –
Whether to search for an input or output grammar file.
By default it is set to True.
• name (str | None) –
The name to be searched in the file names. If None, use the name of the discipline class.
By default it is set to None.
• comp_dir (str | Path | None) –
The directory in which to search the grammar file. If None, use the GRAMMAR_DIRECTORY if any, or the directory of the discipline class module.
By default it is set to None.
Returns
The grammar file path.
Return type
str
check_input_data(input_data, raise_exception=True)
Check the input data validity.
Parameters
• input_data (dict[str, Any]) – The input data needed to execute the discipline according to the discipline input grammar.
• raise_exception (bool) –
Whether to raise on error.
By default it is set to True.
Return type
None
check_jacobian(input_data=None, derr_approx='finite_differences', step=1e-07, threshold=1e-08, linearization_mode='auto', inputs=None, outputs=None, parallel=False, n_processes=2, use_threading=False, wait_time_between_fork=0, auto_set_step=False, plot_result=False, file_path='jacobian_errors.pdf', show=False, fig_size_x=10, fig_size_y=10, reference_jacobian_path=None, save_reference_jacobian=False, indices=None)
Check if the analytical Jacobian is correct with respect to a reference one.
If reference_jacobian_path is not None and save_reference_jacobian is True, compute the reference Jacobian with the approximation method and save it in reference_jacobian_path.
If reference_jacobian_path is not None and save_reference_jacobian is False, do not compute the reference Jacobian but read it from reference_jacobian_path.
If reference_jacobian_path is None, compute the reference Jacobian without saving it.
Parameters
• input_data (dict[str, ndarray] | None) –
The input data needed to execute the discipline according to the discipline input grammar. If None, use the MDODiscipline.default_inputs.
By default it is set to None.
• derr_approx (str) –
The approximation method, either “complex_step” or “finite_differences”.
By default it is set to finite_differences.
• threshold (float) –
The acceptance threshold for the Jacobian error.
By default it is set to 1e-08.
• linearization_mode (str) –
the mode of linearization: direct, adjoint or automated switch depending on dimensions of inputs and outputs (Default value = ‘auto’)
By default it is set to auto.
• inputs (Iterable[str] | None) –
The names of the inputs wrt which to differentiate the outputs.
By default it is set to None.
• outputs (Iterable[str] | None) –
The names of the outputs to be differentiated.
By default it is set to None.
• step (float) –
The differentiation step.
By default it is set to 1e-07.
• parallel (bool) –
Whether to differentiate the discipline in parallel.
By default it is set to False.
• n_processes (int) –
The maximum simultaneous number of threads, if use_threading is True, or processes otherwise, used to parallelize the execution.
By default it is set to 2.
Whether to use threads instead of processes to parallelize the execution; multiprocessing will copy (serialize) all the disciplines, while threading will share all the memory This is important to note if you want to execute the same discipline multiple times, you shall use multiprocessing.
By default it is set to False.
• wait_time_between_fork (float) –
The time waited between two forks of the process / thread.
By default it is set to 0.
• auto_set_step (bool) –
Whether to compute the optimal step for a forward first order finite differences gradient approximation.
By default it is set to False.
• plot_result (bool) –
Whether to plot the result of the validation (computed vs approximated Jacobians).
By default it is set to False.
• file_path (str | Path) –
The path to the output file if plot_result is True.
By default it is set to jacobian_errors.pdf.
• show (bool) –
Whether to open the figure.
By default it is set to False.
• fig_size_x (float) –
The x-size of the figure in inches.
By default it is set to 10.
• fig_size_y (float) –
The y-size of the figure in inches.
By default it is set to 10.
• reference_jacobian_path (str | Path | None) –
The path of the reference Jacobian file.
By default it is set to None.
• save_reference_jacobian (bool) –
Whether to save the reference Jacobian.
By default it is set to False.
• indices (Iterable[int] | None) –
The indices of the inputs and outputs for the different sub-Jacobian matrices, formatted as {variable_name: variable_components} where variable_components can be either an integer, e.g. 2 a sequence of integers, e.g. [0, 3], a slice, e.g. slice(0,3), the ellipsis symbol () or None, which is the same as ellipsis. If a variable name is missing, consider all its components. If None, consider all the components of all the inputs and outputs.
By default it is set to None.
Returns
Whether the analytical Jacobian is correct with respect to the reference one.
check_output_data(raise_exception=True)
Check the output data validity.
Parameters
raise_exception (bool) –
Whether to raise an exception when the data is invalid.
By default it is set to True.
Return type
None
static compute_y_1(x_local, x_shared, y_2)[source]
Evaluate the first coupling equation in functional form.
Parameters
• x_local (numpy.ndarray) – The design variables local to first discipline.
• x_shared (numpy.ndarray) – The shared design variables.
• y_2 (numpy.ndarray) – The coupling variable coming from the second discipline.
Returns
The value of the coupling variable $$y_1$$.
Return type
float
classmethod deactivate_time_stamps()
Deactivate the time stamps.
For storing start and end times of execution and linearizations.
Return type
None
static deserialize(file_path)
Deserialize a discipline from a file.
Parameters
file_path (str | Path) – The path to the file containing the discipline.
Returns
The discipline instance.
Return type
MDODiscipline
execute(input_data=None)
Execute the discipline.
This method executes the discipline:
Parameters
input_data (Mapping[str, Any] | None) –
The input data needed to execute the discipline according to the discipline input grammar. If None, use the MDODiscipline.default_inputs.
By default it is set to None.
Returns
The discipline local data after execution.
Raises
RuntimeError – When residual_variables are declared but self.run_solves_residuals is False. This is not suported yet.
Return type
dict[str, Any]
get_all_inputs()
Return the local input data as a list.
The order is given by MDODiscipline.get_input_data_names().
Returns
The local input data.
Return type
list[Any]
get_all_outputs()
Return the local output data as a list.
The order is given by MDODiscipline.get_output_data_names().
Returns
The local output data.
Return type
list[Any]
get_attributes_to_serialize()
Define the names of the attributes to be serialized.
Returns
The names of the attributes to be serialized.
Return type
list[str]
static get_data_list_from_dict(keys, data_dict)
Filter the dict from a list of keys or a single key.
If keys is a string, then the method return the value associated to the key. If keys is a list of strings, then the method returns a generator of value corresponding to the keys which can be iterated.
Parameters
• keys (str | Iterable) – One or several names.
• data_dict (dict[str, Any]) – The mapping from which to get the data.
Returns
Either a data or a generator of data.
Return type
Any | Generator[Any]
get_disciplines_in_dataflow_chain()
Return the disciplines that must be shown as blocks within the XDSM representation of a chain.
By default, only the discipline itself is shown. This function can be differently implemented for any type of inherited discipline.
Returns
The disciplines shown in the XDSM chain.
Return type
get_expected_dataflow()
Return the expected data exchange sequence.
This method is used for the XDSM representation.
The default expected data exchange sequence is an empty list.
MDOFormulation.get_expected_dataflow
Returns
The data exchange arcs.
Return type
get_expected_workflow()
Return the expected execution sequence.
This method is used for the XDSM representation.
The default expected execution sequence is the execution of the discipline itself.
MDOFormulation.get_expected_workflow
Returns
The expected execution sequence.
Return type
SerialExecSequence
get_input_data(with_namespaces=True)
Return the local input data as a dictionary.
Parameters
with_namespaces
Whether to keep the namespace prefix of the input names, if any.
By default it is set to True.
Returns
The local input data.
Return type
dict[str, Any]
get_input_data_names(with_namespaces=True)
Return the names of the input variables.
Parameters
with_namespaces
Whether to keep the namespace prefix of the input names, if any.
By default it is set to True.
Returns
The names of the input variables.
Return type
list[str]
get_input_output_data_names(with_namespaces=True)
Return the names of the input and output variables.
Args:
with_namespaces: Whether to keep the namespace prefix of the
output names, if any.
Returns
The name of the input and output variables.
Return type
list[str]
get_inputs_asarray()
Return the local output data as a large NumPy array.
The order is the one of MDODiscipline.get_all_outputs().
Returns
The local output data.
Return type
numpy.ndarray
get_inputs_by_name(data_names)
Return the local data associated with input variables.
Parameters
data_names (Iterable[str]) – The names of the input variables.
Returns
The local data for the given input variables.
Raises
ValueError – When a variable is not an input of the discipline.
Return type
list[Any]
get_local_data_by_name(data_names)
Return the local data of the discipline associated with variables names.
Parameters
data_names (Iterable[str]) – The names of the variables.
Returns
The local data associated with the variables names.
Raises
ValueError – When a name is not a discipline input name.
Return type
Generator[Any]
get_output_data(with_namespaces=True)
Return the local output data as a dictionary.
Parameters
with_namespaces
Whether to keep the namespace prefix of the output names, if any.
By default it is set to True.
Returns
The local output data.
Return type
dict[str, Any]
get_output_data_names(with_namespaces=True)
Return the names of the output variables.
Parameters
with_namespaces
Whether to keep the namespace prefix of the output names, if any.
By default it is set to True.
Returns
The names of the output variables.
Return type
list[str]
get_outputs_asarray()
Return the local input data as a large NumPy array.
The order is the one of MDODiscipline.get_all_inputs().
Returns
The local input data.
Return type
numpy.ndarray
get_outputs_by_name(data_names)
Return the local data associated with output variables.
Parameters
data_names (Iterable[str]) – The names of the output variables.
Returns
The local data for the given output variables.
Raises
ValueError – When a variable is not an output of the discipline.
Return type
list[Any]
get_sub_disciplines()
Return the sub-disciplines if any.
Returns
The sub-disciplines.
Return type
is_all_inputs_existing(data_names)
Test if several variables are discipline inputs.
Parameters
data_names (Iterable[str]) – The names of the variables.
Returns
Whether all the variables are discipline inputs.
Return type
bool
is_all_outputs_existing(data_names)
Test if several variables are discipline outputs.
Parameters
data_names (Iterable[str]) – The names of the variables.
Returns
Whether all the variables are discipline outputs.
Return type
bool
is_input_existing(data_name)
Test if a variable is a discipline input.
Parameters
data_name (str) – The name of the variable.
Returns
Whether the variable is a discipline input.
Return type
bool
is_output_existing(data_name)
Test if a variable is a discipline output.
Parameters
data_name (str) – The name of the variable.
Returns
Whether the variable is a discipline output.
Return type
bool
static is_scenario()
Whether the discipline is a scenario.
Return type
bool
linearize(input_data=None, force_all=False, force_no_exec=False)
Execute the linearized version of the code.
Parameters
• input_data (dict[str, Any] | None) –
The input data needed to linearize the discipline according to the discipline input grammar. If None, use the MDODiscipline.default_inputs.
By default it is set to None.
• force_all (bool) –
If False, MDODiscipline._differentiated_inputs and MDODiscipline._differentiated_outputs are used to filter the differentiated variables. otherwise, all outputs are differentiated wrt all inputs.
By default it is set to False.
• force_no_exec (bool) –
If True, the discipline is not re-executed, cache is loaded anyway.
By default it is set to False.
Returns
The Jacobian of the discipline.
Return type
dict[str, dict[str, ndarray]]
notify_status_observers()
Notify all status observers that the status has changed.
Return type
None
remove_status_observer(obs)
Remove an observer for the status.
Parameters
obs (Any) – The observer to remove.
Return type
None
reset_statuses_for_run()
Set all the statuses to MDODiscipline.STATUS_PENDING.
Raises
ValueError – When the discipline cannot be run because of its status.
Return type
None
serialize(file_path)
Serialize the discipline and store it in a file.
Parameters
file_path (str | Path) – The path to the file to store the discipline.
Return type
None
set_cache_policy(cache_type='SimpleCache', cache_tolerance=0.0, cache_hdf_file=None, cache_hdf_node_name=None, is_memory_shared=True)
Set the type of cache to use and the tolerance level.
This method defines when the output data have to be cached according to the distance between the corresponding input data and the input data already cached for which output data are also cached.
The cache can be either a SimpleCache recording the last execution or a cache storing all executions, e.g. MemoryFullCache and HDF5Cache. Caching data can be either in-memory, e.g. SimpleCache and MemoryFullCache, or on the disk, e.g. HDF5Cache.
The attribute CacheFactory.caches provides the available caches types.
Parameters
• cache_type (str) –
The type of cache.
By default it is set to SimpleCache.
• cache_tolerance (float) –
The maximum relative norm of the difference between two input arrays to consider that two input arrays are equal.
By default it is set to 0.0.
• cache_hdf_file (str | Path | None) –
The path to the HDF file to store the data; this argument is mandatory when the MDODiscipline.HDF5_CACHE policy is used.
By default it is set to None.
• cache_hdf_node_name (str | None) –
The name of the HDF file node to store the discipline data. If None, MDODiscipline.name is used.
By default it is set to None.
• is_memory_shared (bool) –
Whether to store the data with a shared memory dictionary, which makes the cache compatible with multiprocessing.
By default it is set to True.
Return type
None
set_disciplines_statuses(status)
Set the sub-disciplines statuses.
To be implemented in subclasses.
Parameters
status (str) – The status.
Return type
None
Set the Jacobian approximation method.
Sets the linearization mode to approx_method, sets the parameters of the approximation for further use when calling MDODiscipline.linearize().
Parameters
• jac_approx_type (str) –
The approximation method, either “complex_step” or “finite_differences”.
By default it is set to finite_differences.
• jax_approx_step (float) –
The differentiation step.
By default it is set to 1e-07.
• jac_approx_n_processes (int) –
The maximum simultaneous number of threads, if jac_approx_use_threading is True, or processes otherwise, used to parallelize the execution.
By default it is set to 1.
Whether to use threads instead of processes to parallelize the execution; multiprocessing will copy (serialize) all the disciplines, while threading will share all the memory This is important to note if you want to execute the same discipline multiple times, you shall use multiprocessing.
By default it is set to False.
• jac_approx_wait_time (float) –
The time waited between two forks of the process / thread.
By default it is set to 0.
Return type
None
set_optimal_fd_step(outputs=None, inputs=None, force_all=False, print_errors=False, numerical_error=2.220446049250313e-16)
Compute the optimal finite-difference step.
Compute the optimal step for a forward first order finite differences gradient approximation. Requires a first evaluation of the perturbed functions values. The optimal step is reached when the truncation error (cut in the Taylor development), and the numerical cancellation errors (round-off when doing f(x+step)-f(x)) are approximately equal.
Warning
This calls the discipline execution twice per input variables.
https://en.wikipedia.org/wiki/Numerical_differentiation and “Numerical Algorithms and Digital Representation”, Knut Morken , Chapter 11, “Numerical Differentiation”
Parameters
• inputs (Iterable[str] | None) –
The inputs wrt which the outputs are linearized. If None, use the MDODiscipline._differentiated_inputs.
By default it is set to None.
• outputs (Iterable[str] | None) –
The outputs to be linearized. If None, use the MDODiscipline._differentiated_outputs.
By default it is set to None.
• force_all (bool) –
Whether to consider all the inputs and outputs of the discipline;
By default it is set to False.
• print_errors (bool) –
Whether to display the estimated errors.
By default it is set to False.
• numerical_error (float) –
The numerical error associated to the calculation of f. By default, this is the machine epsilon (appx 1e-16), but can be higher when the calculation of f requires a numerical resolution.
By default it is set to 2.220446049250313e-16.
Returns
The estimated errors of truncation and cancellation error.
Raises
ValueError – When the Jacobian approximation method has not been set.
store_local_data(**kwargs)
Store discipline data in local data.
Parameters
**kwargs (Any) – The data to be stored in MDODiscipline.local_data.
Return type
None
APPROX_MODES = ['finite_differences', 'complex_step']
AVAILABLE_MODES = ('auto', 'direct', 'adjoint', 'reverse', 'finite_differences', 'complex_step')
AVAILABLE_STATUSES = ['DONE', 'FAILED', 'PENDING', 'RUNNING', 'VIRTUAL']
COMPLEX_STEP = 'complex_step'
FINITE_DIFFERENCES = 'finite_differences'
GRAMMAR_DIRECTORY: ClassVar[str | None] = None
The directory in which to search for the grammar files if not the class one.
HDF5_CACHE = 'HDF5Cache'
JSON_GRAMMAR_TYPE = 'JSONGrammar'
MEMORY_FULL_CACHE = 'MemoryFullCache'
N_CPUS = 2
RE_EXECUTE_DONE_POLICY = 'RE_EXEC_DONE'
RE_EXECUTE_NEVER_POLICY = 'RE_EXEC_NEVER'
SIMPLE_CACHE = 'SimpleCache'
SIMPLE_GRAMMAR_TYPE = 'SimpleGrammar'
STATUS_DONE = 'DONE'
STATUS_FAILED = 'FAILED'
STATUS_PENDING = 'PENDING'
STATUS_RUNNING = 'RUNNING'
STATUS_VIRTUAL = 'VIRTUAL'
activate_cache: bool = True
Whether to cache the discipline evaluations by default.
activate_counters: ClassVar[bool] = True
Whether to activate the counters (execution time, calls and linearizations).
activate_input_data_check: ClassVar[bool] = True
Whether to check the input data respect the input grammar.
activate_output_data_check: ClassVar[bool] = True
Whether to check the output data respect the output grammar.
cache: AbstractCache
The cache containing one or several executions of the discipline according to the cache policy.
property cache_tol: float
The cache input tolerance.
This is the tolerance for equality of the inputs in the cache. If norm(stored_input_data-input_data) <= cache_tol * norm(stored_input_data), the cached data for stored_input_data is returned when calling self.execute(input_data).
Raises
ValueError – When the discipline does not have a cache.
data_processor: DataProcessor
A tool to pre- and post-process discipline data.
property default_inputs: dict[str, Any]
The default inputs.
Raises
TypeError – When the default inputs are not passed as a dictionary.
exec_for_lin: bool
Whether the last execution was due to a linearization.
property exec_time: float | None
The cumulated execution time of the discipline.
This property is multiprocessing safe.
Raises
RuntimeError – When the discipline counters are disabled.
property grammar_type: gemseo.core.grammars.base_grammar.BaseGrammar
The type of grammar to be used for inputs and outputs declaration.
input_grammar: BaseGrammar
The input grammar.
jac: dict[str, dict[str, ndarray]]
The Jacobians of the outputs wrt inputs of the form {output: {input: matrix}}.
property linearization_mode: str
The linearization mode among MDODiscipline.AVAILABLE_MODES.
Raises
ValueError – When the linearization mode is unknown.
property local_data: gemseo.core.discipline_data.DisciplineData
The current input and output data.
property n_calls: int | None
The number of times the discipline was executed.
This property is multiprocessing safe.
Raises
RuntimeError – When the discipline counters are disabled.
property n_calls_linearize: int | None
The number of times the discipline was linearized.
This property is multiprocessing safe.
Raises
RuntimeError – When the discipline counters are disabled.
name: str
The name of the discipline.
output_grammar: BaseGrammar
The output grammar.
re_exec_policy: str
The policy to re-execute the same discipline.
residual_variables: Mapping[str, str]
The output variables mapping to their inputs, to be considered as residuals; they shall be equal to zero.
run_solves_residuals: bool
If True, the run method shall solve the residuals.
property status: str
The status of the discipline.
time_stamps = None
class gemseo.problems.sellar.sellar.Sellar2[source]
The discipline to compute the coupling variable $$y_2$$.
Parameters
• name – The name of the discipline. If None, use the class name.
• input_grammar_file – The input grammar file path. If None and auto_detect_grammar_files=True, look for "ClassName_input.json" in the GRAMMAR_DIRECTORY if any or in the directory of the discipline class module. If None and auto_detect_grammar_files=False, do not initialize the input grammar from a schema file.
• output_grammar_file – The output grammar file path. If None and auto_detect_grammar_files=True, look for "ClassName_output.json" in the GRAMMAR_DIRECTORY if any or in the directory of the discipline class module. If None and auto_detect_grammar_files=False, do not initialize the output grammar from a schema file.
• auto_detect_grammar_files – Whether to look for "ClassName_{input,output}.json" in the GRAMMAR_DIRECTORY if any or in the directory of the discipline class module when {input,output}_grammar_file is None.
• grammar_type – The type of grammar to define the input and output variables, e.g. MDODiscipline.JSON_GRAMMAR_TYPE or MDODiscipline.SIMPLE_GRAMMAR_TYPE.
• cache_type – The type of policy to cache the discipline evaluations, e.g. MDODiscipline.SIMPLE_CACHE to cache the last one, MDODiscipline.HDF5_CACHE to cache them in a HDF file, or MDODiscipline.MEMORY_FULL_CACHE to cache them in memory. If None or if MDODiscipline.activate_cache is True, do not cache the discipline evaluations.
• cache_file_path – The HDF file path when grammar_type is MDODiscipline.HDF5_CACHE.
Return type
None
classmethod activate_time_stamps()
Activate the time stamps.
For storing start and end times of execution and linearizations.
Return type
None
Add inputs against which to differentiate the outputs.
This method updates MDODiscipline._differentiated_inputs with inputs.
Parameters
inputs (Iterable[str] | None) –
The input variables against which to differentiate the outputs. If None, all the inputs of the discipline are used.
By default it is set to None.
Raises
ValueError – When the inputs wrt which differentiate the discipline are not inputs of the latter.
Return type
None
This method updates MDODiscipline._differentiated_outputs with outputs.
Parameters
outputs (Iterable[str] | None) –
The output variables to be differentiated. If None, all the outputs of the discipline are used.
By default it is set to None.
Raises
ValueError – When the outputs to differentiate are not discipline outputs.
Return type
None
Add a namespace prefix to an existing input grammar element.
The updated input grammar element name will be namespace+:data:~gemseo.core.namespaces.namespace_separator+name.
Parameters
• name (str) – The element name to rename.
• namespace (str) – The name of the namespace.
Add a namespace prefix to an existing output grammar element.
The updated output grammar element name will be namespace+:data:~gemseo.core.namespaces.namespace_separator+name.
Parameters
• name (str) – The element name to rename.
• namespace (str) – The name of the namespace.
Add an observer for the status.
Add an observer for the status to be notified when self changes of status.
Parameters
obs (Any) – The observer to add.
Return type
None
auto_get_grammar_file(is_input=True, name=None, comp_dir=None)
Use a naming convention to associate a grammar file to the discipline.
Search in the directory comp_dir for either an input grammar file named name + "_input.json" or an output grammar file named name + "_output.json".
Parameters
• is_input (bool) –
Whether to search for an input or output grammar file.
By default it is set to True.
• name (str | None) –
The name to be searched in the file names. If None, use the name of the discipline class.
By default it is set to None.
• comp_dir (str | Path | None) –
The directory in which to search the grammar file. If None, use the GRAMMAR_DIRECTORY if any, or the directory of the discipline class module.
By default it is set to None.
Returns
The grammar file path.
Return type
str
check_input_data(input_data, raise_exception=True)
Check the input data validity.
Parameters
• input_data (dict[str, Any]) – The input data needed to execute the discipline according to the discipline input grammar.
• raise_exception (bool) –
Whether to raise on error.
By default it is set to True.
Return type
None
check_jacobian(input_data=None, derr_approx='finite_differences', step=1e-07, threshold=1e-08, linearization_mode='auto', inputs=None, outputs=None, parallel=False, n_processes=2, use_threading=False, wait_time_between_fork=0, auto_set_step=False, plot_result=False, file_path='jacobian_errors.pdf', show=False, fig_size_x=10, fig_size_y=10, reference_jacobian_path=None, save_reference_jacobian=False, indices=None)
Check if the analytical Jacobian is correct with respect to a reference one.
If reference_jacobian_path is not None and save_reference_jacobian is True, compute the reference Jacobian with the approximation method and save it in reference_jacobian_path.
If reference_jacobian_path is not None and save_reference_jacobian is False, do not compute the reference Jacobian but read it from reference_jacobian_path.
If reference_jacobian_path is None, compute the reference Jacobian without saving it.
Parameters
• input_data (dict[str, ndarray] | None) –
The input data needed to execute the discipline according to the discipline input grammar. If None, use the MDODiscipline.default_inputs.
By default it is set to None.
• derr_approx (str) –
The approximation method, either “complex_step” or “finite_differences”.
By default it is set to finite_differences.
• threshold (float) –
The acceptance threshold for the Jacobian error.
By default it is set to 1e-08.
• linearization_mode (str) –
the mode of linearization: direct, adjoint or automated switch depending on dimensions of inputs and outputs (Default value = ‘auto’)
By default it is set to auto.
• inputs (Iterable[str] | None) –
The names of the inputs wrt which to differentiate the outputs.
By default it is set to None.
• outputs (Iterable[str] | None) –
The names of the outputs to be differentiated.
By default it is set to None.
• step (float) –
The differentiation step.
By default it is set to 1e-07.
• parallel (bool) –
Whether to differentiate the discipline in parallel.
By default it is set to False.
• n_processes (int) –
The maximum simultaneous number of threads, if use_threading is True, or processes otherwise, used to parallelize the execution.
By default it is set to 2.
Whether to use threads instead of processes to parallelize the execution; multiprocessing will copy (serialize) all the disciplines, while threading will share all the memory This is important to note if you want to execute the same discipline multiple times, you shall use multiprocessing.
By default it is set to False.
• wait_time_between_fork (float) –
The time waited between two forks of the process / thread.
By default it is set to 0.
• auto_set_step (bool) –
Whether to compute the optimal step for a forward first order finite differences gradient approximation.
By default it is set to False.
• plot_result (bool) –
Whether to plot the result of the validation (computed vs approximated Jacobians).
By default it is set to False.
• file_path (str | Path) –
The path to the output file if plot_result is True.
By default it is set to jacobian_errors.pdf.
• show (bool) –
Whether to open the figure.
By default it is set to False.
• fig_size_x (float) –
The x-size of the figure in inches.
By default it is set to 10.
• fig_size_y (float) –
The y-size of the figure in inches.
By default it is set to 10.
• reference_jacobian_path (str | Path | None) –
The path of the reference Jacobian file.
By default it is set to None.
• save_reference_jacobian (bool) –
Whether to save the reference Jacobian.
By default it is set to False.
• indices (Iterable[int] | None) –
The indices of the inputs and outputs for the different sub-Jacobian matrices, formatted as {variable_name: variable_components} where variable_components can be either an integer, e.g. 2 a sequence of integers, e.g. [0, 3], a slice, e.g. slice(0,3), the ellipsis symbol () or None, which is the same as ellipsis. If a variable name is missing, consider all its components. If None, consider all the components of all the inputs and outputs.
By default it is set to None.
Returns
Whether the analytical Jacobian is correct with respect to the reference one.
check_output_data(raise_exception=True)
Check the output data validity.
Parameters
raise_exception (bool) –
Whether to raise an exception when the data is invalid.
By default it is set to True.
Return type
None
static compute_y_2(x_shared, y_1)[source]
Evaluate the second coupling equation in functional form.
Parameters
• x_shared (numpy.ndarray) – The shared design variables.
• y_1 (numpy.ndarray) – The coupling variable coming from the first discipline.
Returns
The value of the coupling variable $$y_2$$.
Return type
float
classmethod deactivate_time_stamps()
Deactivate the time stamps.
For storing start and end times of execution and linearizations.
Return type
None
static deserialize(file_path)
Deserialize a discipline from a file.
Parameters
file_path (str | Path) – The path to the file containing the discipline.
Returns
The discipline instance.
Return type
MDODiscipline
execute(input_data=None)
Execute the discipline.
This method executes the discipline:
Parameters
input_data (Mapping[str, Any] | None) –
The input data needed to execute the discipline according to the discipline input grammar. If None, use the MDODiscipline.default_inputs.
By default it is set to None.
Returns
The discipline local data after execution.
Raises
RuntimeError – When residual_variables are declared but self.run_solves_residuals is False. This is not suported yet.
Return type
dict[str, Any]
get_all_inputs()
Return the local input data as a list.
The order is given by MDODiscipline.get_input_data_names().
Returns
The local input data.
Return type
list[Any]
get_all_outputs()
Return the local output data as a list.
The order is given by MDODiscipline.get_output_data_names().
Returns
The local output data.
Return type
list[Any]
get_attributes_to_serialize()
Define the names of the attributes to be serialized.
Returns
The names of the attributes to be serialized.
Return type
list[str]
static get_data_list_from_dict(keys, data_dict)
Filter the dict from a list of keys or a single key.
If keys is a string, then the method return the value associated to the key. If keys is a list of strings, then the method returns a generator of value corresponding to the keys which can be iterated.
Parameters
• keys (str | Iterable) – One or several names.
• data_dict (dict[str, Any]) – The mapping from which to get the data.
Returns
Either a data or a generator of data.
Return type
Any | Generator[Any]
get_disciplines_in_dataflow_chain()
Return the disciplines that must be shown as blocks within the XDSM representation of a chain.
By default, only the discipline itself is shown. This function can be differently implemented for any type of inherited discipline.
Returns
The disciplines shown in the XDSM chain.
Return type
get_expected_dataflow()
Return the expected data exchange sequence.
This method is used for the XDSM representation.
The default expected data exchange sequence is an empty list.
MDOFormulation.get_expected_dataflow
Returns
The data exchange arcs.
Return type
get_expected_workflow()
Return the expected execution sequence.
This method is used for the XDSM representation.
The default expected execution sequence is the execution of the discipline itself.
MDOFormulation.get_expected_workflow
Returns
The expected execution sequence.
Return type
SerialExecSequence
get_input_data(with_namespaces=True)
Return the local input data as a dictionary.
Parameters
with_namespaces
Whether to keep the namespace prefix of the input names, if any.
By default it is set to True.
Returns
The local input data.
Return type
dict[str, Any]
get_input_data_names(with_namespaces=True)
Return the names of the input variables.
Parameters
with_namespaces
Whether to keep the namespace prefix of the input names, if any.
By default it is set to True.
Returns
The names of the input variables.
Return type
list[str]
get_input_output_data_names(with_namespaces=True)
Return the names of the input and output variables.
Args:
with_namespaces: Whether to keep the namespace prefix of the
output names, if any.
Returns
The name of the input and output variables.
Return type
list[str]
get_inputs_asarray()
Return the local output data as a large NumPy array.
The order is the one of MDODiscipline.get_all_outputs().
Returns
The local output data.
Return type
numpy.ndarray
get_inputs_by_name(data_names)
Return the local data associated with input variables.
Parameters
data_names (Iterable[str]) – The names of the input variables.
Returns
The local data for the given input variables.
Raises
ValueError – When a variable is not an input of the discipline.
Return type
list[Any]
get_local_data_by_name(data_names)
Return the local data of the discipline associated with variables names.
Parameters
data_names (Iterable[str]) – The names of the variables.
Returns
The local data associated with the variables names.
Raises
ValueError – When a name is not a discipline input name.
Return type
Generator[Any]
get_output_data(with_namespaces=True)
Return the local output data as a dictionary.
Parameters
with_namespaces
Whether to keep the namespace prefix of the output names, if any.
By default it is set to True.
Returns
The local output data.
Return type
dict[str, Any]
get_output_data_names(with_namespaces=True)
Return the names of the output variables.
Parameters
with_namespaces
Whether to keep the namespace prefix of the output names, if any.
By default it is set to True.
Returns
The names of the output variables.
Return type
list[str]
get_outputs_asarray()
Return the local input data as a large NumPy array.
The order is the one of MDODiscipline.get_all_inputs().
Returns
The local input data.
Return type
numpy.ndarray
get_outputs_by_name(data_names)
Return the local data associated with output variables.
Parameters
data_names (Iterable[str]) – The names of the output variables.
Returns
The local data for the given output variables.
Raises
ValueError – When a variable is not an output of the discipline.
Return type
list[Any]
get_sub_disciplines()
Return the sub-disciplines if any.
Returns
The sub-disciplines.
Return type
is_all_inputs_existing(data_names)
Test if several variables are discipline inputs.
Parameters
data_names (Iterable[str]) – The names of the variables.
Returns
Whether all the variables are discipline inputs.
Return type
bool
is_all_outputs_existing(data_names)
Test if several variables are discipline outputs.
Parameters
data_names (Iterable[str]) – The names of the variables.
Returns
Whether all the variables are discipline outputs.
Return type
bool
is_input_existing(data_name)
Test if a variable is a discipline input.
Parameters
data_name (str) – The name of the variable.
Returns
Whether the variable is a discipline input.
Return type
bool
is_output_existing(data_name)
Test if a variable is a discipline output.
Parameters
data_name (str) – The name of the variable.
Returns
Whether the variable is a discipline output.
Return type
bool
static is_scenario()
Whether the discipline is a scenario.
Return type
bool
linearize(input_data=None, force_all=False, force_no_exec=False)
Execute the linearized version of the code.
Parameters
• input_data (dict[str, Any] | None) –
The input data needed to linearize the discipline according to the discipline input grammar. If None, use the MDODiscipline.default_inputs.
By default it is set to None.
• force_all (bool) –
If False, MDODiscipline._differentiated_inputs and MDODiscipline._differentiated_outputs are used to filter the differentiated variables. otherwise, all outputs are differentiated wrt all inputs.
By default it is set to False.
• force_no_exec (bool) –
If True, the discipline is not re-executed, cache is loaded anyway.
By default it is set to False.
Returns
The Jacobian of the discipline.
Return type
dict[str, dict[str, ndarray]]
notify_status_observers()
Notify all status observers that the status has changed.
Return type
None
remove_status_observer(obs)
Remove an observer for the status.
Parameters
obs (Any) – The observer to remove.
Return type
None
reset_statuses_for_run()
Set all the statuses to MDODiscipline.STATUS_PENDING.
Raises
ValueError – When the discipline cannot be run because of its status.
Return type
None
serialize(file_path)
Serialize the discipline and store it in a file.
Parameters
file_path (str | Path) – The path to the file to store the discipline.
Return type
None
set_cache_policy(cache_type='SimpleCache', cache_tolerance=0.0, cache_hdf_file=None, cache_hdf_node_name=None, is_memory_shared=True)
Set the type of cache to use and the tolerance level.
This method defines when the output data have to be cached according to the distance between the corresponding input data and the input data already cached for which output data are also cached.
The cache can be either a SimpleCache recording the last execution or a cache storing all executions, e.g. MemoryFullCache and HDF5Cache. Caching data can be either in-memory, e.g. SimpleCache and MemoryFullCache, or on the disk, e.g. HDF5Cache.
The attribute CacheFactory.caches provides the available caches types.
Parameters
• cache_type (str) –
The type of cache.
By default it is set to SimpleCache.
• cache_tolerance (float) –
The maximum relative norm of the difference between two input arrays to consider that two input arrays are equal.
By default it is set to 0.0.
• cache_hdf_file (str | Path | None) –
The path to the HDF file to store the data; this argument is mandatory when the MDODiscipline.HDF5_CACHE policy is used.
By default it is set to None.
• cache_hdf_node_name (str | None) –
The name of the HDF file node to store the discipline data. If None, MDODiscipline.name is used.
By default it is set to None.
• is_memory_shared (bool) –
Whether to store the data with a shared memory dictionary, which makes the cache compatible with multiprocessing.
By default it is set to True.
Return type
None
set_disciplines_statuses(status)
Set the sub-disciplines statuses.
To be implemented in subclasses.
Parameters
status (str) – The status.
Return type
None
Set the Jacobian approximation method.
Sets the linearization mode to approx_method, sets the parameters of the approximation for further use when calling MDODiscipline.linearize().
Parameters
• jac_approx_type (str) –
The approximation method, either “complex_step” or “finite_differences”.
By default it is set to finite_differences.
• jax_approx_step (float) –
The differentiation step.
By default it is set to 1e-07.
• jac_approx_n_processes (int) –
The maximum simultaneous number of threads, if jac_approx_use_threading is True, or processes otherwise, used to parallelize the execution.
By default it is set to 1.
Whether to use threads instead of processes to parallelize the execution; multiprocessing will copy (serialize) all the disciplines, while threading will share all the memory This is important to note if you want to execute the same discipline multiple times, you shall use multiprocessing.
By default it is set to False.
• jac_approx_wait_time (float) –
The time waited between two forks of the process / thread.
By default it is set to 0.
Return type
None
set_optimal_fd_step(outputs=None, inputs=None, force_all=False, print_errors=False, numerical_error=2.220446049250313e-16)
Compute the optimal finite-difference step.
Compute the optimal step for a forward first order finite differences gradient approximation. Requires a first evaluation of the perturbed functions values. The optimal step is reached when the truncation error (cut in the Taylor development), and the numerical cancellation errors (round-off when doing f(x+step)-f(x)) are approximately equal.
Warning
This calls the discipline execution twice per input variables.
https://en.wikipedia.org/wiki/Numerical_differentiation and “Numerical Algorithms and Digital Representation”, Knut Morken , Chapter 11, “Numerical Differentiation”
Parameters
• inputs (Iterable[str] | None) –
The inputs wrt which the outputs are linearized. If None, use the MDODiscipline._differentiated_inputs.
By default it is set to None.
• outputs (Iterable[str] | None) –
The outputs to be linearized. If None, use the MDODiscipline._differentiated_outputs.
By default it is set to None.
• force_all (bool) –
Whether to consider all the inputs and outputs of the discipline;
By default it is set to False.
• print_errors (bool) –
Whether to display the estimated errors.
By default it is set to False.
• numerical_error (float) –
The numerical error associated to the calculation of f. By default, this is the machine epsilon (appx 1e-16), but can be higher when the calculation of f requires a numerical resolution.
By default it is set to 2.220446049250313e-16.
Returns
The estimated errors of truncation and cancellation error.
Raises
ValueError – When the Jacobian approximation method has not been set.
store_local_data(**kwargs)
Store discipline data in local data.
Parameters
**kwargs (Any) – The data to be stored in MDODiscipline.local_data.
Return type
None
APPROX_MODES = ['finite_differences', 'complex_step']
AVAILABLE_MODES = ('auto', 'direct', 'adjoint', 'reverse', 'finite_differences', 'complex_step')
AVAILABLE_STATUSES = ['DONE', 'FAILED', 'PENDING', 'RUNNING', 'VIRTUAL']
COMPLEX_STEP = 'complex_step'
FINITE_DIFFERENCES = 'finite_differences'
GRAMMAR_DIRECTORY: ClassVar[str | None] = None
The directory in which to search for the grammar files if not the class one.
HDF5_CACHE = 'HDF5Cache'
JSON_GRAMMAR_TYPE = 'JSONGrammar'
MEMORY_FULL_CACHE = 'MemoryFullCache'
N_CPUS = 2
RE_EXECUTE_DONE_POLICY = 'RE_EXEC_DONE'
RE_EXECUTE_NEVER_POLICY = 'RE_EXEC_NEVER'
SIMPLE_CACHE = 'SimpleCache'
SIMPLE_GRAMMAR_TYPE = 'SimpleGrammar'
STATUS_DONE = 'DONE'
STATUS_FAILED = 'FAILED'
STATUS_PENDING = 'PENDING'
STATUS_RUNNING = 'RUNNING'
STATUS_VIRTUAL = 'VIRTUAL'
activate_cache: bool = True
Whether to cache the discipline evaluations by default.
activate_counters: ClassVar[bool] = True
Whether to activate the counters (execution time, calls and linearizations).
activate_input_data_check: ClassVar[bool] = True
Whether to check the input data respect the input grammar.
activate_output_data_check: ClassVar[bool] = True
Whether to check the output data respect the output grammar.
cache: AbstractCache
The cache containing one or several executions of the discipline according to the cache policy.
property cache_tol: float
The cache input tolerance.
This is the tolerance for equality of the inputs in the cache. If norm(stored_input_data-input_data) <= cache_tol * norm(stored_input_data), the cached data for stored_input_data is returned when calling self.execute(input_data).
Raises
ValueError – When the discipline does not have a cache.
data_processor: DataProcessor
A tool to pre- and post-process discipline data.
property default_inputs: dict[str, Any]
The default inputs.
Raises
TypeError – When the default inputs are not passed as a dictionary.
exec_for_lin: bool
Whether the last execution was due to a linearization.
property exec_time: float | None
The cumulated execution time of the discipline.
This property is multiprocessing safe.
Raises
RuntimeError – When the discipline counters are disabled.
property grammar_type: gemseo.core.grammars.base_grammar.BaseGrammar
The type of grammar to be used for inputs and outputs declaration.
input_grammar: BaseGrammar
The input grammar.
jac: dict[str, dict[str, ndarray]]
The Jacobians of the outputs wrt inputs of the form {output: {input: matrix}}.
property linearization_mode: str
The linearization mode among MDODiscipline.AVAILABLE_MODES.
Raises
ValueError – When the linearization mode is unknown.
property local_data: gemseo.core.discipline_data.DisciplineData
The current input and output data.
property n_calls: int | None
The number of times the discipline was executed.
This property is multiprocessing safe.
Raises
RuntimeError – When the discipline counters are disabled.
property n_calls_linearize: int | None
The number of times the discipline was linearized.
This property is multiprocessing safe.
Raises
RuntimeError – When the discipline counters are disabled.
name: str
The name of the discipline.
output_grammar: BaseGrammar
The output grammar.
re_exec_policy: str
The policy to re-execute the same discipline.
residual_variables: Mapping[str, str]
The output variables mapping to their inputs, to be considered as residuals; they shall be equal to zero.
run_solves_residuals: bool
If True, the run method shall solve the residuals.
property status: str
The status of the discipline.
time_stamps = None
class gemseo.problems.sellar.sellar.SellarSystem[source]
The discipline to compute the objective and constraints of the Sellar problem.
Parameters
• name – The name of the discipline. If None, use the class name.
• input_grammar_file – The input grammar file path. If None and auto_detect_grammar_files=True, look for "ClassName_input.json" in the GRAMMAR_DIRECTORY if any or in the directory of the discipline class module. If None and auto_detect_grammar_files=False, do not initialize the input grammar from a schema file.
• output_grammar_file – The output grammar file path. If None and auto_detect_grammar_files=True, look for "ClassName_output.json" in the GRAMMAR_DIRECTORY if any or in the directory of the discipline class module. If None and auto_detect_grammar_files=False, do not initialize the output grammar from a schema file.
• auto_detect_grammar_files – Whether to look for "ClassName_{input,output}.json" in the GRAMMAR_DIRECTORY if any or in the directory of the discipline class module when {input,output}_grammar_file is None.
• grammar_type – The type of grammar to define the input and output variables, e.g. MDODiscipline.JSON_GRAMMAR_TYPE or MDODiscipline.SIMPLE_GRAMMAR_TYPE.
• cache_type – The type of policy to cache the discipline evaluations, e.g. MDODiscipline.SIMPLE_CACHE to cache the last one, MDODiscipline.HDF5_CACHE to cache them in a HDF file, or MDODiscipline.MEMORY_FULL_CACHE to cache them in memory. If None or if MDODiscipline.activate_cache is True, do not cache the discipline evaluations.
• cache_file_path – The HDF file path when grammar_type is MDODiscipline.HDF5_CACHE.
Return type
None
classmethod activate_time_stamps()
Activate the time stamps.
For storing start and end times of execution and linearizations.
Return type
None
Add inputs against which to differentiate the outputs.
This method updates MDODiscipline._differentiated_inputs with inputs.
Parameters
inputs (Iterable[str] | None) –
The input variables against which to differentiate the outputs. If None, all the inputs of the discipline are used.
By default it is set to None.
Raises
ValueError – When the inputs wrt which differentiate the discipline are not inputs of the latter.
Return type
None
This method updates MDODiscipline._differentiated_outputs with outputs.
Parameters
outputs (Iterable[str] | None) –
The output variables to be differentiated. If None, all the outputs of the discipline are used.
By default it is set to None.
Raises
ValueError – When the outputs to differentiate are not discipline outputs.
Return type
None
Add a namespace prefix to an existing input grammar element.
The updated input grammar element name will be namespace+:data:~gemseo.core.namespaces.namespace_separator+name.
Parameters
• name (str) – The element name to rename.
• namespace (str) – The name of the namespace.
Add a namespace prefix to an existing output grammar element.
The updated output grammar element name will be namespace+:data:~gemseo.core.namespaces.namespace_separator+name.
Parameters
• name (str) – The element name to rename.
• namespace (str) – The name of the namespace.
Add an observer for the status.
Add an observer for the status to be notified when self changes of status.
Parameters
obs (Any) – The observer to add.
Return type
None
auto_get_grammar_file(is_input=True, name=None, comp_dir=None)
Use a naming convention to associate a grammar file to the discipline.
Search in the directory comp_dir for either an input grammar file named name + "_input.json" or an output grammar file named name + "_output.json".
Parameters
• is_input (bool) –
Whether to search for an input or output grammar file.
By default it is set to True.
• name (str | None) –
The name to be searched in the file names. If None, use the name of the discipline class.
By default it is set to None.
• comp_dir (str | Path | None) –
The directory in which to search the grammar file. If None, use the GRAMMAR_DIRECTORY if any, or the directory of the discipline class module.
By default it is set to None.
Returns
The grammar file path.
Return type
str
check_input_data(input_data, raise_exception=True)
Check the input data validity.
Parameters
• input_data (dict[str, Any]) – The input data needed to execute the discipline according to the discipline input grammar.
• raise_exception (bool) –
Whether to raise on error.
By default it is set to True.
Return type
None
check_jacobian(input_data=None, derr_approx='finite_differences', step=1e-07, threshold=1e-08, linearization_mode='auto', inputs=None, outputs=None, parallel=False, n_processes=2, use_threading=False, wait_time_between_fork=0, auto_set_step=False, plot_result=False, file_path='jacobian_errors.pdf', show=False, fig_size_x=10, fig_size_y=10, reference_jacobian_path=None, save_reference_jacobian=False, indices=None)
Check if the analytical Jacobian is correct with respect to a reference one.
If reference_jacobian_path is not None and save_reference_jacobian is True, compute the reference Jacobian with the approximation method and save it in reference_jacobian_path.
If reference_jacobian_path is not None and save_reference_jacobian is False, do not compute the reference Jacobian but read it from reference_jacobian_path.
If reference_jacobian_path is None, compute the reference Jacobian without saving it.
Parameters
• input_data (dict[str, ndarray] | None) –
The input data needed to execute the discipline according to the discipline input grammar. If None, use the MDODiscipline.default_inputs.
By default it is set to None.
• derr_approx (str) –
The approximation method, either “complex_step” or “finite_differences”.
By default it is set to finite_differences.
• threshold (float) –
The acceptance threshold for the Jacobian error.
By default it is set to 1e-08.
• linearization_mode (str) –
the mode of linearization: direct, adjoint or automated switch depending on dimensions of inputs and outputs (Default value = ‘auto’)
By default it is set to auto.
• inputs (Iterable[str] | None) –
The names of the inputs wrt which to differentiate the outputs.
By default it is set to None.
• outputs (Iterable[str] | None) –
The names of the outputs to be differentiated.
By default it is set to None.
• step (float) –
The differentiation step.
By default it is set to 1e-07.
• parallel (bool) –
Whether to differentiate the discipline in parallel.
By default it is set to False.
• n_processes (int) –
The maximum simultaneous number of threads, if use_threading is True, or processes otherwise, used to parallelize the execution.
By default it is set to 2.
Whether to use threads instead of processes to parallelize the execution; multiprocessing will copy (serialize) all the disciplines, while threading will share all the memory This is important to note if you want to execute the same discipline multiple times, you shall use multiprocessing.
By default it is set to False.
• wait_time_between_fork (float) –
The time waited between two forks of the process / thread.
By default it is set to 0.
• auto_set_step (bool) –
Whether to compute the optimal step for a forward first order finite differences gradient approximation.
By default it is set to False.
• plot_result (bool) –
Whether to plot the result of the validation (computed vs approximated Jacobians).
By default it is set to False.
• file_path (str | Path) –
The path to the output file if plot_result is True.
By default it is set to jacobian_errors.pdf.
• show (bool) –
Whether to open the figure.
By default it is set to False.
• fig_size_x (float) –
The x-size of the figure in inches.
By default it is set to 10.
• fig_size_y (float) –
The y-size of the figure in inches.
By default it is set to 10.
• reference_jacobian_path (str | Path | None) –
The path of the reference Jacobian file.
By default it is set to None.
• save_reference_jacobian (bool) –
Whether to save the reference Jacobian.
By default it is set to False.
• indices (Iterable[int] | None) –
The indices of the inputs and outputs for the different sub-Jacobian matrices, formatted as {variable_name: variable_components} where variable_components can be either an integer, e.g. 2 a sequence of integers, e.g. [0, 3], a slice, e.g. slice(0,3), the ellipsis symbol () or None, which is the same as ellipsis. If a variable name is missing, consider all its components. If None, consider all the components of all the inputs and outputs.
By default it is set to None.
Returns
Whether the analytical Jacobian is correct with respect to the reference one.
check_output_data(raise_exception=True)
Check the output data validity.
Parameters
raise_exception (bool) –
Whether to raise an exception when the data is invalid.
By default it is set to True.
Return type
None
static compute_c_1(y_1)[source]
Evaluate the constraint $$c_1$$.
Parameters
y_1 (numpy.ndarray) – The coupling variable coming from the first discipline.
Returns
The value of the constraint $$c_1$$.
Return type
float
static compute_c_2(y_2)[source]
Evaluate the constraint $$c_2$$.
Parameters
y_2 (numpy.ndarray) – The coupling variable coming from the second discipline.
Returns
The value of the constraint $$c_2$$.
Return type
float
static compute_obj(x_local, x_shared, y_1, y_2)[source]
Evaluate the objective $$obj$$.
Parameters
• x_local (numpy.ndarray) – The design variables local to the first discipline.
• x_shared (numpy.ndarray) – The shared design variables.
• y_1 (numpy.ndarray) – The coupling variable coming from the first discipline.
• y_2 (numpy.ndarray) – The coupling variable coming from the second discipline.
Returns
The value of the objective $$obj$$.
Return type
float
classmethod deactivate_time_stamps()
Deactivate the time stamps.
For storing start and end times of execution and linearizations.
Return type
None
static deserialize(file_path)
Deserialize a discipline from a file.
Parameters
file_path (str | Path) – The path to the file containing the discipline.
Returns
The discipline instance.
Return type
MDODiscipline
execute(input_data=None)
Execute the discipline.
This method executes the discipline:
Parameters
input_data (Mapping[str, Any] | None) –
The input data needed to execute the discipline according to the discipline input grammar. If None, use the MDODiscipline.default_inputs.
By default it is set to None.
Returns
The discipline local data after execution.
Raises
RuntimeError – When residual_variables are declared but self.run_solves_residuals is False. This is not suported yet.
Return type
dict[str, Any]
get_all_inputs()
Return the local input data as a list.
The order is given by MDODiscipline.get_input_data_names().
Returns
The local input data.
Return type
list[Any]
get_all_outputs()
Return the local output data as a list.
The order is given by MDODiscipline.get_output_data_names().
Returns
The local output data.
Return type
list[Any]
get_attributes_to_serialize()
Define the names of the attributes to be serialized.
Returns
The names of the attributes to be serialized.
Return type
list[str]
static get_data_list_from_dict(keys, data_dict)
Filter the dict from a list of keys or a single key.
If keys is a string, then the method return the value associated to the key. If keys is a list of strings, then the method returns a generator of value corresponding to the keys which can be iterated.
Parameters
• keys (str | Iterable) – One or several names.
• data_dict (dict[str, Any]) – The mapping from which to get the data.
Returns
Either a data or a generator of data.
Return type
Any | Generator[Any]
get_disciplines_in_dataflow_chain()
Return the disciplines that must be shown as blocks within the XDSM representation of a chain.
By default, only the discipline itself is shown. This function can be differently implemented for any type of inherited discipline.
Returns
The disciplines shown in the XDSM chain.
Return type
get_expected_dataflow()
Return the expected data exchange sequence.
This method is used for the XDSM representation.
The default expected data exchange sequence is an empty list.
MDOFormulation.get_expected_dataflow
Returns
The data exchange arcs.
Return type
get_expected_workflow()
Return the expected execution sequence.
This method is used for the XDSM representation.
The default expected execution sequence is the execution of the discipline itself.
MDOFormulation.get_expected_workflow
Returns
The expected execution sequence.
Return type
SerialExecSequence
get_input_data(with_namespaces=True)
Return the local input data as a dictionary.
Parameters
with_namespaces
Whether to keep the namespace prefix of the input names, if any.
By default it is set to True.
Returns
The local input data.
Return type
dict[str, Any]
get_input_data_names(with_namespaces=True)
Return the names of the input variables.
Parameters
with_namespaces
Whether to keep the namespace prefix of the input names, if any.
By default it is set to True.
Returns
The names of the input variables.
Return type
list[str]
get_input_output_data_names(with_namespaces=True)
Return the names of the input and output variables.
Args:
with_namespaces: Whether to keep the namespace prefix of the
output names, if any.
Returns
The name of the input and output variables.
Return type
list[str]
get_inputs_asarray()
Return the local output data as a large NumPy array.
The order is the one of MDODiscipline.get_all_outputs().
Returns
The local output data.
Return type
numpy.ndarray
get_inputs_by_name(data_names)
Return the local data associated with input variables.
Parameters
data_names (Iterable[str]) – The names of the input variables.
Returns
The local data for the given input variables.
Raises
ValueError – When a variable is not an input of the discipline.
Return type
list[Any]
get_local_data_by_name(data_names)
Return the local data of the discipline associated with variables names.
Parameters
data_names (Iterable[str]) – The names of the variables.
Returns
The local data associated with the variables names.
Raises
ValueError – When a name is not a discipline input name.
Return type
Generator[Any]
get_output_data(with_namespaces=True)
Return the local output data as a dictionary.
Parameters
with_namespaces
Whether to keep the namespace prefix of the output names, if any.
By default it is set to True.
Returns
The local output data.
Return type
dict[str, Any]
get_output_data_names(with_namespaces=True)
Return the names of the output variables.
Parameters
with_namespaces
Whether to keep the namespace prefix of the output names, if any.
By default it is set to True.
Returns
The names of the output variables.
Return type
list[str]
get_outputs_asarray()
Return the local input data as a large NumPy array.
The order is the one of MDODiscipline.get_all_inputs().
Returns
The local input data.
Return type
numpy.ndarray
get_outputs_by_name(data_names)
Return the local data associated with output variables.
Parameters
data_names (Iterable[str]) – The names of the output variables.
Returns
The local data for the given output variables.
Raises
ValueError – When a variable is not an output of the discipline.
Return type
list[Any]
get_sub_disciplines()
Return the sub-disciplines if any.
Returns
The sub-disciplines.
Return type
is_all_inputs_existing(data_names)
Test if several variables are discipline inputs.
Parameters
data_names (Iterable[str]) – The names of the variables.
Returns
Whether all the variables are discipline inputs.
Return type
bool
is_all_outputs_existing(data_names)
Test if several variables are discipline outputs.
Parameters
data_names (Iterable[str]) – The names of the variables.
Returns
Whether all the variables are discipline outputs.
Return type
bool
is_input_existing(data_name)
Test if a variable is a discipline input.
Parameters
data_name (str) – The name of the variable.
Returns
Whether the variable is a discipline input.
Return type
bool
is_output_existing(data_name)
Test if a variable is a discipline output.
Parameters
data_name (str) – The name of the variable.
Returns
Whether the variable is a discipline output.
Return type
bool
static is_scenario()
Whether the discipline is a scenario.
Return type
bool
linearize(input_data=None, force_all=False, force_no_exec=False)
Execute the linearized version of the code.
Parameters
• input_data (dict[str, Any] | None) –
The input data needed to linearize the discipline according to the discipline input grammar. If None, use the MDODiscipline.default_inputs.
By default it is set to None.
• force_all (bool) –
If False, MDODiscipline._differentiated_inputs and MDODiscipline._differentiated_outputs are used to filter the differentiated variables. otherwise, all outputs are differentiated wrt all inputs.
By default it is set to False.
• force_no_exec (bool) –
If True, the discipline is not re-executed, cache is loaded anyway.
By default it is set to False.
Returns
The Jacobian of the discipline.
Return type
dict[str, dict[str, ndarray]]
notify_status_observers()
Notify all status observers that the status has changed.
Return type
None
remove_status_observer(obs)
Remove an observer for the status.
Parameters
obs (Any) – The observer to remove.
Return type
None
reset_statuses_for_run()
Set all the statuses to MDODiscipline.STATUS_PENDING.
Raises
ValueError – When the discipline cannot be run because of its status.
Return type
None
serialize(file_path)
Serialize the discipline and store it in a file.
Parameters
file_path (str | Path) – The path to the file to store the discipline.
Return type
None
set_cache_policy(cache_type='SimpleCache', cache_tolerance=0.0, cache_hdf_file=None, cache_hdf_node_name=None, is_memory_shared=True)
Set the type of cache to use and the tolerance level.
This method defines when the output data have to be cached according to the distance between the corresponding input data and the input data already cached for which output data are also cached.
The cache can be either a SimpleCache recording the last execution or a cache storing all executions, e.g. MemoryFullCache and HDF5Cache. Caching data can be either in-memory, e.g. SimpleCache and MemoryFullCache, or on the disk, e.g. HDF5Cache.
The attribute CacheFactory.caches provides the available caches types.
Parameters
• cache_type (str) –
The type of cache.
By default it is set to SimpleCache.
• cache_tolerance (float) –
The maximum relative norm of the difference between two input arrays to consider that two input arrays are equal.
By default it is set to 0.0.
• cache_hdf_file (str | Path | None) –
The path to the HDF file to store the data; this argument is mandatory when the MDODiscipline.HDF5_CACHE policy is used.
By default it is set to None.
• cache_hdf_node_name (str | None) –
The name of the HDF file node to store the discipline data. If None, MDODiscipline.name is used.
By default it is set to None.
• is_memory_shared (bool) –
Whether to store the data with a shared memory dictionary, which makes the cache compatible with multiprocessing.
By default it is set to True.
Return type
None
set_disciplines_statuses(status)
Set the sub-disciplines statuses.
To be implemented in subclasses.
Parameters
status (str) – The status.
Return type
None
Set the Jacobian approximation method.
Sets the linearization mode to approx_method, sets the parameters of the approximation for further use when calling MDODiscipline.linearize().
Parameters
• jac_approx_type (str) –
The approximation method, either “complex_step” or “finite_differences”.
By default it is set to finite_differences.
• jax_approx_step (float) –
The differentiation step.
By default it is set to 1e-07.
• jac_approx_n_processes (int) –
The maximum simultaneous number of threads, if jac_approx_use_threading is True, or processes otherwise, used to parallelize the execution.
By default it is set to 1.
Whether to use threads instead of processes to parallelize the execution; multiprocessing will copy (serialize) all the disciplines, while threading will share all the memory This is important to note if you want to execute the same discipline multiple times, you shall use multiprocessing.
By default it is set to False.
• jac_approx_wait_time (float) –
The time waited between two forks of the process / thread.
By default it is set to 0.
Return type
None
set_optimal_fd_step(outputs=None, inputs=None, force_all=False, print_errors=False, numerical_error=2.220446049250313e-16)
Compute the optimal finite-difference step.
Compute the optimal step for a forward first order finite differences gradient approximation. Requires a first evaluation of the perturbed functions values. The optimal step is reached when the truncation error (cut in the Taylor development), and the numerical cancellation errors (round-off when doing f(x+step)-f(x)) are approximately equal.
Warning
This calls the discipline execution twice per input variables.
https://en.wikipedia.org/wiki/Numerical_differentiation and “Numerical Algorithms and Digital Representation”, Knut Morken , Chapter 11, “Numerical Differentiation”
Parameters
• inputs (Iterable[str] | None) –
The inputs wrt which the outputs are linearized. If None, use the MDODiscipline._differentiated_inputs.
By default it is set to None.
• outputs (Iterable[str] | None) –
The outputs to be linearized. If None, use the MDODiscipline._differentiated_outputs.
By default it is set to None.
• force_all (bool) –
Whether to consider all the inputs and outputs of the discipline;
By default it is set to False.
• print_errors (bool) –
Whether to display the estimated errors.
By default it is set to False.
• numerical_error (float) –
The numerical error associated to the calculation of f. By default, this is the machine epsilon (appx 1e-16), but can be higher when the calculation of f requires a numerical resolution.
By default it is set to 2.220446049250313e-16.
Returns
The estimated errors of truncation and cancellation error.
Raises
ValueError – When the Jacobian approximation method has not been set.
store_local_data(**kwargs)
Store discipline data in local data.
Parameters
**kwargs (Any) – The data to be stored in MDODiscipline.local_data.
Return type
None
APPROX_MODES = ['finite_differences', 'complex_step']
AVAILABLE_MODES = ('auto', 'direct', 'adjoint', 'reverse', 'finite_differences', 'complex_step')
AVAILABLE_STATUSES = ['DONE', 'FAILED', 'PENDING', 'RUNNING', 'VIRTUAL']
COMPLEX_STEP = 'complex_step'
FINITE_DIFFERENCES = 'finite_differences'
GRAMMAR_DIRECTORY: ClassVar[str | None] = None
The directory in which to search for the grammar files if not the class one.
HDF5_CACHE = 'HDF5Cache'
JSON_GRAMMAR_TYPE = 'JSONGrammar'
MEMORY_FULL_CACHE = 'MemoryFullCache'
N_CPUS = 2
RE_EXECUTE_DONE_POLICY = 'RE_EXEC_DONE'
RE_EXECUTE_NEVER_POLICY = 'RE_EXEC_NEVER'
SIMPLE_CACHE = 'SimpleCache'
SIMPLE_GRAMMAR_TYPE = 'SimpleGrammar'
STATUS_DONE = 'DONE'
STATUS_FAILED = 'FAILED'
STATUS_PENDING = 'PENDING'
STATUS_RUNNING = 'RUNNING'
STATUS_VIRTUAL = 'VIRTUAL'
activate_cache: bool = True
Whether to cache the discipline evaluations by default.
activate_counters: ClassVar[bool] = True
Whether to activate the counters (execution time, calls and linearizations).
activate_input_data_check: ClassVar[bool] = True
Whether to check the input data respect the input grammar.
activate_output_data_check: ClassVar[bool] = True
Whether to check the output data respect the output grammar.
cache: AbstractCache
The cache containing one or several executions of the discipline according to the cache policy.
property cache_tol: float
The cache input tolerance.
This is the tolerance for equality of the inputs in the cache. If norm(stored_input_data-input_data) <= cache_tol * norm(stored_input_data), the cached data for stored_input_data is returned when calling self.execute(input_data).
Raises
ValueError – When the discipline does not have a cache.
data_processor: DataProcessor
A tool to pre- and post-process discipline data.
property default_inputs: dict[str, Any]
The default inputs.
Raises
TypeError – When the default inputs are not passed as a dictionary.
exec_for_lin: bool
Whether the last execution was due to a linearization.
property exec_time: float | None
The cumulated execution time of the discipline.
This property is multiprocessing safe.
Raises
RuntimeError – When the discipline counters are disabled.
property grammar_type: gemseo.core.grammars.base_grammar.BaseGrammar
The type of grammar to be used for inputs and outputs declaration.
input_grammar: BaseGrammar
The input grammar.
jac: dict[str, dict[str, ndarray]]
The Jacobians of the outputs wrt inputs of the form {output: {input: matrix}}.
property linearization_mode: str
The linearization mode among MDODiscipline.AVAILABLE_MODES.
Raises
ValueError – When the linearization mode is unknown.
property local_data: gemseo.core.discipline_data.DisciplineData
The current input and output data.
property n_calls: int | None
The number of times the discipline was executed.
This property is multiprocessing safe.
Raises
RuntimeError – When the discipline counters are disabled.
property n_calls_linearize: int | None
The number of times the discipline was linearized.
This property is multiprocessing safe.
Raises
RuntimeError – When the discipline counters are disabled.
name: str
The name of the discipline.
output_grammar: BaseGrammar
The output grammar.
re_exec_policy: str
The policy to re-execute the same discipline.
residual_variables: Mapping[str, str]
The output variables mapping to their inputs, to be considered as residuals; they shall be equal to zero.
run_solves_residuals: bool
If True, the run method shall solve the residuals.
property status: str
The status of the discipline.
time_stamps = None
gemseo.problems.sellar.sellar.get_inputs(names=None)[source]
Generate an initial solution for the MDO problem.
Parameters
names (Iterable[str] | None) –
The names of the discipline inputs.
By default it is set to None.
Returns
The default values of the discipline inputs.
Return type
dict[str, ndarray]
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# Power Apps SEQUENCE Function
Description
The SEQUENCE function creates a range of numbers as a single column table. This could be a continuous range like the numbers from 1-to-100 or non-continuous range such as [2, 4, 6, 8 10]. Ranges of dates and letters can be produced as well.
Syntax
Sequence(records, start, step)
• records – Required. Quantity of numbers in the sequence.
• start – Optional. Starting number of the sequence. Default is 1.
• step – Optional. Incremental value. Default is 1
Examples
Numbers
A table with all numbers from 1-to-10
``Sequence(10)``
A table with all multiples of 10 from 0-to-100
``Sequence(10, 0, 10)``
A table with numbers from 10-to-1
``Sequence(10, 10, -1)``
Letters
A table of all letters A-to-Z
``ForAll(Sequence(26, 65, 1), Cha\'r(Value))``
Dates
A table of the next 7 days starting 2020-01-01
``ForAll(Sequence(7), Date(2020, 1, Value))``
A table of the next 5 Saturdays starts 2020-01-04
``ForAll(Sequence(5, 4, 7), Date(2020, 1, Value))``
A table of the first day of each month in the year 2020
``ForAll(Sequence(12), Date(2020, Value, 1))``
A table of the last day of each month in the year 2020
``ForAll(Sequence(12), DateAdd(Date(2020, Value, 1), 1, Months) - 1)``
A table of all of the days in the month of January for the year 2020
``ForAll(Sequence(Day(DateAdd(Date(2020, 1, 1), 1, Months)-1), 1, 1), Date(2020, 1, Value))``
Random
A random number between 1-and-100
``First(Shuffle(Sequence(1, 100, 1))).Value``
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https://www.chegg.com/homework-help/college-students-checking-accounts-typically-write-relativel-chapter-8-problem-20e-solution-9780840068408-exc
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# Student Solutions Manual for Peck/Olsen/Devore's Introduction to Statistics and Data Analysis (4th Edition) View more editions Solutions for Chapter 8 Problem 20EProblem 20E: College students with checking accounts typically write relatively few checks in any given month, whereas nonstudent residents typically write many more checks during a month. Suppose that 50% of a bank’s accounts are held by students and that 50% are held by nonstudent residents. Let x denote the number of checks written in a given month by a randomly selected bank customer. a. Give a sketch of what the probability distribution of x might look like. b. Suppose that the mean value of x is 22.0 and that the standard deviation is 16.5. If a random sample of n = 100 customers is to be selected and denotes the sample average number of checks written during a particular month, where is the sampling distribution of centered, and what is the standard deviation of the distribution? Sketch a rough picture of the sampling distribution. c. Referring to Part (b), what is the approximate probability that is at most 20? at least 25?
• 1191 step-by-step solutions
• Solved by professors & experts
• iOS, Android, & web
Chapter: Problem:
College students with checking accounts typically write relatively few checks in any given month, whereas nonstudent residents typically write many more checks during a month. Suppose that 50% of a bank’s accounts are held by students and that 50% are held by nonstudent residents. Let x denote the number of checks written in a given month by a randomly selected bank customer.
a. Give a sketch of what the probability distribution of x might look like.
b. Suppose that the mean value of x is 22.0 and that the standard deviation is 16.5. If a random sample of n = 100 customers is to be selected and denotes the sample average number of checks written during a particular month, where is the sampling distribution of centered, and what is the standard deviation of the distribution? Sketch a rough picture of the sampling distribution.
c. Referring to Part (b), what is the approximate probability that is at most 20? at least 25?
Step-by-Step Solution:
Chapter: Problem:
• Step 1 of 4
a)
The aim is to sketch the graph of probability distribution.
The graph of probability distribution is shown below:
• Chapter , Problem is solved.
Corresponding Textbook
Student Solutions Manual for Peck/Olsen/Devore's Introduction to Statistics and Data Analysis | 4th Edition
9780840068408ISBN-13: 0840068409ISBN: Authors:
This is an alternate ISBN. View the primary ISBN for: Introduction to Statistics and Data Analysis 4th Edition Textbook Solutions
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Elementary Statistics (12th Edition)
P=0.216. If the P-value is less than the significance level, then this means the rejection of the null hypothesis. Hence:P is more than $\alpha=0.1$, hence we reject the null hypothesis. Hence we can say that there is sufficient evidence to reject that there is independence between the treatment and the fact that the patient stops smoking.
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Lemma 21.30.5. In Situation 21.30.1 if $(V_ n)$ holds, then for $X$ in $\mathcal{C}$ and $\mathcal{F}$ in $\mathcal{A}_ X$ the map $H^{n + 1}_{\tau '}(X, \epsilon _{X, *}\mathcal{F}) \to H^{n + 1}_\tau (X, \mathcal{F})$ is injective with image those classes which become trivial on a $\tau '$-covering of $X$.
Proof. Recall that $\epsilon _ X^{-1}\epsilon _{X, *}\mathcal{F} = \mathcal{F}$ hence the map is given by pulling back cohomology classes by $\epsilon _ X$. The Leray spectral sequence (Lemma 21.14.5)
$E_2^{p, q} = H^ p_{\tau '}(X, R^ q\epsilon _{X, *}\mathcal{F}) \Rightarrow H^{p + q}_\tau (X, \mathcal{F})$
combined with the assumed vanishing gives an exact sequence
$0 \to H^{n + 1}_{\tau '}(X, \epsilon _{X, *}\mathcal{F}) \to H^{n + 1}_\tau (X, \mathcal{F}) \to H^0_{\tau '}(X, R^{n + 1}\epsilon _{X, *}\mathcal{F})$
This is a restatement of the lemma. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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Physics Notes Onerliner Points for SSC Exams
Physics Notes Onerliner Points for SSC Exams
Today is the Part III of oneliner points on Physics for upcoming SSC CHSL and SSC CGL 2018 Exams. These oneliner points will cover Railway NTPC, Apprentice Etc. Here we collect all the major points from which General questions were frequently asked in SSC and Railway. You can notedown these points in your notebook so it would help you during exams to memorize it easily.
• If Vo be the orbital velocity of a satellite in a circular orbit close to the earth’s surface and Ve is the escape velocity for the earth, then the relation between two is Ve=√2Vo.
• The earth revolves around the sun in one year. If the distance between them becomes double, the new period of revolution will be 2√2 years.
• The escape velocity of a particle of mass m varies as m^0.
• A geostationary satellite should be launched such that it moves that west to east in the equatorial plane.
• The plane of the orbit of an earth satellite passes through the center of the earth.
• The tidal waves in the sea are primarily due to the gravitational effect of the moon on the earth.
• If the earth were to shin faster, acceleration due to gravity at poles remains the same.
• The time period of an artificial satellite in a circular orbit is independent of mass of the satellite.
• A ball is dropped from a space craft revolving around the earth. It will continue to move with the same speed along the original orbit of the space craft.
• When a particle executing S.H.M passes through the mean position, it has maximum K.E and minimum P.E.
• The total energy of a particle vibrating in S.H.M is proportional to the square of its amplitude.
• A simple pendulum is oscillating in a lift. If the lift starts moving upwards with a uniform oscillation, the period will be shorter.
• In order to double the period of a simple pendulum, its length should be quadrupled.
• A small drop falls from the rest from a large height h in air. The final velocity is almost independent of h.
• When the terminal velocity is reached, the acceleration of a body moving through a viscous medium is 0.
• The inside pressure inside a drop or a bubble is inversely proportional to its radius.
• When the temperature increases, the angle of contact of liquid decreases.
• Kerosene oil rises up in the wick of lantern because of surface tension.
• Insect can move on the surface of water without sinking due to surface tension of the water.
• When a body is immersed in a liquid, the force acting on it is upthrust.
• Water drops do not stick to the oily surface due lack of adhesive force.
• Newton’s first law is also known as law of inertia.
• A man inside an artificial satellite feels weightlessness due to force of attraction due to earth is zero.
• Fat can be separated from milk in a cream separator because of centrifugal force.
• If a body moves in constant speed in a circle, then no work is done on it.
• The process of combining string is known as concatenation.
• A bullet hits and gets embedded in a solid block resting on a horizontal frictionless table. Only momentum is conserved in this process.
• The unit of Plank’s constant is JS.
• Kilowatt is a unit of power.
• Lightyear is measurement of stellar distances.
• Very small time intervals can be accurately measured by Atomic Clocks.
• The conservation of momentum helps a rocket to work.
• The value of g on the earth’s surface changes due to both rotation and shape of earth.
• Heat Energy of an object is total energy of the molecules of the object.
• Galileo developed the early thermometers.
• Heat is associated with the K.E of random motion of molecules.
• SI unit of temperature is Kelvin.
• Centigrade and Fahrenheit temperatures are same at 40Degree Celsius.
• The absolute zero on Celsius Scale is 273.15 Degree Celsius.
• For the measurement of temperature of the order of 400degree Celsius, we will prefer Thermocouple.
• The temperature of a gas is measured with a Pyrometer.
• The normal temperature of human body is 37Degree Celsius.
• SI unit of heat is Joule.
• When an object is heated, the molecules of that object began to move faster.
• At 0K the velocity of the molecules is 0.
• Coefficient of linear expansion always remains with the increase in temperature.
• Mercury Thermometers can measure temperature up to 360 Degree Celsius.
Study Materials and Important Notifications
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# What means the velocity constant of the integration in PI controller formulas?
I have an equation of a PI controller. What means here the velocity constant "R" exactly?
$$m(t)=K(e(t)+R{\int}e(t)dt)$$
$$R=1sec^{-1}$$
• See, how beautiful formula I've made for you. :-) Jun 28, 2016 at 8:53
• Thank you very much. But can you explain me what it means? Jun 28, 2016 at 9:30
• Yes, but not on the level of a professional. I've learned it in the University around 15 years ago. But it is a wonderful thing, control theory is much more wonderful as you could have ever imagined. Really, it is not about math. Really, it is not about machines. Really, it is about the life. The whole life is control theory. Jun 28, 2016 at 10:07
• If you don't get answer in some hours, ping me (write a comment beginning with "@peterh"), and I will explain what I can. Jun 28, 2016 at 10:08
• @gamlielbasha It seems that you are mixing the formula in the time domain ($t$) and in the Laplace/frequency domain ($s$). Jun 28, 2016 at 15:05
The $$R$$ in the PI controller equation would best be described as a "weighting factor" that determines the influence of the integral term relative to the proportional term. The larger $$R$$ gets, the more the integrated error affects the output of the controller compared to the instantaneous error.
# More explanation
Often PI controllers are expressed in the following form:
$$u(t) = K e(t) + K_I \int e(t)dt$$
where $$K$$ is called the proportional gain, and $$K_I$$ is called the integral gain. You tune the controller by changing $$K$$ and $$K_I$$ until the system behaves the way you want.
The equation given in the question expresses $$K_I$$ as a multiple of $$K$$, i.e. $$K_I = RK$$. Therefore if you substitute it back into the controller equation:
$$u(t) = K e(t) + RK \int e(t)dt = K (e(t) + R \int e(t)dt)$$
Now, instead of tuning $$K$$ and $$K_I$$ separately, you can look at the process as tuning the proportional gain $$K$$, then deciding how much influence the integral term has by adjusting $$R$$. This way of expressing the gains might be useful depending on how you approach the problem of tuning the controller. Really it's just mathematical semantics, you can use whichever form makes more sense to you.
# Aside
$$R$$ has units of $$\text{s}^{-1}$$ (inverse seconds) in order to make the equation have consistent units. For example if $$e(t)$$ is a distance, then $$\int e(t) dt$$ has units of distance * time. Therefore, $$R$$ must have units of 1/time in order to cancel the time in the integral term.
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# math
number sense what division sentence means the same as the following subtraction sentences?
12-4=8
8-4=4
4-4=0
1. 👍
2. 👎
3. 👁
4. ℹ️
5. 🚩
1. answer to question what division sentence means the same as the following subtraction sentences?
1. 👍
2. 👎
3. ℹ️
4. 🚩
2. 12-4=8 8-4=4 4-4=0
1. 👍
2. 👎
3. ℹ️
4. 🚩
## Similar Questions
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1. Complete two column proof. Given: x/6+2=15 Prove: x=78 x/6+2=15 A:______ x/6=13 B:______ x=78 C:_______ A:a. Given b. Subtraction Property of Equality c. Division Property of Equality B:a. Given b. Addition Property of Equality
2. ### English
In 3–5 sentences, write down what you told someone in that conversation. Once you write your sentences, go back and identify the sentence type of each one. They will be one of these three options: simple sentence, compound
1. For the equation 17h+6=40, select the appropriate property of equality to move the constant term to the right side of the equation. A. Multiplication Property of Equality B. Addition Property of Equality***** C.Subtraction
4. ### Algebra
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2. ### Algebra
Which operation should be performed first in the expression? 5+7÷7x4-17 Addition Subtraction Multiplication Division***** **Is my answer
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Solve this division problem using repeated subtraction. Show all of your steps! 27/9 = ________________ ps... the slash ( / ) means divided by.... 27 divided by 9
2. ### Mathematics
32 and 1 over 4 Division sign 1 over 8 equals 258 Part A: Create a story or context for this number sentence. Part B: Rewrite this number sentence using multiplication. Part C: Give a verbal explanation that describes how these
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Expedition Impossible
Episode Report Card
Pablo G: C | Grade It Now!
Camels! Again!
Commercials
In the night, the wind has picked up greatly. It's a baby sandstorm and some teams find it hard to sleep. The next day, Salmoni states the obvious -- the weather will be a factor and he reiterates today is elimination day. Do we even need this guy? Today's itinerary will have the teams hiking along the nearby river into the wind to a checkpoint where some ancient pillars stand (or were erected for this competition, making them totally not ancient at all). The Moustaches and Rednecks start out simultaneously because they finished thusly yesterday. Somewhat Limited, Akbar And The Players go next, followed by Finding Nino, The S.W.A.T. and Team Gay Angst. Salmoni takes this opportunity to remind Mac how bad of shape she was in yesterday and asks if she's good to go. She says she's OK. NY'sWCE, Grandpa's Last Expedition, WGIYNIKA and Valley Girls bring up the rear. TH for Brittany of Valley Girls rehashes the storyline that the girl teams don't want another girls team going home.
I'm sure this sandstorm is relatively mild as far as sandstorms go, but wow. This thing is rowdy. Low visibility and the teams are having trouble just walking. Kelsey asks Mac how she's doing. She doesn't want to talk about it or think about it. Snip snap. The visibility is so bad The S.W.A.T. is relying almost entirely on GPS to navigate. The Moustaches arrive first. Each "ancient" pillar has a combination lock (yup) on it. The combination is written in code on the scrolls the teams have been carrying since the first checkpoint. To decipher the code, the scroll needs to be wrapped around the pillar candy cane style so the numbers will be visible in the correct order. There is no huge disparity in when the teams arrive at this challenge. Amazingly, The Rednecks figure out the puzzle first. All of the teams start trying to wrap the pillar correctly. Jeff politely explains what's happening to Blind Guy Erik here. Seriously heartwarming -- the bond these guys are sharing through this expedition. Despite that, twice their deciphered code doesn't open the lock. Again, The Rednecks have a breakthrough first. From here, they'll hike one mile to the next checkpoint at a foot bridge where they'll find instructions to the finish line. The rankings at this point are Rednecks, Somewhat Limited, Moustaches, Akbar, NY'sWCE, Team Gay Angst, S.W.A.T., Valley Girls, Grandpa, WGIYNIKA and Finding Nino -- in that order. All teams solve the puzzle pretty much simultaneously except for Finding Nino. They are completely flummoxed and are left standing alone in the blowing sand.
Expedition Impossible
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## [sword finger Offer] sort
Sword finger Offer 40. Minimum number of k Enter the integer array arr to find the minimum number of k. For example, if you enter 8 numbers: 4, 5, 1, 6, 2, 7, 3 and 8, the minimum 4 numbers are 1, 2, 3 and 4. Example 1: Input: arr = [3,2,1], k = 2 Output:[1,2] perhaps [2,1] Example 2: Input: arUTF-8...
Posted by geethalakshmi on Sat, 25 Sep 2021 07:55:20 +0530
## [graphic C language] quick sorting, really good and quick
Quick sort is an exchange sort method of binary tree structure proposed by Hoare in 1962. It is worth mentioning that bigwigs are still active in academia. Introduction to quick sort Let's first get a rough idea of quick sort: The basic idea is: take any element in the element sequence to be soUTF-8...
Posted by Ibnatu on Sat, 25 Sep 2021 23:16:16 +0530
## [data structure and algorithm] search
1, Basic introduction In java, there are four common Searches: 1) Sequential (linear) lookup 2) Binary search / half search 3) Interpolation lookup 4) Fibonacci search 2, Sequential (linear) lookup **Question: * * there is a sequence: {1,9,11, - 1,34,89}. Judge whether the sequence contains a vUTF-8...
Posted by Garrett on Sat, 25 Sep 2021 23:52:30 +0530
## Data structure knowledge points - stack and queue
Stack Definition: a linear table that can only be inserted and deleted at one end Logical structure: it is the same as linear table and still has a one-to-one relationship Storage structure: sequential stack or chain stack can be used for storage, but sequential stack is more common Operation rUTF-8...
Posted by jbruns on Sun, 26 Sep 2021 08:32:52 +0530
## [graphic C language] is just a simple sort
Today we will introduce several common sorting methods: Insert sort Direct insert sort The idea of direct insertion sorting is relatively simple. We divide the array into two parts: ordered area and unnecessary area. We continuously insert the first element of the disordered region into the ordUTF-8...
Posted by Akira on Sun, 26 Sep 2021 12:20:48 +0530
Introduction to linked list ·A linked list is an ordered list ·Linked lists are stored as nodes ·Each node contains the data field and the next field ·The nodes of the linked list are not necessarily stored continuously ·The linked list is divided into the linked list with the leading node and UTF-8...
Posted by Fantast on Sun, 26 Sep 2021 15:27:50 +0530
## Seven sorting algorithms (bubble sort, select sort, insert sort, heap sort, Hill sort, merge sort, quick sort)
Bubble sorting Bubble sorting is the simplest sorting algorithm, called bubble: in fact, it is really like a bubble emerging from the bottom of the water. As the position of the bubble gets closer and closer to the water surface, the bubble will get bigger and bigger. Bubble sorting is to compaUTF-8...
Posted by spivey on Mon, 27 Sep 2021 13:52:13 +0530
## Data structure -- time complexity and space complexity
1, Time complexity The number of basic operations in the algorithm is called the time complexity of the algorithm, which is expressed by T(n). T(n) = O(f(n)), that is, the increase of function time complexity tends to f(n) at most. Large "O" progressive representation (1) Replace all constant UTF-8...
Posted by koughman on Tue, 28 Sep 2021 06:44:38 +0530
## Data structure note 1 - linear table
The pictures in the notes are from the classroom PPT Learning, if there are mistakes, welcome to correct! 1, Properties of linear table 1. There is only one start node, and the start node has no forward node. 2. There is only one end node, and the end node has no successor node. 3. Other nodes UTF-8...
Posted by jamesm6162 on Wed, 29 Sep 2021 01:12:49 +0530
## Java data structures and algorithm prefix, suffix, and suffix expressions (inverse Polish expressions)
@TOP Prefix expression (Polish expression) Prefix expressions, also known as Polish, have operators that precede operands Examples show that the prefix expression corresponding to (3+4)*5-6 is-*3 4 5 6 General expression prefix expression: it is easy to write without details Start at the top, tUTF-8...
Posted by shanx24 on Wed, 29 Sep 2021 21:51:35 +0530
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# cooking yield calculator
Cooking Yield CalculatorIn the culinary world, precise measurements and calculations are essential for ensuring consistent results and efficient operations.One of the key calculations that professional chefs and cooks often encounter is determining the cooking yield.
Published on 01/01/2023Sébastien Vassaux
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Cooking Yield Calculator
In the culinary world, precise measurements and calculations are essential for ensuring consistent results and efficient operations. One of the key calculations that professional chefs and cooks often encounter is determining the cooking yield. Cooking yield refers to the amount of food obtained after cooking, taking into account factors such as moisture loss, shrinkage, and waste.
## Understanding Cooking Yield
To accurately determine the cooking yield, it is important to consider various factors that can affect the final amount of food. These factors include:
• Moisture loss: During the cooking process, moisture evaporates, resulting in a reduction in the overall weight of the food.
• Shrinkage: Certain ingredients, such as meats, shrink when cooked due to the loss of fat and water content.
• Waste: Trimmings, bones, and other inedible parts contribute to the overall waste generated during cooking.
### Calculating Cooking Yield
Calculating the cooking yield requires a systematic approach to account for the aforementioned factors. Here's a step-by-step guide:
### Step 1: Determine the Raw Weight
Weigh the raw ingredient or dish before cooking. This will serve as the starting point for calculating the cooking yield.
### Step 2: Record the Cooked Weight
Once the cooking process is complete, weigh the food again to obtain the cooked weight.
### Step 3: Account for Moisture Loss
To determine the amount of moisture lost during cooking, subtract the cooked weight from the raw weight. This will provide you with the moisture loss in grams or ounces.
### Step 4: Calculate Shrinkage Percentage
Shrinkage percentage can be calculated by dividing the moisture loss by the raw weight and multiplying by 100. This will give you the shrinkage percentage as a decimal or percentage.
### Step 5: Subtract Waste and Trim Weight
If there is any waste or trimmings generated during the cooking process, subtract that weight from the cooked weight. This will give you the net yield, accounting for any inedible parts.
### Step 6: Determine the Cooking Yield Percentage
To calculate the cooking yield percentage, divide the net yield by the raw weight and multiply by 100. This will provide you with the final cooking yield percentage as a decimal or percentage.
## Importance of Cooking Yield Calculation
Accurate cooking yield calculations are crucial for several reasons:
• Recipe Scaling: Knowing the cooking yield allows chefs to adjust recipes based on the desired final quantity, ensuring consistent taste and portions.
• Cost Control: Understanding the cooking yield helps in accurately estimating the cost of ingredients required for a specific yield, preventing unnecessary waste and expenses.
• Menu Planning: By considering cooking yield, chefs can plan menus effectively, ensuring they have sufficient quantities of ingredients for anticipated demand.
• Consistency: Calculating cooking yield allows chefs to maintain consistency in portion sizes and quality, enhancing the dining experience for customers.
## Tips for Maximizing Cooking Yield
To optimize cooking yield and minimize waste, consider the following tips:
• Proper Trimming: Ensure that excessive fat, gristle, and unwanted parts are trimmed before cooking to reduce waste.
• Efficient Cooking Techniques: Utilize cooking techniques that minimize moisture loss, such as steaming or braising, to retain the natural juices and flavors.
• Accurate Measurements: Use precise measurements when portioning ingredients to achieve consistent results and minimize variations in cooking yield.
• Quality Ingredients: Select high-quality ingredients that have minimal moisture loss and shrinkage to maximize the cooking yield.
• Inventory Management: Regularly monitor inventory levels to prevent overstocking or wastage of perishable ingredients.
By implementing these tips, chefs and cooks can optimize their cooking yield and achieve greater efficiency in their culinary operations.
## Conclusion
Calculating cooking yield is a crucial skill for professionals in the culinary industry. By understanding the factors that impact cooking yield and following a systematic approach, chefs can ensure consistency, control costs, and optimize their operations. Maximizing cooking yield not only benefits the business but also enhances the dining experience for customers.
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Subscribe to our disrupting service to boost your productivity and profitability
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Related Articles
Mersenne Prime
• Difficulty Level : Easy
• Last Updated : 26 Mar, 2021
Mersenne Prime is a prime number that is one less than a power of two. In other words, any prime is Mersenne Prime if it is of the form 2k-1 where k is an integer greater than or equal to 2. First few Mersenne Primes are 3, 7, 31 and 127.
The task is print all Mersenne Primes smaller than an input positive integer n.
Examples:
```Input: 10
Output: 3 7
3 and 7 are prime numbers smaller than or
equal to 10 and are of the form 2k-1
Input: 100
Output: 3 7 31 ```
The idea is to generate all the primes less than or equal to the given number n using Sieve of Eratosthenes. Once we have generated all such primes, we iterate through all numbers of the form 2k-1 and check if they are primes or not.
Below is the implementation of the idea.
## C++
`// Program to generate mersenne primes``#include``using` `namespace` `std;` `// Generate all prime numbers less than n.``void` `SieveOfEratosthenes(``int` `n, ``bool` `prime[])``{`` ``// Initialize all entries of boolean array`` ``// as true. A value in prime[i] will finally`` ``// be false if i is Not a prime, else true`` ``// bool prime[n+1];`` ``for` `(``int` `i=0; i<=n; i++)`` ``prime[i] = ``true``;` ` ``for` `(``int` `p=2; p*p<=n; p++)`` ``{`` ``// If prime[p] is not changed, then it`` ``// is a prime`` ``if` `(prime[p] == ``true``)`` ``{`` ``// Update all multiples of p`` ``for` `(``int` `i=p*2; i<=n; i += p)`` ``prime[i] = ``false``;`` ``}`` ``}``}` `// Function to generate mersenne primes less``// than or equal to n``void` `mersennePrimes(``int` `n)``{`` ``// Create a boolean array "prime[0..n]"`` ``bool` `prime[n+1];` ` ``// Generating primes using Sieve`` ``SieveOfEratosthenes(n,prime);` ` ``// Generate all numbers of the form 2^k - 1`` ``// and smaller than or equal to n.`` ``for` `(``int` `k=2; ((1<
## Java
`// Program to generate``// mersenne primes``import` `java.io.*;` `class` `GFG {`` ` ` ``// Generate all prime numbers`` ``// less than n.`` ``static` `void` `SieveOfEratosthenes(``int` `n,`` ``boolean` `prime[])`` ``{`` ``// Initialize all entries of`` ``// boolean array as true. A`` ``// value in prime[i] will finally`` ``// be false if i is Not a prime,`` ``// else true bool prime[n+1];`` ``for` `(``int` `i = ``0``; i <= n; i++)`` ``prime[i] = ``true``;`` ` ` ``for` `(``int` `p = ``2``; p * p <= n; p++)`` ``{`` ``// If prime[p] is not changed`` ``// , then it is a prime`` ``if` `(prime[p] == ``true``)`` ``{`` ``// Update all multiples of p`` ``for` `(``int` `i = p * ``2``; i <= n; i += p)`` ``prime[i] = ``false``;`` ``}`` ``}`` ``}`` ` ` ``// Function to generate mersenne`` ``// primes lessthan or equal to n`` ``static` `void` `mersennePrimes(``int` `n)`` ``{`` ``// Create a boolean array`` ``// "prime[0..n]"`` ``boolean` `prime[]=``new` `boolean``[n + ``1``];`` ` ` ``// Generating primes`` ``// using Sieve`` ``SieveOfEratosthenes(n, prime);`` ` ` ``// Generate all numbers of`` ``// the form 2^k - 1 and`` ``// smaller than or equal to n.`` ``for` `(``int` `k = ``2``; (( ``1` `<< k) - ``1``) <= n; k++)`` ``{`` ``long` `num = ( ``1` `<< k) - ``1``;`` ` ` ``// Checking whether number is`` ``// prime and is one less then`` ``// the power of 2`` ``if` `(prime[(``int``)(num)])`` ``System.out.print(num + ``" "``);`` ``}`` ``}`` ` ` ``// Driven program`` ``public` `static` `void` `main(String args[])`` ``{`` ``int` `n = ``31``;`` ``System.out.println(``"Mersenne prime"``+`` ``"numbers smaller than"``+`` ``"or equal to "``+n);`` ` ` ``mersennePrimes(n);`` ``}``}` `// This code is contributed by Nikita Tiwari.`
## Python3
`# Program to generate mersenne primes` `# Generate all prime numbers``# less than n.``def` `SieveOfEratosthenes(n, prime):` ` ``# Initialize all entries of boolean`` ``# array as true. A value in prime[i]`` ``# will finally be false if i is Not`` ``# a prime, else true bool prime[n+1]`` ``for` `i ``in` `range``(``0``, n ``+` `1``) :`` ``prime[i] ``=` `True` ` ``p ``=` `2`` ``while``(p ``*` `p <``=` `n):`` ` ` ``# If prime[p] is not changed,`` ``# then it is a prime`` ``if` `(prime[p] ``=``=` `True``) :`` ` ` ``# Update all multiples of p`` ``for` `i ``in` `range``(p ``*` `2``, n ``+` `1``, p):`` ``prime[i] ``=` `False`` ` ` ``p ``+``=` `1`` ` `# Function to generate mersenne``# primes less than or equal to n``def` `mersennePrimes(n) :` ` ``# Create a boolean array`` ``# "prime[0..n]"`` ``prime ``=` `[``0``] ``*` `(n ``+` `1``)` ` ``# Generating primes using Sieve`` ``SieveOfEratosthenes(n, prime)` ` ``# Generate all numbers of the`` ``# form 2^k - 1 and smaller`` ``# than or equal to n.`` ``k ``=` `2`` ``while``(((``1` `<< k) ``-` `1``) <``=` `n ):`` ` ` ``num ``=` `(``1` `<< k) ``-` `1` ` ``# Checking whether number`` ``# is prime and is one`` ``# less then the power of 2`` ``if` `(prime[num]) :`` ``print``(num, end ``=` `" "` `)`` ` ` ``k ``+``=` `1`` ` `# Driver Code``n ``=` `31``print``(``"Mersenne prime numbers smaller"``,`` ``"than or equal to "` `, n )``mersennePrimes(n)` `# This code is contributed``# by Smitha`
## C#
`// C# Program to generate mersenne primes``using` `System;` `class` `GFG {`` ` ` ``// Generate all prime numbers less than n.`` ``static` `void` `SieveOfEratosthenes(``int` `n,`` ``bool` `[]prime)`` ``{`` ` ` ``// Initialize all entries of`` ``// boolean array as true. A`` ``// value in prime[i] will finally`` ``// be false if i is Not a prime,`` ``// else true bool prime[n+1];`` ``for` `(``int` `i = 0; i <= n; i++)`` ``prime[i] = ``true``;`` ` ` ``for` `(``int` `p = 2; p * p <= n; p++)`` ``{`` ` ` ``// If prime[p] is not changed,`` ``// then it is a prime`` ``if` `(prime[p] == ``true``)`` ``{`` ` ` ``// Update all multiples of p`` ``for` `(``int` `i = p * 2; i <= n; i += p)`` ``prime[i] = ``false``;`` ``}`` ``}`` ``}`` ` ` ``// Function to generate mersenne`` ``// primes lessthan or equal to n`` ``static` `void` `mersennePrimes(``int` `n)`` ``{`` ` ` ``// Create a boolean array`` ``// "prime[0..n]"`` ``bool` `[]prime = ``new` `bool``[n + 1];`` ` ` ``// Generating primes`` ``// using Sieve`` ``SieveOfEratosthenes(n, prime);`` ` ` ``// Generate all numbers of`` ``// the form 2^k - 1 and`` ``// smaller than or equal to n.`` ``for` `(``int` `k = 2; (( 1 << k) - 1) <= n; k++)`` ``{`` ``long` `num = ( 1 << k) - 1;`` ` ` ``// Checking whether number is`` ``// prime and is one less then`` ``// the power of 2`` ``if` `(prime[(``int``)(num)])`` ``Console.Write(num + ``" "``);`` ``}`` ``}`` ` ` ``// Driven program`` ``public` `static` `void` `Main()`` ``{`` ``int` `n = 31;`` ` ` ``Console.WriteLine(``"Mersenne prime numbers"`` ``+ ``" smaller than or equal to "` `+ n);`` ` ` ``mersennePrimes(n);`` ``}``}` `// This code is contributed by nitin mittal.`
## PHP
``
## Javascript
``
Output:
```Mersenne prime numbers smaller than or equal to 31
3 7 31 ```
References:
https://en.wikipedia.org/wiki/Mersenne_prime
This article is contributed by Rahul Agrawal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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# Stuck on equating co-efficents! DiffEQ
I'm confused on how to equate coniffecents. What i'm doing is, finding a particular solution for just the polynomial then i'm going to find it just for the exponential and add them together rather then putting it all together and making it a mess, but i'm stuck when i try to add the Co-efficents, can someone explain that process to me? Thanks!
Here is my work!
http://suprfile.com/src/1/1ta5k1/lastscan.jpg [Broken]
Last edited by a moderator:
## Answers and Replies
Related Calculus and Beyond Homework Help News on Phys.org
quasar987
Homework Helper
Gold Member
I kinda see.
Fact: Two polynomials are equal iff their respective coefficients are the same. (You might want to try proving this (hint set x=0. u get 1 relation, differentiate, set x=0, u get a second equality, differentiate, set x=0, etc))
This is what you have here. So you want to rewrite the LHS as
$$-At^2 + (2A-B)t + (12A+B-C) = t^2 + t$$
Now what the theorem/fact stated above is telling you is that
-A=1
2A-B=1
12A+B-C=0
It's a linear system of equation. Start row-reducin' friend!
THanks man worked great!!
benorin
Homework Helper
Gold Member
There is another method: just plug-in three different values of t (like, say t=0,1,2,) into the given equation to generate three equations in A,B, and C.
shweet.
thanks for the tip
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# (9,7),(8,-3),and(2,-6) find the perimeter - Samsung Cell Phones
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From the three points you give, the perimeter appears to be the three sides of a triangle going from (9, 7) to (8, -3) to (2, -6) and back to (9, 7).
Or are you actually asking for something else? The length, perhaps? That would be about 31.5 .
Posted on Feb 13, 2014
Hi,
a 6ya expert can help you resolve that issue over the phone in a minute or two.
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Posted on Jan 02, 2017
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## Related Questions:
### How do you figure the perimeter of a rectangle when l = 4 and w = 10angle formula perimeter
Add together the lengths of the four sides. Two sides are 4 each and the other two are 10 each, so the total perimeter is 28.
May 02, 2014 | Computers & Internet
### What is the greatest perimeter that a 48 meter squared rectangle could have?
Infinite.
A one meter by 48 meter rectangle has an area of 48 square meters and a perimeter of 98 meters. A one centimeter by 4800 meter rectangle has the same area and a perimeter of 9600.02 meters. A one millimeter by 48,000 meter rectangle has the same area and a perimeter of 96,000.002 meters. A rectangle one micron by 48,000,000 meters has an area of 48 square meters and a perimeter of 96,000,000.000002 meters. Keep making the rectangle skinnier and skinnier without changing the area, and the perimeter keeps getting longer and longer.
May 01, 2014 | Computers & Internet
### If the ratio of the corresponding side lengths of two similar polygons is 5:9, what is the ratio of their perimeters?
This works for any polygon:
Say you have 2 squares - the sides are 5 and 9
Perimeter 1 = 20
Perimeter 2 = 36
20:36 = 5:9
Apr 10, 2014 | Texas Instruments TI-83 Plus Calculator
### Formula for perimeter
A square has four equal sides. If the perimeter is 180 yards then each side is 180/4 or 45 yards.
Apr 08, 2014 | Computers & Internet
### What is the formula for finding the perimeter of a square box with each side measuring 3 metres
Since a square has four equal sides, the perimeter is four times the length of one side. If a square has a side of 3 metres then the perimeter is 12 metres.
Jan 26, 2014 | Perimeter Pant
### Perimeter of a rectangle
1 tile by 20 tiles - perimeter of 1+ 20 +1 + 20 = 42
2 tiles by 10 tiles - perimeter of 2 + 10 + 2 + 10 = 24
(can't make a rectangle with 3 rows)
4 tiles by 5 tiles - perimeter of 4 + 5 + 4 + 5 = 18
5 tiles by 4 tiles - perimeter of 5 + 4 + 5 + 4 = 18
(can't make a rectangle with 6, 7, 8 or 9 rows)
10 tiles by 2 tiles - perimeter of 10 + 2 + 10 + 2 = 24
20 tiles by 1 tile - perimeter of 20 + 1 + 20 + 1 = 42
So 6 different rectangles. Smallest perimeter is 18.
Jan 08, 2014 | Computers & Internet
### A rectan Dear Madam Please look at the math questions below. Please help me solve the questions Question 1 : A rectangular tablecloth is 2.5 m long. Trim is to be sewn around the perimeter of the cloth...
1. Well this one is pretty simple. Since you are not supposed to trim the length trim the width instead.
to make the perimeter 0.4 m shorter cut the cloth by 0.2 m width the picture will make it clear.
2. let the length of rectangle be x cm
now original perimeter = 2(x+20) cm
new length = x-30 cm
breadth remains same = 20 cm
new perimeter = 2(x-30+20) cm = 2(x-10) cm
According to question
new perimeter = old perimeter /2
2(x-10) = 2(x+20) /2
2x -20 = x + 20
x = 40 cm
hence length of longer side 40 cm.
If you want to directly comm ankitsharma.220@gmail.com
Feb 13, 2011 | Vivendi Excel@ Mathematics Study Skills...
### Area and perimeter
Area= width * length
Perimeter= 2 * width + 2* length
Are you looking for an Excel function?
There are no Area or Perimeter functions I am aware of in Excel.
Mar 03, 2008 | Microsoft Office Standard for PC
## Open Questions:
#### Related Topics:
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# Boat Stereo Wiring Diagram
Do you require a Boat Stereo Wiring Diagram? The Boat Stereo Wiring Diagram, ideas, and frequently asked questions are all available here. We created this page for people searching for a Boat Stereo Wiring Diagram. Our short article will assist you in fixing your issue.
## Boat Stereo Wiring Diagram
A wiring diagram will reveal you where the wires need to be connected, so you do not have to presume.
You do not have to presume, a wiring diagram will show you exactly how to attach the wires.
See the Boat Stereo Wiring Diagram images below
## Tips and tricks for reading wiring diagrams
• When taking a look at a wiring diagram, don’t try to concentrate on the whole page all at once. It’s overwhelming. Put a blank sheet of paper beside the wiring diagram and just draw the simple circuit. Focus on the simple part and follow the current flow from power to ground or from ground to power. All complex wiring diagrams are just a series of simple diagrams, and it makes it difficult to look at if you do not limit to the circuit that you’re doing.
• Print the wiring diagram off and use highlighters to trace the circuit. When you use your finger or follow the circuit with your eyes, it’s simple to mistrace the circuit. One technique that I use is to print the very same wiring diagram off twice.
• To appropriately check out a wiring diagram, one has to understand how the components in the system operate. If a module is powered up and it sends out a signal of half the voltage and the technician does not understand this, he would think he has a problem, as he would expect a 12V signal. Following diagrams is fairly simple, but using it within the scope of how the system operates is a various matter. My finest suggestions is not only look at the diagram, however understand how the elements run when in use.
• Read wiring diagrams from negative to positive and redraw the circuit as a straight line. All circuits are the same– voltage, ground, single component, and switches.
• Prior to checking out a schematic, get familiar and comprehend all the signs. Read the schematic like a roadmap. I print the schematic and highlight the circuit I’m detecting to ensure I’m staying on the ideal path.
33 Marine Stereo Wiring Diagram – Wiring Diagram List
### What are the types of wiring diagram?
• Schematic Diagrams.
• Wiring diagrams.
• Block diagrams.
• Pictorial diagrams.
### Where is a wiring diagram utilized?
Wiring diagrams are primarily utilized when trying to reveal the connection system in a circuit. It is majorly utilized by building coordinators, architects, and electrical experts to present the wiring connections in a building, a room, or perhaps a simple gadget.
### Why is wiring diagram crucial?
It reveals the parts of the circuit as streamlined shapes, and how to make the connections in between the devices. A wiring diagram typically provides more information about the relative position and arrangement of devices and terminals on the devices.
### Why do we require wiring diagrams?
A wiring diagram is typically used to repair issues and to make sure that all the connections have been made which everything exists.
### What is the distinction between a schematic and wiring diagram?
A wiring diagram is a generalized pictorial representation of an electrical circuit. The elements are represented utilizing streamlined shapes in wiring diagrams.
Wiring in 3rd amp, need help/input – Stereo Info & How-To
How to Install A Radio and Speakers In A Boat
Wiring the Stereo This Weekend – Boats, Accessories & Tow Vehicles
### Are all wiring diagrams the same?
Wiring diagrams might follow various requirements depending upon the nation they are going to be utilized. They might have different designs depending on the business and the designer who is designing that. They likewise might be drawn by various ECAD software such as EPLAN or AutoCAD electrical.
### What is an architectural wiring diagram?
Architectural wiring diagrams show the approximate places and affiliations of receptacles, lighting, and permanent electrical services in a building.
### How are wiring diagrams read?
The electrical schematics are read from left to right, or from top to bottom. This is necessary to get right, as the signal direction indicates the flow of current in the circuit. It is then simple for a user to comprehend when there is a modification in the course of the circuit.
### How do you read electrical wire numbers?
An electrical cable is categorized by two numbers separated by a hyphen, such as 14-2. The first number denotes the conductor’s gauge; the second represents the variety of conductors inside the cable. For example, 14-2 has two 14-gauge conductors: a hot and a neutral.
### How do you read vehicle wiring diagrams?
A car wiring diagram is a map. To read it, identify the circuit in question and starting at its power source, follow it to the ground. Utilize the legend to comprehend what each symbol on the circuit indicates.
### How do you read wire size charts?
Wire gauges range from low numbers to high numbers, with smaller numbers describing smaller diameters and larger numbers representing larger sizes. AWG 4 is 0.2043 inches in diameter, and AWG 40 is. 0031 inches in size.
### How is wire numbered?
American Wire Gauge (AWG) is the basic way to represent wire size in The United States and Canada. In AWG, the larger the number, the smaller the wire diameter and thickness. The biggest basic size is 0000 AWG, and 40 AWG is the tiniest standard size.
### What is the schematic format?
A schematic, or schematic diagram, is a representation of the aspects of a system utilizing abstract, graphic symbols rather than realistic photos.
### What should a schematic consist of?
Schematics need to include the complete description and locations of all developing code elements, such as the heating/ventilation/air conditioning (likewise referred to as HVAC), plumbing, and electrical systems. Schematic designs are only a basic layout to communicate a style scheme to the owner.
### Is AWG aluminum or copper?
The AWG requirement includes copper, aluminum and other wire products. Normal home copper wiring is AWG number 12 or 14. Telephone wire is normally 22, 24, or 26. The higher the gauge number, the smaller the diameter and the thinner the wire.
### Can you touch a live black wire?
If you are available in contact with an energized black wire– and you are also in contact with the neutral white wire– current will pass through your body. You will get an electrical shock. You will get a shock if you touch two wires at various voltages at the same time.
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0
How many square inches is an 8 X 13 square feet room?
Updated: 8/20/2019
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10y ago
14,976 square inches.
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Related questions
How many square feet in a room that has 26904 square inches in it?
186.83 square feet.
How many square feet are in a room that is 110 inches by 220 inches?
Approximately 168.056 square feet.
A room 12 feet 10 inches by 10 feet 7 inches how many square feet is this?
135.82 square feet
How many square feet is a room that is 10 feet 8 inches by 12 feet 11 inches?
137.8 square feet.
How many square feet is a room that measures 179 inches x 192 inches?
238.6 square feet.
How many square feet in a room 10 feet 7 inches by 10 feet five inches?
Approximately 110.24 square feet.
How many square feet are in a room that is 160 inches by 220 inches?
160 inches times 220 inches are 35200 square inches. 35200 square inches are 244.444444 square feet.
How many square feet is 7125 square inches on a room floor?
Approximately 49.48 square feet.
How many inches are in a room measuring 100 square feet?
That depends on the length and width of the room - there are many different rectangles that have the same surface, in this case, 100 square feet. 100 square feet = 14,400 square inches
How many square feet are in a room that measures 15 feet 7 inches by 12 feet 7 inches?
Approximately 196.1 square feet.
How many square inches are in a room measuring 12feet by 16 feet?
12 feet x 16 feet = 27,648 square inches
How many square feet are there in a room that measures 15 feet 4 inches by 14 feet half an inch?
215.3 square feet for this room.
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# 1.07 Simplifying numerical expressions involving integers
Lesson
In life, the order in which we do things is important. For example, we put on socks then shoes, rather than shoes and then socks.
The same goes for evaluating expressions in math with more than one operation. There are a number of rules which need to be followed in order to solve these problems correctly. The order goes:
Order of operations
1. Complete all operations within grouping symbols such as parentheses (...) or absolute values |...|. If there are grouping symbols within other grouping symbols, do the innermost operation first.
2. Evaluate all exponential expressions, such as squares and cubes.
3. Multiply and/or divide in order from left to right.
4. Add or subtract in order from left to right.
#### Worked examples
##### question 1
Evaluate: $5\times\left(6+6\right)$5×(6+6)
Think: Remember the order of operations. Firstly, we'll evaluate what's inside the parentheses, then we'll evaluate the multiplication.
Do
$5\times\left(6+6\right)$5×(6+6) $=$= $5\times12$5×12 Add within the grouping symbol $=$= $60$60 Multiply
Reflect: How would this be different if there were no parentheses?
##### question 2
Evaluate: $100-9\times6+18\div6$1009×6+18÷6
Think: There are no grouping symbols in this question, so firstly we'll evaluate the multiplication and division (going from left to right), then we will evaluate the addition and subtraction (going from left to right).
Do
$100-9\times6+18\div6$100−9×6+18÷6 $=$= $100-54+3$100−54+3 Evaluating $9\times6=54$9×6=54 and $18\div6=3$18÷6=3 $=$= $46+3$46+3 Evaluating $100-54$100−54 $=$= $49$49 Simplifying
The two examples above used only whole numbers. However, we can work through order of operations problems with any type of real number such as integers, fractions or decimals. Let's look at some involving integers.
##### Question 3
Evaluate: $\left(48\div\left(-12\right)-35\div5\right)\times3^2$(48÷(12)35÷5)×32
Think: We need to simplify the problem by using our order of operation rules.
• Firstly, we evaluate what's inside the parentheses - division first and then subtraction.
• Then, evaluate the exponents.
• Then, we evaluate any other multiplication or division, working from left to right.
Do:
$\left(48\div\left(-12\right)-35\div5\right)\times3^2$(48÷(−12)−35÷5)×32 $=$= $\left(-4-7\right)\times3^2$(−4−7)×32 Within the parentheses we start with $48\div\left(-12\right)=-4$48÷(−12)=−4 and $35\div5=7$35÷5=7 $=$= $-11\times3^2$−11×32 Now within the parentheses we can evaluate $-4-7$−4−7 $=$= $-11\times9$−11×9 Next we can evaluate the exponent $3^2=9$32=9 $=$= $-99$−99 Perform the multiplication
Reflect: How might a number line help us to simplify this question?
#### Practice questions
##### Question 4
Evaluate $-18+21\div7$18+21÷7
##### Question 5
Evaluate $\left(-14\right)\div2-10\div\left(-2\right)$(14)÷210÷(2)
##### Question 6
Evaluate $9^2+6$92+6
### Outcomes
#### 7.NS.1
Apply and extend previous understandings of addition and subtraction to add and subtract integers and other rational numbers; represent addition and subtraction on a horizontal or vertical number line diagram.
#### 7.NS.1d
d. Apply properties of operations as strategies to add and subtract rational numbers.
#### 7.NS.2
Apply and extend previous understandings of multiplication and division and of fractions to multiply and divide integers and other rational numbers.
#### 7.NS.2c
c. Apply properties of operations as strategies to multiply and divide rational numbers.
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# Working with European Decimal Numbers in Excel
How to work with European decimal numbers in text boxes on User Form in Excel. Watch the video below:
Europeans use the comma instead of the dot in decimal numbers. One of the main reasons is that their numerical keypad has only a comma. Therefore, finance professionals who tend to enter decimal values via the numerical keyboard use the comma.
If we have huge European decimal numbers like statistical data, we can easily convert them to standard dot decimal numbers using the ‘NUMBERVALUE’ formula as shown below:
Another way to handle European decimal numbers in an Excel worksheet is to disable the ‘Use system separators’ by clicking on File –> Options –> Advanced as shown in the images below:
For example, if we create a workbook in the USA and share it with someone in Denmark then the user in Denmark will see the numbers according to his Regional settings in Windows.
If the display of the decimal numbers doesn’t happen automatically then it only means that we have configured Excel to ignore the ‘Use system separators’ as shown below.
Now handling European decimal numbers in text boxes on a user-form and transferring them to an Excel worksheet becomes very interesting, especially if we wish to perform calculations with the European decimal numbers. We’ll use the concepts above in our macro that automates the complete process of transferring European decimal numbers from UserForm text boxes into an Excel worksheet:
Private Sub Workbook_Activate()
Application.UseSystemSeparators = False
End Sub
Private Sub Workbook_Deactivate()
Application.UseSystemSeparators = True
End Sub
Sub ShowUserForm()
UserForm1.Show
End Sub
Private Sub CommandButton1_Click()
Dim CommaPos As Long, NumCharsAfterComma As Long
CommaPos = InStr(TextBox1.Text, “,”)
‘MsgBox CommaPos
NumCharsAfterComma = Len(Mid(TextBox1, CommaPos + 1))
If NumCharsAfterComma > 2 Then
MsgBox “You’ve entered too many numbers after the commain TextBox1. Only two numbers allowed.”
Exit Sub
End If
Range(“B1”) = TextBox1
Range(“B1”) = Range(“B1”) * 0.01
CommaPos = InStr(TextBox1.Text, “,”)
NumCharsAfterComma = Len(Mid(TextBox2, CommaPos + 1))
If NumCharsAfterComma > 2 Then
MsgBox “You’ve entered too many numbers after the comma in TextBox2. Only two numbers allowed.”
Exit Sub
End If
Range(“B2”) = TextBox2
Range(“B2”) = Range(“B2”) * 0.01
Range(“B3”) = Range(“B1”) * Range(“B2”)
Range(“B1:B3”).NumberFormat = “0.00”
TextBox3 = Range(“B3”).Text
End Sub
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# Decimal to roman numeral converter in c
C program for Decimal to roman numeral converter. Here mentioned other language solution.
``````// Include header file
#include <stdio.h>
#include <stdlib.h>
// C program for
// Conversion from Decimal to roman number
// Display roman value of n
void result (int n)
{
switch (n)
{
// Test Cases
case 1:
printf("I");
break;
case 4:
printf("IV");
break;
case 5:
printf("V");
break;
case 9:
printf("IX");
break;
case 10:
printf("X");
break;
case 40:
printf("XL");
break;
case 50:
printf("L");
break;
case 90:
printf("XC");
break;
case 100:
printf("C");
break;
case 400:
printf("CD");
break;
case 500:
printf("D");
break;
case 900:
printf("DM");
break;
case 1000:
printf("M");
break;
}
}
long selecting(long number, int collection[], int size)
{
int n = 1;
int i = 0;
for (i = 0; i < size; i++)
{
if (number >= collection[i])
{
n = collection[i];
}
else
{
break;
}
}
result(n);
// Reduce the value of number
return number - n;
}
void romanNo (long number)
{
if (number <= 0)
{
// When is not a natural number
return;
}
// Base case collection
int collection[] =
{1, 4, 5, 9, 10, 40, 50, 90, 100, 400, 500, 900, 1000};
// Get the size of collection
int size = sizeof(collection)/sizeof(collection[0]);
printf(" %ld : " , number );
while(number > 0)
{
number = selecting(number, collection, size);
}
// Add new line
printf("\n");
}
int main()
{
// Test Case
romanNo(10);
romanNo(18);
romanNo(189);
romanNo(604);
romanNo(982);
romanNo(3000);
return 0;
}``````
Output
`````` 10 : X
18 : XVIII
189 : CLXXXIX
604 : DCIV
982 : DMLXXXII
3000 : MMM``````
## Comment
Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.
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Place the indicated product in the proper location on the grid. a 2b (3a 2 + 4ab 2 )
a^2b (3a^2 + 4ab^2 ) = a^2b*3a^2 + a^2b*4ab^2 = 3a^4b + 4a^3b^3
Question
Updated 2/3/2015 2:11:32 AM
Rating
3
a^2b (3a^2 + 4ab^2 )
= a^2b*3a^2 + a^2b*4ab^2
= 3a^4b + 4a^3b^3
Confirmed by yumdrea [2/3/2015 1:18:36 PM]
Questions asked by the same visitor
Write the standard form of the line with the given information. Include your work in your final answer. Type your answer in the box provided or use the upload option to submit your solution. The line containing a slope of 3 and a y-intercept of -2. User: The equations of two lines are x - 3y = 6 and y = 3x + 2. Determine if the lines are parallel, perpendicular or neither. parallel perpendicular neither
Question
Updated 1/28/2015 8:21:50 PM
The slope of the line passing through the points (2, 7) and (-4, 8) is slope = (8 - 7)/(-4 - 2) = 1/-6 = -1/6
Confirmed by Andrew. [1/29/2015 5:13:34 AM]
(-1, -6) does not lie on the line whose equation is x + y = 7.
Confirmed by Andrew. [1/29/2015 5:13:50 AM]
6x - 2y = 5 3x - y = 10 Solve the system of equations. (3, -1) no solution
Question
Updated 1/28/2015 7:31:44 PM
6x - 2y = 5 (equ. 1)
3x - y = 10 multiply 2 on both sides the result is:
6x - 2y = 20 (equ. 2)
subtracting equ. 1 and 2 the result is:
0 = -15 which is false
There is no solution for the system 6x - 2y = 5 3x - y = 10.
Confirmed by Andrew. [1/29/2015 4:00:13 AM]
Place the indicated product in the proper location on the grid. 6xy (1/2x 2 - 1/3xy + 1/6y 2 )
Updated 2/2/2015 8:32:57 AM
6xy (1/2x^2 - 1/3xy + 1/6y^2 ) = 3x^3y - 2x^2y^2 + xy^3
Confirmed by jeifunk [2/2/2015 8:41:59 AM]
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# Nested roots
• Nov 7th 2009, 05:41 AM
shane99
Nested roots
Determine the exact value of √(10 + √(10 + √(10 + ....))) it goes on forever
Show all steps and explain most of them
• Nov 7th 2009, 07:03 AM
ramiee2010
let $\displaystyle x= \sqrt{10+\sqrt{10+\sqrt{10+......}}}$
therefore
$\displaystyle x= \sqrt{10+x}$
taking square both side
$\displaystyle x^2= (10+x)$
$\displaystyle x^2-x-10=0$
now solve the quadratic equation .........
• Nov 7th 2009, 12:08 PM
mr fantastic
Quote:
Originally Posted by shane99
Determine the exact value of √(10 + √(10 + √(10 + ....))) it goes on forever
Show all steps and explain most of them
No, that's not how it works. You do not make demands like that. You show what you can do and then say where you're stuck.
I see you've been given a massive start on how to do this question. So now it's your turn. If you need more help, say what you've done and where you get stuck.
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# Exact differential
by zeshkani
Tags: differential, exact
Sci Advisor HW Helper PF Gold P: 4,771 The definition of the total differential of a function f(x,y) is the expression $$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$$ Now with the z you're given, is the expression for dz correct? That is to say, is $$\frac{\partial z}{\partial x}=(y+0)$$ and $$\frac{\partial z}{\partial y}=(x-1)$$ ??
Math Emeritus Sci Advisor Thanks PF Gold P: 39,491 Yes. z=xy-y+lnx+2 so dz= (y+ 1/x)dx + (x-1)dy And that is an "exact" differential precisely because it is the differential of z. Now suppose you were given the differential form (y+ 1/x)dx+ (x- 1)dy without having been given z. How would you determine whether it was an exact differential? If $$(y+ 1/x)= \frac{\partial z}{\partial x}$$ for some function z and $$x-1= \frac{\partial z}{\partial y}$$ then $$\frac{\partial(y+ 1/x)}{\partial y}= \frac{\partial^2 z}{\partial x\partial y}$$ and $$\frac{\partial(x-1)}{\partial x}= \frac{\partial^2 z}{\partial y\partial x}$$ and so must be equal. Fortunately, it is easy to see that each second derivative is 1 and so they are in fact equal. Okay, now, how would we find z? Knowing that $$x-1= \frac{\partial z}{\partial y}$$ integrating with respect to y (treating x as a constant) we get z= xy- y+ C, except that, since we are treating x as a constant, that "constant of integration", C, may depend on x: z= xy- y+ C(x). Differentiating that with respect to x, $$\frac{\partial z}{\partial x}= y+ C'(x)= y+ 1/x$$ Notice that the "y" terms cancel (it was the "cross condition" above that guarenteed that) and so we have C'(x)= 1/x. Then C(x)= ln(x)+ C where "C" now really is a constant. Any z(x,y)= xy- y+ ln(x)+ C satisfies dz= (y+ 1/x)dx+ (x-1)dy.
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+0
# Suppose I have 6 shirts, 4 ties, 3 pairs of pants
0
64
1
Suppose I have 6 shirts, 4 ties, 3 pairs of pants, and 2 pairs of shoes. If an outfit requires a shirt and pants and shoes, and can either have a tie or not have a tie, how many outfits can I make?
Nov 14, 2020
#1
+112016
+1
Just treat it as 5 ties. The fifth one is inviable.
I will simplify the question.
If you have 3 shirts and 2 pairs of pants. How many outfits can you make? (using 1 shirt and one pair of pantspants)
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# If an ant runs randomly through an enclosed circular field of radius 2 feet with an inner circle of radius 1 foot, what is the probability that the ant will be in the inner circle at any one time? a.1/8 b 1/6 c 1/4 d.1/2 e 1
1
by Diepenbrock580
## Answers
2014-06-02T23:57:01-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
The area of a circle is π (radius²)
The area of the outer (2-ft) circle is π (2²) = 4 π square feet.
The area of the inner (1-ft) circle is π (1²) = 1 π square feet.
The inner circle covers 1/4 of the area of the outer circle.
So if the ant wanders around totally aimlessly and randomly, and there's no way to
know where he came from, where he is now, or where he's going next, and there's
an equal chance of him being anywhere in the big circle at any time, then there's a
25% chance of him being inside the small circle at any time, because 1/4 of the
total area is in there.
That's choice c).
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# Confusion about quantum walks and the quantum walk operator
I am looking at the Quantum Signal Processing paper by GH Low and IL Chuang here. One step that they used was Child's quantum walks. They constructed a walk operator, $$W \left | u_\lambda \right > = e^{i\theta_\lambda} \left | u_\lambda \right >$$, from a Hamiltonian matrix $$\hat{H}$$, with two oracles $$O_H$$ and $$O_F$$. $$O_H$$ gives $$\left < j \right | \hat{H} \left | k \right>$$ where $$j$$ and $$k$$ are the row and column. $$O_F$$ gives the $$l$$th non-zero element of any row $$j$$.
My question is: What exactly do these oracles look like? For example, if I have a hamiltonian matrix
$$\begin{bmatrix} 2 & 3-2i & 2i \\ 3+2i & 3 & 6 \\ -2i & 6 & 5 \end{bmatrix}$$
How would I calculate the oracles $$O_H$$ and $$O_F$$, and how would I use these to construct the quantum walk operator $$W$$?
## 1 Answer
For these problems we're not usually given a matrix that is small enough that we can write down explicitly, as is done in the question. Rather, these oracles $$O_H$$ and $$O_F$$ are most useful to consider when the Hermitian matrix is too large to write down, and instead we rely on these oracles to describe the sparse operator implicitly.
For example one operator that I like to think a lot about is the adjacency matrix $$A$$ of the Rubik's cube with quarter-turn or half-turn twists of each of the faces. This is a huge matrix! But, we can still have operators $$O_H$$ and $$O_F$$ defined in a straightforward and meaningful manner.
In more detail index the rows and columns of the adjacency matrix $$A$$ by one of the $$43,252,003,274,489,856,000$$ positions of the cube. Set entries $$A_{jk}$$ of the matrix to $$1$$ if there is a clockwise or counterclockwise rotation of one of the faces from position $$j$$ to position $$k$$, and set $$A_{jk}$$ to $$0$$ otherwise. This matrix has only $$\{0,1\}$$ entries and is symmetric about the diagonal; hence it is Hermitian and accordingly $$\exp(-iAt)$$ is unitary.
Such a matrix is very sparse - as most entries are $$0$$. Furthermore, given any two scrambles of the cube $$j$$ and $$k$$, we can quickly answer whether they are adjacent or not after a single twist of one of the faces; hence we have an oracle for $$\langle k|A|j\rangle$$. Additionally given any position of the cube $$j$$, we can just iterate through all of the twists to list all of the immediate neighbors to find the $$l$$th non-zero column in the matrix (in this case $$l\le 18$$, as there are $$3\times 6=18$$ twists that can conventionally be done).
It may also be helpful to think of these oracles $$O_F$$ and $$O_H$$ as procedures or subroutines that correspond to a sequence of instructions/sequence of gates which return the respective answer. For the Rubik's cube, the oracles themselves are not information-theoretically as large as a $$43,252,003,274,489,856,000^2$$ matrix; rather they encode the matrix itself in a much smaller form.
But, for the particular matrix in the question, we may have $$O_H(2,2)=3$$, because the entry of $$\hat H_{2,2}=3$$. Similarly, $$O_F(3,2)=6$$ because the second non-zero entry on the third row is $$6$$. But again the goal is not to explicitly describe the matrix and then construct the oracles; rather the goal is to implicitly describe a large matrix with the oracles, so as to avoid having to explicitly write down the matrix.
• Hi, thank you for such a good answer. I am just confused on two things. 1. Do we assume the Hamiltonian is described using these oracles? For example, in QSP, why can we assume that the person using the algorithm has access to these oracles if they are hard to compute. 2. Will the oracles not be $n \times n$ where n is the dimension of the matrix? Because it still needs to encode all the data points. In your rubik's cube example, each state can compute the next state (adjacent states), but how would that be done for indices in a Hamiltonian matrix? Sep 22, 2022 at 23:23
• Updated answer but briefly, yes the Hamiltonian could be described with the oracles, and the oracles are assumed to be easy to compute. You can think of the oracles as just snippets of code that succinctly encode or describe the entire matrix. Sep 23, 2022 at 13:10
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# Quick Answer: How Do You Simplify A Ratio?
## How do you simplify a ratio in simplest form?
How to Simplify a Ratio A : B when A and B are both whole numbersList the factors of A.List the factors of B.Find the greatest common factor of A and B, GCF(A, B)Divide A and B each by the GCF.Use the whole number results to rewrite the ratio in simplest form..
## What is the ratio of 5 to 3?
For example, if we have a ratio 250 to 150, we can simplify it by dividing both numbers by 10 and then by 5 to get 5 to 3: 250 : 150 25 : 15 5:3 . The ratio 5 to 3 is the simplest form of the ratio 250 to 150, and all three ratios are equivalent.
## What is ratio formula?
When we compare the relationship between two numbers dealing with a kind, then we use the ratio formula. It is denoted as a separation between the number with a colon (:). Sometimes a division sign is also used to express ratios.
## What is the ratio of 2 to 4?
1:2Multiplying or dividing each term by the same nonzero number will give an equal ratio. For example, the ratio 2:4 is equal to the ratio 1:2.
## What is the ratio of 5 to 7?
20 : 285 : 7 = 20 : 28. And 7 is a fourth of 28.
## What does a ratio of 4 to 1 mean?
The relationship between two measures, The taller one is obviously four times taller than the other, so we say thier heights are in the ratio of four to one. … What this means is that for every one unit of height on the left bar, the tall bar has four height units. Hence “four to one”.
## How do you solve 3 ratios?
How to Calculate Ratios of 3 NumbersStep 1: Find the total number of parts in the ratio by adding the numbers in the ratio together.Step 2: Find the value of each part in the ratio by dividing the given amount by the total number of parts.Step 3: Multiply the original ratio by the value of each part.
## Do you need to simplify ratios?
Many ratios can be written with smaller numbers. This is called writing ratios in their simplest form , or simplifying ratios. Simplifying ratios makes them easier to work with. … As the two examples below show, you simplify ratios by dividing the number on each side by their greatest common factor.
## What is the ratio of 2 to 5?
The ratio is 2 to 5 or 2:5 or 2/5. All these describe the ratio in different forms of fractions. The ratio can consequently be expressed as fractions or as a decimal. 2:5 in decimals is 0.4.
## What is the ratio of 1 to 5?
A ratio of 1:5 says that the second quantity is five times as large as the first.
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## Set the Stage Examples.doc - Section 1: Set the Stage
Set the Stage Examples.doc
Unit 11: Statistics
Lesson 7 of 17
## Big Idea: You want me to subtract and square how many numbers?!? Make the process of managing data less crazy using a frequency distribution.
Print Lesson
1 teacher likes this lesson
Standards:
Subject(s):
Math, Statistics, Algebra II, master teacher project, 11th Grade
55 minutes
### Merrie Rampy
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# Why complex signal has no imaginary spectrum
I am learning about complex sampling.
I am confused why $$~e^{ j 2\pi f~ n}~$$ has only a real spectrum. I would have thought the $$j ~\sin(2 \pi f n)$$ would produce a single spike in imaginary spectrum just like there is a single spike in real axis from $$\cos(2 \pi f n)$$.
I understand that the spectrum is one sided because the negative complex exponentials cancel out, but why is there not a one sided real and imaginary spectrum?
Many thanks
The imaginary part of the spectrum corresponds to the odd part of the time-domain sequence. Since the given signal is even, i.e.,
$$x[n]=e^{jn\omega_0}=x^*[-n]=e^{-j(-n)\omega_0}=e^{jn\omega_0}$$
the imaginary part of the spectrum is zero, i.e., the spectrum is purely real-valued.
Note that for complex-valued signals, even in this context means $$x[n]=x^*[-n]$$, i.e., its real-part is even and its imaginary part is odd.
In sum, if a sequence $$x[n]$$ satisfies
$$x[n]=x^*[-n]$$
(i.e., the sequence is even and, consequently, its odd part is zero), then its DTFT is purely real-valued.
• Hi @MattL. I do not doubt you are right, my script shows its imag spectrum is zero. You say the entire signal is even, but it has an odd component from the sine and I thought the sum of an even function and an odd function is neither odd nor even. – Natalie Johnson Jan 13 at 21:04
• @NatalieJohnson: This is about complex-valued sequences, for which the odd part is defined by $\frac12\big(x[n]-x^*[-n]\big)$. So in this sense, your sequence is even (because its odd part is zero). – Matt L. Jan 14 at 9:09
Your last paragraph seems to have the concept exactly backwards.
A strictly real cos(w) or a purely imaginary i*sin(w) in the time domain is two sided in the frequency domain, where the two sides cancel out the other component.
Otherwise exp(i*w)m alone is a spiral in complex space with its excursions into both real and imaginary space being non-zero (e.g. see one of Euler's identities). You need a mirror image spiral to cancel out the twist of one spiral with its opposite rotation.
A complex signal can have either a strictly real or strictly imaginary (or any ratio of the two) spectrum (depending on phase), as long as it does not have a equal magnitude mirror image, complex conjugated, to completely cancel out the imaginary component.
• The spectrum of $~e^{ j 2\pi f~ n}~$ is one sided with a single the positive frequency that is not mirrored in the negative frequency... I see this in my python script. "The negative exponetials cancel out" - what I meant was when you write cos + isin in the complex exponential form, the negative frequency exponentials cancel out leaving just positive frequency so the spectrum of this complex is one sided (only positive frequency). Maybe we mean the same thing and I was not clear... – Natalie Johnson Jan 13 at 20:57
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111
Numbers to 10
Workbook 1A Chapter 1 Numbers to 10
Workbook 1A Chapter 1 Numbers to 10
G1S1W1-MW Numbers to 10
112
Number Bonds
Workbook 1A Chapter 2 Number Bonds
Workbook 1A Chapter 2 Number Bonds
G1S1W2-MW Number Bonds
113
Workbook 1A Chapter 3 Addition Facts to 10
Workbook 1A Chapter 3 Addition within 10
114Subtraction Facts to 10Workbook 1A Chapter 4 Subtraction Facts to 10Workbook 1A Chapter 4 Subtraction within 10
G1S1W4-MW Subtraction to 10 part 1
115
G1S1W5-MW Subtraction to 10 part 2
116Numbers to 20Workbook 1A Chapter 7 Numbers to 20Workbook 1A Chapter 6 Numbers to 20
G1S1W6-MW Numbers to 20_Counting and Place Value
117
G1S1W7-MW Numbers to 20_Comparing and Number Pattern
118Addition within 20Workbook 1A Chapter 8 Addition and Subtraction Facts to 20
119
1110
G1S1W10-MW Addition to 20_Derived Facts Doubles
1111
1112Subtraction within 20
G1S1W12-MW Subtraction to 20_Counting back_Decomposing to 10
1113
G1S1W13-MW Relationship between Addition and Subtraction
1114
G1S1W14-MW Decomposing to ten_relationship between addition and subtraction
1115
G1S1W15-MW Subtraction to 20_Decomposing to 10_Subracting one and tens
G1S1W16-MW Addition and Subtraction Word Problems
1117
G1S1W17-MW Addition and Subtraction Word Problems
1118
*Ordinal Numbers and Positions
Workbook 1A Chapter 6 Ordinal Numbers and Positions
Workbook 1A Chapter 5 Ordinal Numbers
G1S1W18-MW Ordinal Numbers and Position
121ShapesWorkbook 1A Chapter 5 ShapesWorkbook 1A Chapter 7 Shapes, Workbook 1A Chapter 14 Halves and Fourths
G1S2W1-MW Shapes 1
122
G1S2W2-MW Shapes 2
123Numbers to 40Workbook 1B Chapter 12 Numbers to 40Workbook 1B Chapter 11 Numbers to 40
G1S2W3-MW1 Numbers to 40_Counting and Place Value
123
G1S2W3-MW2 Numbers to 40_Comparing and Number Pattern
125
G1S2W5-MW Addition to 40_With Regroup_Part 1
126Subtraction within 40
G1S2W6-MW Subtraction to 40_Without Regroup
127
G1S2W7-MW Subtraction to 40_With Regroup
128
Workbook 1B Chapter 3 Numbers to 20
129Measurements: LengthWorkbook 1A Chapter 9 Length
G1S2W9-MW1 Length
129
G1S2W9-MW2 Length
1210Measurements: Time, Data and Graphs
Workbook 1B Chapter 15 Calendar and Time
Workbook 1B Chapter 15 Time
G1S2W10-MW Time
1210
Workbook 1B Chapter 11 Picture Graphs and Bar Graphs
Workbook 1B Chapter 15 Graphs
G1S2W10-MW2 Picture Graphs and Bar Graphs
1211
Mental Math Strategies
Workbook 1B Chapter 14 Mental Math StrategiesWorkbook 1B Chapter 16 Numbers to 120
G1S2W11-MW Mental Math Strategies
1212Numbers to 100, Addition within 100, Subtract within 100
G1S2W12-MW Numbers to 100
1213Workbook 1B Chapter 16 Numbers to 100, Workbook 1B Chapter 17 Addition and Subtraction to 100
1213
1214
G1S2W14-MW1 Subtraction to 100_without regroup
1214
G1S2W14-MW2 Subtraction to 100_with regroup
1215*Money (Optional)Workbook 1B Chapter 19 MoneyWorkbook 1B Chapter 16 Money
G1S2W15-MW Money_coins and bills_WithSolutions
1216
G1S2W16-MW Money_coins and bills_WithSolutions
1217
*Multiplication and Division (Optional)
Workbook 1B Chapter 18 Multiplication and Division
Workbook 1B Chapter 12 Mulitplication, Workbook 1B Chapter 13 Division
#N/A#N/A
211Numbers to 1,000Workbook 2A Chapter 1 Numbers to 1,000Workbook 2A Chapter 1 Numbers to 1,000
G2S1W1-MW Numbers to 1000 Counting and Place Value
212
G2S1W2-MW Numbers to 1000 Comparing Numbers and Number Pattern
213Addition to 1,000Workbook (2A) Chapter 2 Addition to 1,000Workbook (2A) Chapter 2 Numbers to 1,000
G2S1W3-MW1 Addition to 1,000 Without Regroup
213
G2S1W3-MW2 Addition to 1,000 Regroup in Ones
214
G2S1W4-MW1 Addition to 1,000 Regroup in Tens
214
G2S1W4-MW2 Addition to 1,000 Regroup in Ones and Tens
215Subtraction to 1,000Workbook (2A) Chapter 3 Subtraction to 1,000Workbook (2A) Chapter 1 Subtraction to 1,000
G2S1W5-MW1 Subtraction to 1000 No Regroup
215
G2S1W5-MW2 Subtraction to 1000 Regroup in Tens and Ones
216
G2S1W6-MW1 Subtraction to 1,000 Regrouping in Hundreds and Tens
216
G2S1W6-MW2 Subtraction to 1,000 Regrouping in Hundreds, Tens and Ones
217
G2S1W7-MW Subtraction tp 1,000 Subtraction Across Zeros
218Using Bar Models: Addition and SubtractionWorkbook (2A) Chapter 4 Using Bar Models: Addition and SubtractionWorkbook (2A) Chapter 1 Subtraction to 1,000
G2S1W8-MW1 Bar Models Part-Part-Whole Model
218
G2S1W8-MW2 Bar Models Comparison Model
219
G2S1W9-MW Bar Models Two Step Problems
2110Multiplication and Division
Workbook (2A) Chapter 5 Multiplication and Division
Workbook (2A) Chapter 4 Multiplication and Division
G2S1W10-MW Multiplication and Division
2111Workbook (2A) Chapter 6 Multiplication Tables for 2, 5 and 10Workbook (2A) Chapter 4 Multiplication and Division for 2
Workbook (2B) Chapter 7 Multiplication and Division for 5 and 10
G2S1W11-MW Multiplication and Division Word Problems
2112Multiplication Tables of 2, 5 and 10
G2S1W12-MW Multiplication Tables of 2
2113
G2S1W13-MW Multiplication Tables of 5
2114
G2S1W14-MW Multiplication Tables of 10
2115LengthWorkbook (2A) Chapter 7 Metric Measurement of Length
Workbook (2A) Chapter 4 Customary Measurement of Length
Workbook (2A) Chapter 3 Length
G2S1W15-MW Length_Measuring and comparing length
2116
G2S1W16-MW Length_Measuring and comparing length
2117
Mass and Volume(Optional)
Workbook (2A) Chapter 8 Mass
G2S1W17-MW Mass
221Mental Math EstimationWorkbook (2B) Chapter 10 Mental Math and EstimationWorkbook (2B) Chapter 6 Addition and Subtraction
G2S2W1-MW Mental Math and Estimation
222
G2S2W2-MW Mental Math and Estimation_Subtraction_Estimation
223MoneyWorkbook (2B) Chapter 11 MoneyWorkbook (2B) Chapter 8 Money
G2S2W3-MW Money_coins and bills
224
G2S2W4-MW Money_comparing and word problem
225FractionsWorkbook (2B) Chapter 12 FractionsWorkbook (2B) Chapter 9 Fractions
use G3S2W6-MW1 Understanding Fractions (with answers)
225
use G3S2W6-MW2 Fractions on the number line (with answers)
226
use G3S2W7-MW2 Compare Fractions With Same Numerator Or Denominator
227
227
G2S2W7-MW2 Fractions_Subtracting like Fractions_Sum less than one
228TimeWorkbook (2B) Chapter 14 TimeWorkbook (2B) Chapter 10 Time
G2S2W8-MW1 Time
228
G2S2W8-MW2 Time
229Multiplication Tables of 3 and 4Workbook (2B) Chapter 15 Multiplication Tables for 3 and 4Workbook (2A) Chapter 5 Multiplication and Division for 3
Workbook (2B) Chapter 7 Multiplication and Division for 4
G2S2W9-MW1 Multiplication Tables of 3
2210
G2S2W10-MW1 Multiplication Tables of 4
2211
G2S2W11-MW1 Multiplication Tables of 3 and 4
2212Using Bar Models: Multiplication and DivisionWorkbook (2B) Chapter 17 Picture GraphsWorkbook (2B) Chapter 11 Tables and Graphs
G2S2W12-MW Bar Model Multiplication Division_WithSolutions
2213
G2S2W13-MW Bar Model Multiplication Division_WithSolutions
2214Picture GraphsWorkbook (2B) Chapter 17 Picture GraphsWorkbook (2B) Chapter 11 Tables and Graphs
G2S2W14-MW Bar Graphs and Line Plots
2215
G2S2W15-MW Bar Graphs and Line Plots
2216Geometry – Lines and Surfaces, Shapes and PatternsWorkbook (2B) Chapter 18 Lines and Surfaces
Workbook (2B) Chapter 19 Shapes and Patterns
Workbook (2B) Chapter 12 Geometry
G2S2W16-MW Shapes and Patterns_with partial solution
2217
G2S2W17-MW Shapes and Patterns
311Numbers to 10,000Workbook (3A) Chapter 1 Numbers to 10, 000Workbook (3A) Chapter 1 Numbers to 10, 000
G3S1W1-MW Numbers to 10,000 Counting and Place Value
312
G3S1W2-MW Numbers to 10,000 Comparing Numbers and Ordering Numbers
313Mental Math and EstimationWorkbook (3A) Chapter 2 Mental Math and EstimationWorkbook (3A) Chapter 2 Addition and Subtraction
G3S1W3-MW Mental Math and Estimation_Mental Add and Subtract
314
G3S1W4-MW Mental Math and Estimation_Estimation
G3S1W5-MW1 Addition to 10,000 No Regroup
315
G3S1W5-MW2 Addition to 10,000 Regroup in Hundreds
315
G3S1W5-MW3 Addition to 1,000 Regroup in Ones, Tens, Hundreds
316Subtraction to 10,000Workbook (3A) Chapter 4 Subtraction to 10, 000Workbook (3A) Chapter 2 Addition and Subtraction
G3S1W6-MW1 Subtraction to 10,000 No Regroup
316
G3S1W6-MW2 Subtraction to 10,000 Regroup in Hundreds and thousands
317
G3S1W7-MW1 Subtraction to 10,000 Regroup in Ones, Tens, Hundreds, Thousand
317
G3S1W7-MW2 Subtraction to 10,000 Across Zeros
318
Using Bar Models: Addition and Subtraction
Workbook (3A) Chapter 5 Using Bar Models: Addition and Subtraction
Workbook (3A) Chapter 2 Addition and Subtraction
G3S1W8-MW Bar Modeling
319Multiplication Tables of 6, 7, 8, and 9Workbook (3A) Chapter 6 Multiplication Tables of 6, 7, 8, and 9Workbook (3A) Chapter 4 Multiplication Tables of 6, 7, 8, and 9
G3S1W9-MW1 Multiplication Tables of 6
319
G3S1W9-MW2 Multiplication Tables of 7
3110
G3S1W10-MW1 Multiplication Tables of 8
3110
G3S1W10-MW2 Multiplication Tables of 9
3111
Workbook (3A) Chapter 8 Division
Workbook (3A) Chapter 3 Multiplication and Division
G3S1W11-MW Division
3112MultiplicationWorkbook (3A) Chapter 7 Multiplication
G3S1W12-MW Multiplication No Regroup
3113
G3S1W13-MW Multiplication With Regroup
3114DivisionWorkbook (3A) Chapter 8 Division
G3S1W14-MW Division Quotient and remainder
3115
G3S1W15-MW Division with regroup
3116Using Bar Models: Multiplication and DivisionWorkbook (3A) Chapter 9 Using Bar Models: Multiplication and
Division
G3S1W16-MW Bar Models Multiplication and Division
3117
G3S1W17-MW Bar Models Division
321MoneyWorkbook (3B) Chapter 10 MoneyWorkbook (3B) Chapter 8 Money
322
G3S2W2-MW Money Subtraction
323Metric Length, Mass, and Volume/Real-World ProblemsWorkbook (3B) Chapter 11 Metric Length, Mass, and Volume
Workbook (3B) Chapter 12 Real-World Problems: Measurement
Workbook (3B) Chapter 6 Mass and Weight
Workbook (3B) Chapter 7 Capacity
G3S2W3-MW Metric Length
324
G3S2W4-MW1 Metric Mass and Volume
324
G3S2W4-MW2 Metric Mass and Volume word problem
325
Bar Graphs and Line Plots
Workbook (3B) Chapter 13 Bar Graphs and Line Plots
Workbook (3B) Chapter 11 Data Analysis
G3S2W5-MW Bar Graphs and Line Plots
326FractionsWorkbook (3B) Chapter 14 FractionsWorkbook (3B) Chapter 9 Fractions
326
G3S2W6-MW2 Fractions on the number line (with answers)
327
G3S2W7-MW1 Equivalent Fractions
327
G3S2W7-MW2 Compare Fractions With Same Numerator Or Denominator
327
G3S2W7-MW3 Compare Fractions With Different Numerator And Denominator
328
G3S2W8-MW1 Adding like Fractions (Sum Less Than 1)
328
G3S2W8-MW2 Subtracting like Fractions (Sum Less Than 1)
328
G3S2W8-MW3 Fraction of a Set Part 1
329Time and TemperatureWorkbook (3B) Chapter 16 Time and TemperatureWorkbook (3B) Chapter 10 Time
G3S2W9-MW1 Time
329
G3S2W9-MW2 Time
3210
Angles and Lines
Workbook (3B) Chapter 17 Angles and Lines
Workbook (3B) Chapter 12 Geometry
G3S2W10-MW2 Angles
3211Two-Dimensional ShapesWorkbook (3B) Chapter 18 Two–Dimensional ShapesWorkbook (3B) Chapter 12 Geometry
G3S2W11-MW1 Two-dimensional Shapes
3212
G3S2W12-MW Two-dimensional Shapes
3213Area and PerimeterWorkbook (3B) Chapter 19 Area and PerimeterWorkbook (3B) Chapter 12 Area and Perimeter
G3S2W13-MW Area and Perimeter
3214
G3S2W14-MW Area and Perimeter
3215
G3S2W15-MW Area and Perimeter
411Numbers to 100,000Workbook (4A) Chapter 1 Place Values of Whole NumbersWorkbook (4A) Chapter 1 Numbers to 10, 000
G4S1W1-MW Numbers to 100,000 Counting and Place Value
412
G4S1W2-MW Numbers to 100,000 Comparing Numbers and Ordering Numbers_WithSolutions
413Estimation and Number TheoryWorkbook (4A) Chapter 2 Estimation and Number TheoryWorkbook (4A) Chapter 1 Numbers to 10, 000
G4S1W3-MW Estimation
414
G4S1W4-MW Factors
415
G4S1W5-MW Multiples
416Whole Number Multiplication and DivisionWorkbook (4A) Chapter 3 Whole Number Multiplication and DivisionWorkbook (4A) Chapter 2 The Whole Operations of Whole Numbers
G4S1W6-MW Multiply by a 1-digit Number
417
G4S1W7-MW Multiply by a 2-digit Number
418
G4S1W8-MW Modeling Division with regroup
419
G4S1W9-MW Dividing by a 1-digit Number
4110
G4S1W10-MW Multiplication Division word problem
4111FractionsWorkbook (4A) Chapter 5 Fractions and Mixed NumbersWorkbook (4A) Chapter 3 Fractions
Workbook (4A) Chapter 4 Operations on Fractions
4111
G4S1W11-MW2 Subtracting Unlike Fractions
4112
G4S1W12-MW Mixed Numbers
4113
G4S1W13-MW Mixed numbers to improper fractions
4114
G4S1W14-MW Fraction of a set (Part 2)
4115
G4S1W15-MW Word Problems_Add Subtract Fraction_Fraction of a set
4116Tables and Line GraphsWorkbook (4A) Chapter 4 Tables and Line GraphsWorkbook (4B) Chapter 10 Bar Graphs and Line Plots
G4S1W16-MW Tables and Lines
4117
G4S1W17-MW Tables and Lines
421Decimals/Adding and Subtracting DecimalsWorkbook (4B) Chapter 7 Decimals, Chapter 8 Adding and Subtracting DecimalsWorkbook (4B) Chapter 6 Decimals, Chapter 4 Operations of Decimals
G4S2W1-MW Decimals_Understanding tenths and hundredths
422
G4S2W2-MW Decimals_Comparing and Rounding
423
G4S2W3-MW Decimals_Fractions and Decimals
424
425
G4S2W5-MW Decimals_Subtracting Decimals
426AnglesWorkbook (4B) Chapter 9 AnglesWorkbook (4B) Chapter 8 Geometry
G4S2W6-MW Angles
427
G4S2W7-MW Angles
428Perpendicular and Parallel Line SegmentsWorkbook (4B) Chapter 10 Perpendicular and Parallel Line SegmentsWorkbook (4B) Chapter 8 Geometry
G4S2W8-MW Perpendicular and Parallel Line Segments
429
G4S2W9-MW Perpendicular and Parallel Line Segments
4210Squares and RectanglesWorkbook (4B) Chapter 11 Squares and RectanglesWorkbook (4B) Chapter 8 Geometry
G4S2W10-MW Squares and Rectangles - With Solutions
4211
G4S2W11-MW Squares and Rectangles_WithSolutions
4212Area and PerimeterWorkbook (4B) Chapter 12 Area and PerimeterWorkbook (4B) Chapter 8 Area and Perimeter
G4S2W12-MW Area and Perimeter_WithSolutions
4213
G4S2W13-MW Area and Perimeter_WithSolutions
4214SymmetryWorkbook (4B) Chapter 13 SymmetryWorkbook (4B) Chapter 8 Geometry
G4S2W14-MW Symmetry_WithSolutions
4215
511Numbers to 10,000,000Workbook (5A) Chapter 1 Whole NumbersWorkbook (5A) Chapter 1 Whole Numbers
G5S1W1-MW Numbers to 10,000,000 Counting and Place Value_WithSolutions
512
G5S1W2-MW Numbers to 10,000,000 Comparing Numbers and Ordering Numbers_WithSolutions
513Whole Number Multiplication and DivisionWorkbook (5A) Chapter 2 Whole Number Multiplication and DivisionWorkbook (5A) Chapter 1 Numbers to 10, 000
G5S1W3-MW Whole Number Multiplication
514
G5S1W4-MW Whole Number Division
515
G5S1W5-MW Whole Number Order of Operations
516
G5S1W6-MW Whole Number Word Problems Bar Modeling
517Fractions and Mixed NumbersWorkbook (5A) Chapter 3 Fractions and Mixed NumbersWorkbook (5A) Chapter 3 Fractions
517
G5S1W7-MW2 Subtracting Unlike Unrelated Fractions
518
G5S1W8-MW Fractions_Division Expressions as Decimals
519
G5S1W9-MW Add and Subtract Mixed Numbers
5110Multiplying and Dividing Fractions and Mixed NumbersWorkbook (5A) Chapter 4 Multiplying and Dividing Fractions and Mixed
Numbers
Workbook (5A) Chapter 3 Fractions
Workbook (5A) Chapter 4 Multiply and Divide Fractions
G5S1W10-MW Multiply Proper Fractions
5111
G5S1W11-MW Multiply Mixed Numbers
5112
G5S1W12-MW Dividing Fraction by a whole number
5113
G5S1W13-MW Multiplying Dividing Fraction word problem
5114Area of a TriangleWorkbook (5A) Chapter 6 Area of a TriangleWorkbook (5A) Chapter 5 Perimeter and Area
G5S1W14-MW Area of a triangle_WithSolutions
5115
G5S1W15-MW Area of a triangle_WithSolutions
5116*Algebra(Optional)Workbook (5A) Chapter 5 Algebra
G5S1W16-MW Algebra_using letters as numbers
5117
G5S1W17-MW Algebra_using letters as numbers
521DecimalsWorkbook (5B) Chapter 8 DecimalsWorkbook (5B) Chapter 7 Decimals
G5S2W1-MW Decimals
522
G5S2W2-MW Decimals_Rounding and Fractions
523Multiplying and Dividing DecimalsWorkbook (5B) Chapter 9 Multiplying and Dividing DecimalsWorkbook (5B) Chapter 7 Decimals
Workbook (5B) Chapter 8 More Calculations
G5S2W3-MW Multiplying Decimals
524
G5S2W4-MW Dividing Decimals
525
G5S2W5-MW Estimating Decimals
526
G5S2W6-MW Decimals Word Problems
527PercentWorkbook (5B) Chapter 10 PercentWorkbook (5B) Chapter 7 Decimals
G5S2W7-MW Percent
528
G5S2W8-MW Percent
529
G5S2W9-MW Percent Problem Solving
5210Properties of Triangles and Four-sided FiguresWorkbook (5B) Chapter 13 Properties of Triangles and Four-sided FiguresWorkbook (5B) Chapter 7 Decimals
G5S2W10-MW Properties of Triangles and 4-sided figures
5211
G5S2W11-MW1 Properties of Triangles and 4-sided figures
5211
G5S2W11-MW2 Properties of Triangles and 4-sided figures
5212
G5S2W12-MW1 Properties of Triangles and 4-sided figures_WithSolutions
5212
G5S2W12-MW2 Properties of Triangles and 4-sided figures_WithSolutions
5213
Three Dimensional Shapes
Math in Focus Workbook (5B) Chapter 14 Three Dimensional Shapes, Chapter 15 Surface Area and Volume,Primary Mathematics Workbook (5B) Chapter 9 Volume
G5S2W13-MW Three Dimensional Figures_WithSolutions
5214Surface Area and Volume
G5S2W14-MW Surface Area and Volume_With Solutions
5215
G5S2W15-MW Surface Area and Volume_WithSolutions
5216
G5S2W16-MW Surface Area and Volume_WithSolutions
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http://pari.math.u-bordeaux.fr/archives/pari-users-0706/msg00011.html
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Bill Allombert on Mon, 11 Jun 2007 19:01:31 +0200
Re: random sequence reversal
```On Mon, Jun 11, 2007 at 04:04:59PM +0200, Alain SMEJKAL wrote:
> Hello all,
>
> Is there a way to reverse a number sequence generated by the random function
> ?
> More precisely my problem is to retrieve the first generated numbers when
> many random(k) iterations were done with the same k and the only last random
> numbers are known (and maybe useful, the used Pari session is still alive).
You can do something like:
back(k)=local(a,n);a=random();setrand(1);n=0;while(random()!=a,n++);setrand(1);for(i=1,n-k,random())
back(n) will set the generator n step backward.
Note that the generator initial seed is the same in all PARI sessions for
a given version.
> ps : Pari version is 2.3.2
In that case you can do a true reversion:
rev(rand)=542687373*(rand-12347)%2^31
if rand is the last random() value, this return the previous one,
that you can feed to setrand().
? random()
%1 = 1000288896
? random()
%2 = 768462011
? random()
%3 = 892785826
? random()
%4 = 739165157
? rev(%)
%5 = 892785826
? rev(%)
%6 = 768462011
? rev(%)
%7 = 1000288896
? setrand(%)
%8 = 1000288896
? random()
%9 = 768462011
? random()
%10 = 892785826
? random()
%11 = 739165157
? random()
%12 = 1874708212
Note: random(N) actually compute several value of random() and combine
them, and the number of random() step used by a random(N) call depend
on the PARI version.
Cheers,
Bill.
```
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| 2.578125
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https://www.kapom.org/big-family/how-big-a-tax-return-will-q-family-of-six-get-with-a-48k-income.html
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# How Big A Tax Return Will Q Family Of Six Get With A 48k Income?
## How much will I get back in taxes if I make 45000?
Income Tax Calculator California If you make \$45,000 a year living in the region of California, USA, you will be taxed \$9,044. That means that your net pay will be \$35,956 per year, or \$2,996 per month. Your average tax rate is 20.1% and your marginal tax rate is 27.5%.
## How much will I get back in taxes if I make 50000?
If you make \$50,000 a year living in the region of California, USA, you will be taxed \$10,417. That means that your net pay will be \$39,583 per year, or \$3,299 per month. Your average tax rate is 20.8% and your marginal tax rate is 33.1%.
You might be interested: Quick Answer: What Is The Best Way For A Very Ery Large Family To Enjoy Croatia?
## How much will I get back in taxes if I make 40000?
If you make \$40,000 a year living in the region of California, USA, you will be taxed \$7,672. That means that your net pay will be \$32,328 per year, or \$2,694 per month. Your average tax rate is 19.2% and your marginal tax rate is 27.5%.
## How much is 48k a year after taxes?
Income Tax Calculator California If you make \$48,000 a year living in the region of California, USA, you will be taxed \$9,868. That means that your net pay will be \$38,132 per year, or \$3,178 per month. Your average tax rate is 20.6% and your marginal tax rate is 27.5%.
## How much will I get back in taxes if I make 35000?
If you make \$35,000 a year living in the region of California, USA, you will be taxed \$6,366. That means that your net pay will be \$28,634 per year, or \$2,386 per month. Your average tax rate is 18.2% and your marginal tax rate is 26.1%.
## How much will I get back in taxes if I make 20000?
If you make \$20,000 a year living in the region of California, USA, you will be taxed \$2,756. That means that your net pay will be \$17,244 per year, or \$1,437 per month. Your average tax rate is 13.8% and your marginal tax rate is 22.1%.
## How much will I get back in taxes if I make 60000?
If you make \$60,000 a year living in the region of California, USA, you will be taxed \$14,053. That means that your net pay will be \$45,947 per year, or \$3,829 per month. Your average tax rate is 23.4% and your marginal tax rate is 40.2%.
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## How much do I owe in taxes if I made 30000?
If you make \$30,000 a year living in the region of California, USA, you will be taxed \$5,103. That means that your net pay will be \$24,897 per year, or \$2,075 per month. Your average tax rate is 17.0% and your marginal tax rate is 25.3%.
## How much does Head of Household get back in taxes?
Heads of household can claim a 50% larger standard tax deduction than single filers. They also benefit from wider tax brackets on lower income levels, among other benefits. Suspecting abuse, Congress recently required tax preparers to get documentation that qualifies someone to be a head of household.
## How much do you get back in taxes for 2 dependents?
The child tax credit is worth up to \$2,000 for the 2020 tax year, for those who meet its requirements. Having dependent children may also allow you to claim other significant tax credits, including the earned income credit (EIC).
## Do you get a bigger tax refund if you make less money?
Having less taken out will give you bigger paychecks, but a smaller tax refund (or potentially no tax refund or a tax bill at the end of the year). Any additional income tax you would like withheld from each paycheck.
## Can you live off of 50k a year?
Earning \$50,000 a year should be plenty to live on in America. The nation’s median income is just over \$60,000, meaning that \$50,000 per annum is the sort of salary that should clearly secure the basics, at the least. However, depending on where you live, that’s not always true.
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## How much is \$50 000 a year hourly?
An average person works about 40 hours per week, which means if they make \$50,000 a year, they earn \$24.04 per hour.
## How much is 48000 a year after tax?
If your salary is £ 48,000, then after tax and national insurance you will be left with £36,280. This means that after tax you will take home £3,023 every month, or £698 per week, £139.60 per day, and your hourly rate will be £23.08 if you’re working 40 hours/week.
## What salary do I need to take home 48000?
On a £ 48,000 salary, your take home pay will be £36,302 after tax and National Insurance. This equates to £3,025 per month and £698 per week. If you work 5 days per week, this is £140 per day, or £17 per hour at 40 hours per week.
## What Is The Best Large Dog Breed For A Family With Small Children?What Is The Best Large Dog Breed For A Family With Small Children?
Contents1 What big dogs are good with small kids?2 What is the best big dog to have around children?3 What is the friendliest big dog?4 What is the calmest large
big family
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# 使用規則運算式中 NPSUse Regular Expressions in NPS
## 模式比參考資料Pattern-matching reference
`\` 標記的下一個字元符合字元。Marks the next character as a character to match. `/n/ matches the character "n". The sequence /\n/ matches a line feed or newline character.`
`^` 符合輸入或一行的開頭。Matches the beginning of the input or line.
`\$` 符合輸入或一行的結尾。Matches the end of the input or line.
`*` 符合上述字元或多個時間。Matches the preceding character zero or more times. `/zo*/ matches either "z" or "zoo."`
`+` 符合上述字元一或多個時間。Matches the preceding character one or more times. `/zo+/ matches "zoo" but not "z."`
`?` 比對前一個字元零或一次。Matches the preceding character zero or one times. `/a?ve?/ matches the "ve" in "never."`
`.` 符合新行字元以外的任何一個字元。Matches any single character except a newline character.
`( pattern )` 符合「模式」，並會記住相符項目。Matches "pattern" and remembers the match. `To match ( ) (parentheses), use "\(" or "\)".`
' x`x Y 'y ` 符合 x 或 y。Matches either x or y.
`{ n }` 符合完全 n 次 \（n 是 non\ 負 integer\）。Matches exactly n times (n is a non-negative integer). `/o{2}/ does not match the "o" in "Bob," but matches the first two instances of the letter o in "foooood."`
`{ n ,}` 符合至少 n 時間 \（n 是 non\ 負 integer\）。Matches at least n times (n is a non-negative integer). `/o{2,}/ does not match the "o" in "Bob" but matches all of the instances of the letter o in "foooood." /o{1,}/ is equivalent to /o+/.`
`{ n , m }` 符合至少 n 和最 m 時間 \（m 和 n 是 non\ 負 integers\）。Matches at least n and at most m times (m and n are non-negative integers). `/o{1,3}/ matches the first three instances of the letter o in "fooooood."`
`[ xyz ]` 比對任何一個括號字元 (a character set)。Matches any one of the enclosed characters (a character set). `/[abc]/ matches the "a" in "plain."`
`[^ xyz ]` 比對任何不包含的字元 \ (負字元 set)。Matches any characters that are not enclosed (a negative character set). `/[^abc]/ matches the "p" in "plain."`
`\b` 符合 word 邊界 \ (例如，space)。Matches a word boundary (for example, a space). `/ea*r\b/ matches the "er" in "never early."`
`\B` 符合非邊界。Matches a nonword boundary. `/ea*r\B/ matches the "ear" in "never early."`
`\d` 符合數字字元 \（相當於數字 0 到 9\）。Matches a digit character (equivalent to digits from 0 to 9).
`\D` 符合非數字字元 \ (相當於`[^0-9]`)。Matches a nondigit character (equivalent to `[^0-9]`).
`\f` 相符項目換字元。Matches a form feed character.
`\n` 相符項目換字元。Matches a line feed character.
`\r` 比對換字元。Matches a carriage return character.
`\s` 符合空間，索引標籤，然後送紙包括任何空格字元 \ (相當於`[ \f\n\r\t\v]`)。Matches any white space character including space, tab, and form feed (equivalent to `[ \f\n\r\t\v]`).
`\S` 比對任何非空格字元 \ (相當於`[^ \f\n\r\t\v]`)。Matches any non-white space character (equivalent to `[^ \f\n\r\t\v]`).
`\t` 符合] 索引標籤的字元。Matches a tab character.
`\v` 符合垂直] 索引標籤的字元。Matches a vertical tab character.
`\w` 比對任何文字的字元，包括底線 \ (相當於`[A-Za-z0-9_]`)。Matches any word character, including underscore (equivalent to `[A-Za-z0-9_]`).
`\W` 比對任何 non\ 字字元，不含底線 \ (相當於`[^A-Za-z0-9_]`)。Matches any non-word character, excluding underscore (equivalent to `[^A-Za-z0-9_]`).
`\ num` 指向記憶相符項目 \ (`?num`、num 位置是正 integer)。Refers to remembered matches (`?num`, where num is a positive integer). 可以使用這個選項只在取代設定屬性操作時的文字方塊。This option can be used only in the Replace text box when configuring attribute manipulation. `\1` 將會取代項目儲存在第一次記憶相符項目。replaces what is stored in the first remembered match.
`/ n /` 讓 ASCII 代碼插入規則運算式 \ (`?n`、n 是進位、十六進位或小數點 esc 鍵 value)。Allows the insertion of ASCII codes into regular expressions (`?n`, where n is an octal, hexadecimal, or decimal escape value).
## 網路原則屬性範例Examples for network policy attributes
• 若要指定的 899 區碼在所有的電話號碼，語法為：To specify all phone numbers within the 899 area code, the syntax is:
`899.*`
• 若要指定 192.168.1 開始 IP 位址，語法為：To specify a range of IP addresses that begin with 192.168.1, the syntax is:
`192\.168\.1\..+`
## 範例操作領域中的使用者名稱屬性名稱Examples for manipulation of the realm name in the User Name attribute
• 尋找：@microsoft\.comFind: @microsoft\.com
• 取代：Replace:
user@example.microsoft.comexample.microsoft.com\userTo replace user@example.microsoft.com with example.microsoft.com\user
• 尋找：Find:`(.*)@(.*)`
• 取代：Replace:`\$2\\$1`
• 尋找：Find:`(.*)\\(.*)`
• 取代：specific_domainReplace: specific_domain`\\$2`
• 尋找：Find:`\$`
• 取代：@specific_domainReplace: @specific_domain
## 範例 RADIUS 郵件轉寄 proxy 伺服器Example for RADIUS message forwarding by a proxy server
• NetBIOS 名稱 **:NetBIOS name**: `WCOAST`
• 模式 **:Pattern**: `^wcoast\\`
• NetBIOS 名稱 **:NetBIOS name**: `WCOAST`
• UPN 尾碼 **:UPN suffix**: `wcoast.microsoft.com`
• 模式 **:Pattern**: `^wcoast\\|@wcoast\.microsoft\.com\$`
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String Pattern Matching: The Knuth-Morris-Pratt Algorithm C/C++ Help
Ideally, we would like an algorithm that works in O(LenglhP+ LengthS) time. This is optimal.for this problem, all in the worst case it is necessary to look at all characters in the pattern and string at least once. V(e want to search the string for the pattern without moving backwards in the string. That is, if a mismatch occurs we want to use our knowledge of the characters in the pattern and the position in the pattern, where the mismatch occurred to determine where we should continue the search. Knuth, Morris, and Pratt have developed a pattern-matching algorithm that works in this way and has linear complexity. Using their example, suppose
pat = ‘a b c a b c a e a b’
We observe that the search for a match can proceed by comparing S; +4 and the second character in pal, b. This is the first place a partial match can occur by sliding the pattern pat towards the right. Thus, by knowing the characters in the pattern and the position in the pattern where a mismatch occurs with a character in s. we can determine where in the pattern to continue the search for a match without moving backwards in s. To formalize this. we define a failure function for a pattern. .
From the definition of the failure function. we arrive at the following rule for pattern matching: If a partial match is found such that and then matching may be resumed by comparing s, and P1(j_1)+1 then we may continue by comparing Si+1 and Po. This pattern-matching rule translates to function FastFind (Program 2.15). FastFind uses an array of lnts. f, to represent the failure function a private data member of String. .
Analysis of FastFind: The correctness of FastFind follows from the definition of the .:. failure function. To determine the computing time, we observe that lines 8 and II can be executed for a total of at most LengthS times, since in each iteration PosS is incremented by I but PosS is never decremented in the algorithm. As a result, PosP can move right on pat at most LengthS times (line 8). Since each execution of the else clause in line 12 moves PosP left 011 pat, it follows that this clause can be executed at most LengthS times, since otherwise PosP becomes less than O. Consequently, the maximum number of iterations of the while loop is LengthS, and the computing time of FastFind is O(LengthS).
From the analysis of FastFind, it follows that if we can compute the failure function in O(LengthP) time. then the entire pattern-matching process will have a computing time proportional to the sum of the lengths of the string and pattern. Fortunately, there is a fast way to compute the failure function. This is based upon the following restatement of the failure function:
Analysis of fail: In each iteration of the while loop the value of i decreases (by the definition of). The variable i is reset at the beginning of each iteration of the Cor loop. However, it is either reset to -1 (when j = 1 when the previous iteration went through line II), or it is reset to a value 1 greater than its terminal value on the previous iteration i.e when the previous iteration went through line 10). Since only LengthP – 1 executions of line 8 are made, the value of i therefore has a total increment of at most LengthP – 1. Hence it cannot be decremented more than LengthP – 1 times. Consequently, the while loop is iterated at most LengthP – 1 times over the whole algorithm, and the computing time of/ail is O(LengthP).
Note that when the failure function is not known in advance, pattern matching can be carried out in time O(LengthP + LengthS) by first computing the failure function using function and then performing a pattern match using function FastFind.
EXERCISES
1. Write a function String::frequency that determines the frequency of occurrence of each of the distinct characters in the string. Test your function using suitable data.
2. Write a function, Delete, that accepts two integers, star and length. The function computes a new string that is equivalent to the original string, except that length characters beginning at start have been removed.
3. Write a function, String::CharDelete, that accepts a character c. The function returns the string with all occurrences of c removed.
Related C/C++ Assignments
Posted on November 11, 2015 in ARRAYS
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Which of the following statements contains a quotient? A. 7 6 = 42 B. 7 6 = 1 C. 7 6 = 13 D. 42 6 = 7
42 / 6 = 7 contains a quotient.
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# Minh Tan Trinh
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# Physics
posted by Jon
1)As a sound source moves away from a stationary observer, the number of waves will.
A)increase
B)decrease
C)remain the same
D)need to know the speed of the source
(answer:decrease.) As the detector recedes from the source, the relative velocity is smaller, resulting in a decrease in the wave crests reaching the detector each second. (thats from the book)
2)How fast should a car move toward you for the car's horn to sound 2.88% higher in frequency than when the car is stationary? The speed of sound is 343 m/s
1. bobpursley
I agree with the book, the number of waves per second (which is frequency) decreases. I am still not certain of the poor wording of the question.
2. I will be happy to critique your calcs.
2. Jon
2.88% times 343
= 9.8784 I don't understand this one b/c the doppler equation doesnt work
3. bobpursley
actually, it does. And, you just used the approximate solution in your argument.
F/fo= 343/(343-v)= 1/(1-v/343) or
1-v/343=1/1.0288
solve for v. As I recall, this is one of your choices.
4. Jon
drwls already worked it out for me,I found it on another page. answer is 9.6 m/s
5. bobpursley
good. I hope you understand how it is worked.
## Similar Questions
1. ### DrBob222
As a sound source moves away from a stationary observer, the number of waves will. decrease
2. ### physics
1)As a sound source moves away from a stationary observer, the number of waves will. decrease 2)How fast should a car move toward you for the car's horn to sound 2.88% higher in frequency than when the car is stationary?
3. ### physics
1)As a sound source moves away from a stationary observer, the number of waves will. decrease 2)How fast should a car move toward you for the car's horn to sound 2.88% higher in frequency than when the car is stationary?
4. ### physics
1)As a sound source moves away from a stationary observer, the number of waves will. A)increase B)decrease C)remain the same D)need to know the speed of the source (answer:decrease.) As the detector recedes from the source, the relative …
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In this equation, vd is the speed of the detector, and vs is the speed of the source. Also, fs is the frequency of the source, and fd is the frequency of the detector. If the detector moves toward the source, vd is positive. If the …
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# A curve y = f(x) passing through the point (1,2) satisfies the differential equation x dy/dx + y = bx4 such that
more_vert
A curve y = f(x) passing through the point (1,2) satisfies the differential equation $x{dy\over dx}+y=bx^4$ such that $\int\limits_1^2f(y)dy={62\over5}$. The value of b is
(1) 10
(2) 11
(3) $32\over5$
(4) $62\over5$
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Questions?
See the FAQ
or other info.
# Polytope of Type {6,44}
Atlas Canonical Name : {6,44}*528b
if this polytope has a name.
Group : SmallGroup(528,160)
Rank : 3
Schlafli Type : {6,44}
Number of vertices, edges, etc : 6, 132, 44
Order of s0s1s2 : 33
Order of s0s1s2s1 : 4
Special Properties :
Compact Hyperbolic Quotient
Locally Spherical
Non-Orientable
Flat
Related Polytopes :
Facet
Vertex Figure
Dual
Facet Of :
{6,44,2} of size 1056
Vertex Figure Of :
{2,6,44} of size 1056
Quotients (Maximal Quotients in Boldface) :
11-fold quotients : {6,4}*48b
22-fold quotients : {3,4}*24
Covers (Minimal Covers in Boldface) :
2-fold covers : {6,44}*1056
3-fold covers : {18,44}*1584b, {6,132}*1584d
Permutation Representation (GAP) :
```s0 := ( 2, 3)( 6, 7)(10,11)(14,15)(18,19)(22,23)(26,27)(30,31)(34,35)(38,39)
(42,43);;
s1 := ( 3, 4)( 5,41)( 6,42)( 7,44)( 8,43)( 9,37)(10,38)(11,40)(12,39)(13,33)
(14,34)(15,36)(16,35)(17,29)(18,30)(19,32)(20,31)(21,25)(22,26)(23,28)
(24,27);;
s2 := ( 1, 8)( 2, 7)( 3, 6)( 4, 5)( 9,44)(10,43)(11,42)(12,41)(13,40)(14,39)
(15,38)(16,37)(17,36)(18,35)(19,34)(20,33)(21,32)(22,31)(23,30)(24,29)(25,28)
(26,27);;
poly := Group([s0,s1,s2]);;
```
Finitely Presented Group Representation (GAP) :
```F := FreeGroup("s0","s1","s2");;
s0 := F.1;; s1 := F.2;; s2 := F.3;;
rels := [ s0*s0, s1*s1, s2*s2, s0*s2*s0*s2, s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1,
s2*s0*s1*s2*s0*s1*s0*s1*s2*s0*s1*s2*s0*s1*s0*s1,
s0*s1*s2*s1*s0*s1*s2*s1*s0*s1*s2*s1*s0*s1*s2*s1,
s1*s2*s1*s2*s1*s2*s1*s2*s1*s0*s2*s1*s0*s2*s1*s0*s2*s1*s2*s1*s2*s1*s2*s1*s2 ];;
poly := F / rels;;
```
Permutation Representation (Magma) :
```s0 := Sym(44)!( 2, 3)( 6, 7)(10,11)(14,15)(18,19)(22,23)(26,27)(30,31)(34,35)
(38,39)(42,43);
s1 := Sym(44)!( 3, 4)( 5,41)( 6,42)( 7,44)( 8,43)( 9,37)(10,38)(11,40)(12,39)
(13,33)(14,34)(15,36)(16,35)(17,29)(18,30)(19,32)(20,31)(21,25)(22,26)(23,28)
(24,27);
s2 := Sym(44)!( 1, 8)( 2, 7)( 3, 6)( 4, 5)( 9,44)(10,43)(11,42)(12,41)(13,40)
(14,39)(15,38)(16,37)(17,36)(18,35)(19,34)(20,33)(21,32)(22,31)(23,30)(24,29)
(25,28)(26,27);
poly := sub<Sym(44)|s0,s1,s2>;
```
Finitely Presented Group Representation (Magma) :
```poly<s0,s1,s2> := Group< s0,s1,s2 | s0*s0, s1*s1, s2*s2,
s0*s2*s0*s2, s0*s1*s0*s1*s0*s1*s0*s1*s0*s1*s0*s1,
s2*s0*s1*s2*s0*s1*s0*s1*s2*s0*s1*s2*s0*s1*s0*s1,
s0*s1*s2*s1*s0*s1*s2*s1*s0*s1*s2*s1*s0*s1*s2*s1,
s1*s2*s1*s2*s1*s2*s1*s2*s1*s0*s2*s1*s0*s2*s1*s0*s2*s1*s2*s1*s2*s1*s2*s1*s2 >;
```
References : None.
to this polytope
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### ADDITIVE COMBINATORICS TERENCE TAO PDF
Buy Additive Combinatorics (Cambridge Studies in Advanced Mathematics) on ✓ FREE Terence Tao (Author), Van H. Vu (Contributor). out of . Additive Combinatorics has 7 ratings and 1 review. Pietro said: The material is brilliantly motivated, and intuition all but oozes out of its pages. (One. Advanced Topics in Discrete Mathematics: Additive Combinatorics. MW: Baker Hall B, PM Terence Tao’s Lecture Notes on Additive.
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Thanks for the corrections! I am getting back to my thesis, and was wondering about the solution of exercise 2.
After the second display, and should be and respectively, and similarly on the third display. Retrieved from ” https: Dear Terry, I was trying to do the following: In the last sentence of the proof of Theorem 9.
Terenve 42, Exercise 1. Thus it remains to verify the S 1.
### Additive Combinatorics – Terence Tao, Van H. Vu – Google Books
Tao, Sorry, I made a typo too, that is Ex: I think that in Exercise 4. I cannot find additive structure of these sets. I think should read as an estimate on the binomial coefficient.
HEAT EXCHANGER DESIGN BY ARTHUR P.FRAAS PDF
Writing up the results, and exploring negative t Career advice The uncertainty principle A: Show that has cardinalitybut has cardinality at least. Alongkorn Chaisen rated it it was amazing Feb 14, This particular display is just a consequence of the monotonicity properties of stated at the beginning of the paragraph.
One way to proceed is to first use the lossy argument, apply Exercise 2. ST 6 non-technical admin 43 advertising 30 diversions 4 media 12 journals 3 obituary 12 opinion 30 paper book 17 Companion 13 update 19 question polymath 83 talk 64 DLS 19 teaching A — Real analysis 11 B — Real analysis 21 C — Real analysis 6 A — complex analysis 9 C — complex analysis 5 A — analytic prime number theory 16 A — ergodic theory 18 A — Hilbert’s fifth problem 12 A — Incompressible fluid equations 5 A — random matrices 14 B — expansion in groups 8 B — Higher order Fourier analysis 9 B — incompressible Euler equations 1 A — probability theory 6 G — poincare conjecture 20 Logic reading seminar 8 travel Oh, I get it!
In the paragraph after 1. In the last line, should be. Thanks for pointing out the issue!
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0
# How do you convert 3.75 into a fraction?
Updated: 4/28/2022
Wiki User
10y ago
3.75 = 375/100 which can be simplified to 15/4.
3.75 = 375/100 which can be simplified to 15/4.
3.75 = 375/100 which can be simplified to 15/4.
3.75 = 375/100 which can be simplified to 15/4.
Wiki User
10y ago
Wiki User
10y ago
3.75 = 375/100 which can be simplified to 15/4.
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Cody
# Problem 233. Reverse the vector
Solution 276905
Submitted on 9 Jul 2013 by Dariusz
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
This solution is outdated. To rescore this solution, sign in.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 1; y_correct = 1; assert(isequal(reverse(x),y_correct))
2 Pass
%% x = -10:1; y_correct = 1:-1:-10; assert(isequal(reverse(x),y_correct))
3 Pass
%% x = 'able was i ere i saw elba'; y_correct = 'able was i ere i saw elba'; assert(isequal(reverse(x),y_correct))
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# How the Building Exterior Affects HVAC Energy Efficiency
Roofs, walls, windows: All need to be designed to reduce demand on HVAC.
By Miles Martschink
Energy efficient HVAC systems require a building’s envelope (the walls, windows, and roof) to be designed to constantly reduce overall demand on the system. Traditional building envelope analysis involves calculating the U-value of all the various components of the envelope to establish how much heat flows through each surface. Often this analysis focuses on the most extreme temperatures the building will experience, based on its location. However, buildings do not operate continuously in extreme conditions, which means that accurate optimization of the building envelope must consider all conditions that occur throughout the year. The use of computer-aided energy modeling allows for this type of detailed analysis to be quickly and accurately performed. The process involves creating a 3D model of the entire building, which is then simulated with the building’s local weather conditions and the internal loads created by the people and equipment inside the building, for an entire year. The results of this process deliver detailed information on both thermal conditions inside each space (and their associated surfaces), as well as the energy consumption of all the building systems. This process reveals a complex system of heat being absorbed, being rejected, and flowing through a building, making energy modeling the ideal measurement tool to maximize the energy savings provided by the building.
What’s the Weather?
Extreme weather conditions are the maximum temperatures that an HVAC system is designed to be capable of handling. But how often do these maximum temperatures occur? The published ASHRAE Climatic Design conditions offer answers of 99.6 percent and 0.4 percent design temperatures for heating and cooling, respectively.
As a real-world example, the published values for 2017 in Charleston, S.C., are a 27.5 F dry bulb temperature for heating and a 94.3 F dry bulb temperature for cooling. This means that for 99.6 percent of the year, the temperature in Charleston will be higher than 27.5 F, and for 0.4 percent of the year the temperature will exceed 94.3 F. This equates to roughly 70 hours of the entire year, where the temperatures are equal to or more extreme than the design values.
But what happens for the remaining 8,690 hours of the year? Energy models use the typical meteorological year (TMY3) weather data which is based on historical weather data for a location. An analysis of the TMY3 weather data for Charleston presents a very different perspective, as summarized in the following table:
The temperatures greater than 56.1 F and less than or equal to 84.8 F represent the part-load conditions when the HVAC system must fluctuate between heating and cooling operation. In Charleston, part-load conditions account for 64.4 percent of the entire year. Since these are the typical conditions the building will endure, it is important for the design team to understand how the building’s envelope will interact with the HVAC system during this time.
Impact of Windows
Windows are inherent weak spots in the building envelope as they allow higher levels of heat conduction and solar radiation to penetrate into the building. The amount of solar radiation that a window allows into a space is expressed by the solar heat gain coefficient (SHGC). In the winter, the solar radiation helps to heat the building and reduces the heating load on the HVAC system. Conversely, during warmer months, all the heat entering through a window places a higher load on the HVAC system. Through the use of energy modeling, architects and engineers can analyze multiple types of windows to find the precise balance between reduced cooling loads and free heating energy. The size, placement, and geometry of each window affects how much heat is gained or lost through each surface. Electrochromic windows offer the ability to alter the SHGC in response to the ambient conditions, while traditional (less expensive) windows have a fixed SHGC. Energy modeling allows multiple layouts and types of windows to be analyzed, quickly showing the overall impact on the HVAC system’s energy consumption.
Continue Reading: Thermal Massing
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View more editions
# TEXTBOOK SOLUTIONS FOR A Problem Solving Approach to Mathematics for Elementary School Teachers 11th Edition
• 2588 step-by-step solutions
• Solved by publishers, professors & experts
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Chapter: Problem:
a. A box contains three white balls and two black balls. A ball is drawn at random from the box and not replaced. Then a second ball is drawn from the box. Draw a tree diagram for this experiment and find the probability of the event that the two balls are of different colors.
b. Suppose that a ball is drawn at random from the box in part (a), its color is recorded, and then the ball is put back in the box. Draw a tree diagram for this experiment and find the probability of the event that the two balls are of different colors.
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 2
(a) Two balls are drawn from a box, one after another without replacement. A tree diagram for this experiment is given below.
Where denotes white and denotes black.
Therefore, probability of the event that the two balls are of different colours is .
Hence, the required probability is .
• Step 2 of 2
(b) If two balls are drawn from the same box, one after another, with replacement after denoting its colour, then tree diagram of this experiment is given below
Where denotes white and denotes black.
Therefore, the probability of the event that the two balls are of different colour is .
Hence, the required probability is .
Corresponding Textbook
A Problem Solving Approach to Mathematics for Elementary School Teachers | 11th Edition
9780321756664ISBN-13: 0321756665ISBN: Authors:
Alternate ISBN: 9780321237170, 9780321783264, 9780321783325, 9780321828026, 9780321840127, 9780321843135, 9780321849250
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# The Doppler Effect
A powerpoint presentation explaining the Doppler effect including 2 questions with detailed solutions.
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### The Doppler Effect
1. 1. The Doppler Effect According to Sheldon Cooper of the Big Bang Theory, the Doppler effect is “the apparent change in the frequency of a wave caused by relative motion between the source of the wave and the observer.” It affects any type of wave including light and sound. The Doppler effect only applies when the motion is directly towards or away between the source and the observer.
2. 2. The Doppler Effect As illustrated by this image, when an object emitting waves moves, it changes the frequency. The waves in front get pushed closer together while the ones behind get more spread out. Image Source: http://images.tutorvista.com/cms/images/83/doppler-effect-image.PNG
3. 3. The Doppler Effect You can see that the wavelength in front of the motion of the object decreases. Although the object is moving, the sound speed itself does not change. If the medium stays the same, the speed also stays constant. Thus, the frequency increases as a result. It’s the opposite for the waves behind the motion of the object. The wavelength increases and the frequency decreases. This relationship can be seen in the following equation: v=fλ
4. 4. Equations The relationship between the frequencies of the receiver and the source of the waves is denoted by the following equation where v is the speed of sound, vr is the speed of the receiver, vs is the speed of the source and fr and fs are the frequencies of the source and the receiver respectively.
5. 5. Equations Numerator: + is used when the receiver moves towards the source - is used when the receiver moves away from the source Denominator: - is used when the source moves towards the receiver + is used when the source moves away from the receiver Tip: the sign on top always relates to motion towards, bottom sign relates to motion away
6. 6. Question You robbed a bank and speed away in a car at 80 m/s. A police car is chasing you from behind at 95 m/s. Its siren, sounding at a frequency of 775 Hz, makes you anxious. Make a prediction. Do you expect the frequency you hear to be higher or lower? Calculate the frequency will you hear due to the motion of the cars. 95 m/s 775 Hz 80 m/s Image Source: http://fc05.deviantart.net/fs71/f/2012/088/2/8/bmw_car_chase_by_domino3d-d4ub8uw.jpg
7. 7. Solution You should expect to hear a higher frequency because the motion of the police car is headed towards you, pushing the wave fronts of sound closer together and increasing the frequency that you’ll hear them at. Looking at the question, you are given vr= 80 m/s, vs= 95 m/s, and fs= 775 Hz, and you know the speed of sound in air is 343 m/s. The easy part is plugging them into the equation. Then you just have to decide on whether to use + or – signs.
8. 8. Solution Since you are moving away from the police car, you use a – sign in the numerator, and since the police car is moving towards you, you use a – sign in the denominator. Plugging in the numbers gives you: So you get fr= 822 m/s, which is the frequency you hear from the siren which is higher than 775 m/s as predicted.
9. 9. Question You are standing beside a lake when suddenly a scary looking goose comes running through the grass towards you at a constant speed, honking angrily. You hear a frequency of 84.0 Hz. The goose as it turns out, is not mad at you, but rather at the kid chasing its goslings, and runs straight through your legs. After it passes, you hear a frequency of 56.0 Hz. What is the speed of the goose? Image Source: http://hollypointassociation.org/a_angry_goose_400p.jpg
10. 10. Solution We know 2 frequencies: As the goose comes towards us and as it goes away from us. So, we need to use 2 equations. We don’t know the actual frequency. With what we know, these are the 2 equations we get: We have a – in the 1st equation because the goose is moving towards us, + in the 2nd because it moves away
11. 11. Solution There are two unknowns but that’s okay because the unknowns are the same in both equations. The actual frequency stays the same and the speed does too since the goose is moving at a constant velocity. We divide the equations by each other so we can cancel out several parts of them.
12. 12. Solution 84/56 is 3/2. fs and the 343 m/s in the numerator of the top and the bottom equations cancel out leaving you with:
13. 13. Clarification If you are confused about the previous slide, this is where I will explain. If not, skip this slide. These are equivalent to each other because when you divide by a fraction, you simply multiply by the reciprocal (flip the fraction). Since 343 m/s cancels out, you are left with:
14. 14. Solution Cross multiply, distribute, gather the like terms on separate sides to isolate the variable vs. Then solve for vs. Final answer: The goose was running at 68.6 m/s
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A powerpoint presentation explaining the Doppler effect including 2 questions with detailed solutions.
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# LETREC + CALL/CC = SET! even in a limited setting
57 views
### Alan Bawden
Mar 2, 1989, 11:27:00 AM3/2/89
to
The following function, which I mailed out to this last last February, was
intended to demonstrate how CALL-WITH-CURRENT-CONTINUATION could be used to
expose the SET! hidden in the definition of LETREC:
(define (make-cell)
(call-with-current-continuation
(lambda (return-from-make-cell)
(letrec ((state
(call-with-current-continuation
(lambda (return-new-state)
(return-from-make-cell
(lambda (op)
(case op
((set)
(lambda (value)
(call-with-current-continuation
(lambda (return-from-access)
(return-new-state
(list value return-from-access))))))
((get) (car state)))))))))
Here is another way to do it that has an additional interesting property:
(define (make-cell initial-value)
(letrec ((state (call-with-current-continuation
(lambda (return-new-state)
(list initial-value return-new-state #F)))))
(lambda (op)
(case op
((get) (car state))
((set)
(lambda (new-value)
(call-with-current-continuation
(lambda (return-from-set)
Notice that the variable STATE does not occur anywhere in the right hand
side of its own definition in the LETREC. Thus you might think that you
could optimize this into an instance of LET. But that optimization would
be incorrect (assuming that the expansion of LETREC given in R^3RS really
-is- its definition), because then this code will stop producing
side-effectable cells.
A simpler case of this peculiarity of LETREC:
(define (test)
(letrec ((x (call-with-current-continuation
(lambda (c)
(list #T c)))))
(if (car x)
((cadr x) (list #F (lambda () x)))
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Time Left - 12:00 mins
# BIHAR PRIMARY TEACHER 2023 Mini Mock - 06
Attempt now to get your rank among 91 students!
Question 1
A certain sum of money is divided into three parts in the proportion of 11 : 8 : 5. If third part is Rs.530, what is the sum (in Rs.) of the other two parts?
Question 2
A pen was sold for ₹28.75 at a profit of 15%. If it was sold for ₹25.75, then what would have been the percentage of profit?
Question 3
Calculate the 3rd proportional to 14 and 28 .
Question 4
The value of:
[(5+3÷5x5)÷(3÷3 of 6) of (4x4÷4 of 4 + 4÷4x4)] is:
Question 5
A series is given with one term missing. Select the correct alternative from the given ones that will complete the series.
LMB, NKD, PIF, RGH, ?
Question 6
In a code language, 'AMIT' is coded as 86 and 'VIJAY' is coded as 134. How will 'GARIMA' be coded in the same language?
Question 7
Ashok's brother's sister is the wife of Ram's son Rahul. How is Ashok related to Rahul's daughter?
Question 8
‘Nuakhai’ is the greatest harvesting festival of Odisha, celebrated one day after which of the following festivals?
Question 9
Divya Kakran is associated to which of the following sports?
Question 10
Collagen is a type of _______
Question 11
Which subject is dealt with in 'Sushruta Samhita'?
Question 12
Vernacular press Act was enacted in which year?
Question 13
The Ganga plain extends between which two rivers?
Question 14
The Finance Commission is constituted by the President ______.
Question 15
The boundary layer of the forest at which the energy exchange occurs and some insolation is returned directly to space is
• 91 attempts
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Waves and the Electromagnetic Spectrum
Presentation on theme: "Waves and the Electromagnetic Spectrum"— Presentation transcript:
Waves and the Electromagnetic Spectrum
What is a wave? A rhythmic disturbance that carries energy but not matter It is able to make matter move This is called a mechanical wave- it uses matter to trasfer energy
Types of Mechanical waves
Transverse waves- moves at a right angle to the direction of the wave Like up and down Side to side They create peaks and troughs Like a rope Radio wave
Compressional waves Moves forward and backward in the direction of the wave Like an accordion Sound waves are compression waves Every sound is made by a vibration
How to measure waves Amplitude- the amount of energy a wave has
The distance of the crest or the trough of a wave to the midline. This is the amplitude of a wave Measure in decibels- how loud it is.
Wavelength describes frequency, or how fast a wave is.
Frequency= number of wavelengths per second Measured in hertz It also tells us where a wave will fall on the electromagnetic spectrum A view of all the wavelengths from very fast gamma rays To very slow radio waves The measure from one crest to another.
Wavelength does not have to be equal to amplitude, they are separate
Color and pitch are based on wavelength Faster wavelength= faster vibration and higher pitch Blue light has a faster frequency than red light This is what it used for photosynthesis, more energy
Sound Waves travel better through dense mediums like solids.
Electromagnetic waves, like light travel best in open space This is why we get light from the sun so fast But it changes direction as it hits the atmosphere
Reflection and refraction
reflection is when a wave bounces off an object and changes direction Echo and mirrors Refraction is when the wave slows down and changes its angle as it passes from one medium to the other. A glass with water in it
refraction
Doppler effect- sound waves get compressed causing a high pitch as the object gets closer to you. as the object moves past the sound waves get longer and the tone gets lower. Like a siren Blue-red shift- as light moves away from you it get red, because red has a longer wave length As light come toward you it is blue, blue has a shorter wave length
Use Doppler effect to Check speed for police radar
Check position of satellites Get information over stars and other planets Are they moving away or towards us, we can tell by the wavelength we receive
Lenses and mirrors Concave and convex Transparent, translucent, opaque
Draw and describe the differences Transparent, translucent, opaque
Diffraction Bending of waves around a barrier Like a rainbow
Like water waves moving around an object They separate and move around the object Remember, a wave is energy and it will keep moving untie all energy is released
Wave Interference When 2 waves combine to form 1 wave
Constructive interference- amplitude increases
Destructive interference- amplitude decreases
Occurs when trough and crest meet Used in noise canceling headphones
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# The solution of geometrical exercises, explained and illustrated; with a complete key to the School Euclid
### Hva folk mener -Skriv en omtale
Vi har ikke funnet noen omtaler på noen av de vanlige stedene.
### Innhold
Del 1 1 Del 2 3 Del 3 18 Del 4 40
Del 5 55 Del 6 95 Del 7 102 Del 8 103
### Populære avsnitt
Side 25 - The straight line joining the middle points of two sides of a triangle is parallel to the third side, and equal to half of it.
Side 29 - Divide a given straight line into two parts such that the square on one of them may be double the square on the other.
Side 20 - In any triangle, if a straight line be drawn from the vertex to the middle of the base, twice the square of this line, together with twice the square of half the base, is equivalent to the sum of the squares of the other two sides of the triangle. Let...
Side 39 - Show that the locus of a point such that the sum of the squares of its distances from two fixed points is constant, is a circle.
Side 24 - ... exterior of two concentric circles, two straight lines be drawn touching the interior and meeting the exterior ; the distance between the points of contact will be half that between the points of intersection.
Side 31 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Side 22 - ... line is bisected at the middle point of the first. 37. If through any point equidistant from two parallel straight lines, two straight lines be drawn cutting the parallel straight lines, they will intercept equal portions of these parallel straight lines. 38. If the straight line bisecting the exterior angle of a triangle be parallel to the base, shew that the triangle is isosceles. 39. Find a point B in a given straight line CD, such that if AB be drawn to B from a given point A, the angle ABC...
Side 27 - Prove that the square on any straight line drawn from the vertex of an isosceles triangle to the base, is less than the square on a side of the triangle by the rectangle contained by the segments of the base : and conversely.
Side 29 - The straight line drawn from the vertex of a triangle to the middle point of the base is less than half the sum of the remaining sides.
Side 9 - The line which joins the middle points of two sides of a triangle is parallel to the third side, and is equal to half the third side.
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Computation of cross price elasticity of demand
Assignment Help Macroeconomics
Reference no: EM1315645
For each of the following pairs of goods, would you expect the cross-elasticity of demand to be positive or negative? Large (in absolute value) or small? Defend your answers:
a. Computer hardware and computer software.
b. Antibiotics and over the counter decongestants.
c. Gasoline and automobile repairs.
Questions Cloud
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Pricing decision on the basis of elasticity of demand
A tariff I ssimply a tax on imports. Use our model of the excise tax (with diagram) to expain why domistic firms request that tariff? Consider both the domestic and the foreign country in your answer
Macro variables in a closed economy
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Discuss how absolute advantage and comparative advantage differ? Kyle can read 20 pages of the economics in an hour. He can also read 50 pages of history in an hour. He spends 5 hours pre day studying. Draw Kyle's production possibilities fron..
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Assume W = 10 000. Draw the aggregate expenditure function on a scale diagram along with 45°line. What is the equilibrium level of national income?
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A profit-maximizing monopolist never produces in the inelastic part of a linear demand curve. The short-run supply curve of a competitive firm is its MC curve.
Answer the following questions as these general questions pertain to the specific issue selected.The questions that you will cover with respect to your choice of broad social issue in the paper are given.
Comparing total cost of watching movie
What might be included in the "total cost" of acquiring and watching movie on DVD? What about the "total cost" of seeing a movie at the multiplex?
Evaluate total revenue-marginal revenue-total cost
Compute total revenue, marginal revenue, total cost and profit at each quantity. What quantity would a profit-maximizing publisher choose? What price would it charge?
The requirement is:- term paper on International Business from economic view point. The topic is effect of corruption on Chinese and Indian economy and how India's IT sector.
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handout10
# handout10 - Lecture 10 Gene and Genotypic Frequencies At...
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1 Lecture 10 Gene and Genotypic Frequencies At Equilibrium Within a generation there is a relationship between gene and genotypic frequencies such that if the f(A) is p then f(AA) is p 2 . Across generations, gene and genotypic frequencies remain constant. At Equilibrium (Autosomal Locus) f(A) = p f(a) = q AA = p 2 Aa = 2pq aa = q 2
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2 Gene and Genotypic Frequency (Sex Linked) Sex Linked Frequency Genotype Population Within sex In the population, 1/2 of the individuals are male and 1/2 are females; therefore, frequencies are halved. The frequency within sex are conditional; for example, the frequency of X A Y given male is P(X A Y/male) = P(X A Y) = p/2 = p P(male) 1/2 Sex Linked Frequency Genotype Population Within sex males X A Y p/2 p X a Y q/2 q
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# 1 Ml To Tbsp
1 Milliliters to Tablespoons calculator converts 1 mL into tbsp quickly.
## How many tablespoons made up 1 milliliters?
Simply divide 1 mL by 14.78676 to convert it into tablespoons.
## What is the value of 1 milliliters in tablespoons?
1 mL is equal to 0.0676 tbsp.
## 1 Milliliters Conversion
US Tablespoons 0.0676 US Liquid Gallons 0.000264201 US Fluid Ounces 0.0338135 US Cups 0.00421941 US Liquid Quarts 0.00201613 US Liquid Pints 0.00211416 US Legal Cups 0.00416667 US Teaspoons 0.202881 Imperial Gallons 0.000219974 Imperial Quarts 0.000879507 Imperial Pints 0.00176056 Imperial Cups 0.00352113 Imperial Fluid Ounces 0.0351952 Imperial Tablespoons 0.0563126 Imperial Teaspoons 0.168947
## Relevant Calculators
1 mL to tbsp calculator converts 1 milliliters into tablespoon, liters, ounces and more simultaneously.
## Other Lengths
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# Where did I go wrong with my odd proof that $\frac{3dx}{3x} = \frac{5dx}{5x} \iff 3=5$?
I don't know where I went wrong, but it's interesting for me. Please check where my fault is! It is obvious that the below equation is correct:
$$\frac{3dx}{3x}=\frac{5dx}{5x}$$ $$u=3x$$and$$v=5x$$ $$\frac{du}{u}=\frac{dv}{v}$$ integrate both sides: $$ln(u)=ln(v)$$ $$u=v$$ $$3x=5x$$ so, $$3=5$$
• When you integrate both sides you need your constant of integration. "+C" – Doug M Aug 10 '18 at 20:25
• Forgot the constant of integration – Niall Aug 10 '18 at 20:25
• An excellent example of the importance of remembering the constant of integration. I will certainly use this as an example when I teach integration .... – Bruce Aug 10 '18 at 20:27
• 3 does equal 5. ... plus a constant. – fleablood Aug 10 '18 at 20:28
• @fleablood In this case multiplied by a constant. – Serge Seredenko Aug 10 '18 at 21:32
Integrating we obtain
$$\ln(u)=\ln(v)+ C$$
• Where $C = \ln u - \ln v = \ln \frac{u}{v}$ So, $\ln u = \ln \left(v \cdot \frac{u}{v}\right)$, giving us $3 = 5 \cdot \frac{3}{5}$..Spelling out what you were implying. – Davislor Aug 11 '18 at 1:16
• @Davislor I'm not implying nothing! I just gave the answer to the OP that is "I don't know where I went wrong". You are free to add your own answer here below with all the details you like. Morover try to be polite when you formulate any kind of request. – gimusi Aug 11 '18 at 6:38
• Sorry to have sounded impolite! I just wanted to acknowledge that you were aware of the steps you left out. – Davislor Aug 11 '18 at 7:22
• @Davislor Thanks for the clarification. I would like to proceed as in the answer here below exponentiating both sides to obtain $u=kv$ wiith $k=e^C=3/5$. Let add your own answer if you like to suggest another way. Bye – gimusi Aug 11 '18 at 7:33
If you're taking an indefinite integral, then you need to include constants. Integrating, we get $\ln(u)+C_1 = ln(v)+C_2$. We can collect $C_1$ and $C_2$ into one constant by setting $C_3 = C_2-C_1$, getting $\ln(u) = ln(v)+C_3$. We can then exponentiate both sides, getting $u=ve^{C_3}$. Setting $C_4=e^{C_3}$, this becomes $u=C_4v$. In this case, $C_4$ is $\frac35$.
If you're taking definite integrals, you need to include the limits. Suppose we integrate starting from $x = 1$. Then $u$ starts at $3$. So when we integrate $\frac{du}u$, we have $\ln(u)$ at the upper limit, but we have to subtract $\ln$ of the lower limit, and the lower limit is $3$. So we have that the definite integral is $\ln(u)-\ln(3)=\ln(\frac u3)$. Similarly, for $v$, we get that the definite integral is $\ln(\frac v 5)$. Plugging in $u=3x$ and $v=5x$, we get $\ln(\frac {3x}3)=\ln(\frac {5x} 5)$, which simplifies to $\ln(x)=\ln(x)$.
• After setting $C_3$, the LHS is just $ln(u)$ – CDspace Aug 10 '18 at 21:33
To turn my comment into an answer, your proof is correct up to the step $$\frac{\mathrm{d} u}{u} = \frac{\mathrm{d} v}{v}$$
At that point, as others have said, you forget to add the constants of integration when you integrate:
$$\ln u + \mathrm{C_1} = \ln v + \mathrm{C_2}$$
Substituting $\mathrm{C} = \mathrm{C}_2 - \mathrm{C}_1$ gives us:
\begin{align} \ln u &= \ln v + \mathrm{C} \\ \ln u - \ln v &= \mathrm{C} \\ \end{align}
Substituting for $\mathrm{C}$ in the first equation just gets us $\ln u = \ln u$, which is clearly no contradiction. A little more interesting, perhaps, is to turn our identity for $\mathrm{C}$, $\ln u - \ln v$, into $\ln \frac{u}{v}$ and then substitute that into the other equation above to get:
\begin{align} \ln u &= \ln v + \ln \frac{u}{v} \\ &= \ln \left( v \cdot \frac{u}{v} \right) \end{align}
Now take the exponential of both sides.
\begin{align} u &= v \cdot \frac{u}{v} \\ 3x &= 5x \cdot \frac{3x}{5x} \\ 3 &= 5 \cdot \frac{3}{5} \end{align}
This is presumably what Serge Seredenko meant when he said in his comment that 3 is equal to 5, “[i]n this case multiplied by a constant.”
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# Some optimization tale
Bengt Richter bokr at oz.net
Mon Dec 29 23:12:16 CET 2003
```On Sat, 27 Dec 2003 15:23:34 -0500, "Terry Reedy" <tjreedy at udel.edu> wrote:
>
>"Stephan Diehl" <stephan.diehlNOSPAM at gmx.net> wrote in message
>news:bskcb0\$36n\$00\$1 at news.t-online.com...
>> All of this can be found at
>> http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/252177
>>
>> What I was most surprised at was the inefficency of the trivial solutions
>> (and that the right algorithm makes indeed a difference).
>
>A common surprise. The science of algorithms (including empirical testing)
>gives real benefits.
>
>> --
>os.path.commonprefix --------------------------------------------------
>>
>> def f1(m):
>> "Given a list of pathnames, returns the longest common leading
>> component"
>> if not m: return ''
>> prefix = m[0]
>
> prefix = m.pop() # avoids comparing prefix to itself as first item
>
>> for item in m:
>> for i in range(len(prefix)):
>> if prefix[:i+1] != item[:i+1]:
>
>I am 99% sure that above is a typo and performance bug that should read
> if prefix[i:i+1] != item[i:i+1]:
>since all previous chars have been found equal in previous iteration.
>
>> prefix = prefix[:i]
>> if i == 0:
>> return ''
>> break
>> return prefix
>
>Perhaps you could retest this with the two suggested changes. It will be
>somewhat faster.
>
>> The problem with this algorithm is the copying of all those small
>strings.
>
>Since Python does not have pointers, comparing characters within strings
>requires pulling out the chars as separate strings. This is why using
>C-coded comparison functions may win even though more comparisons are done.
>
>The reason f1uses slicing (of len 1 after my correction) instead of
>indexing is to avoid exceptions when len(item) < len(prefix). However, all
>the +1s have a cost (and I suspect slicing does also), so it may pay to
>truncate prefix to the length of item first. The simplest fix for this
>(untested) gives
>
>def f9(m): # modified f1 == os.path.commonprefix
> "Given a list of strings, returns the longest common prefix"
> if not m: return ''
> prefix = m.pop()
> for item in m:
> prefix = prefix[:len(item)]
> for i in range(len(prefix)):
> if prefix[i] != item[i]:
> if not i: return ''
> prefix = prefix[:i]
> break
> return prefix
>
>> This can be easily fixed
>> --- optimized
>os.path.commonprefix ----------------------------------------
>>
>> def f2(m):
>> "Given a list of pathnames, returns the longest common leading
>> component"
>> if not m: return ''
>> if len(m) == 1: return m[0]
>> prefix = m[0]
>> for i in xrange(len(prefix)):
>> for item in m[1:]:
>> if prefix[i] != item[i]:
>> return prefix[:i]
>> return prefix[:i]
>
>and easily messed up;-) If len(item) < len(prefix), item[i] throws
>exception. For this approach to work, prefix should be set as shortest
>string of m in preliminary loop. Also, you reslice m several times. Do it
>once before outer loop.
>
>> It is just not nessesary to compare all strings in the list.
>
>Every string has to be compared to something at least once.
>
>> It is enough to sort the list first
>> and then compare the first and the last element.
>
>Sorting compares all strings in the list to something at least once, and
>most more than once.
>
>> Even though the 'sort' algorithm is coded in C and is therefore quite
>fast,
>> the order of runtime has changed.
>
>The C part is what makes f3 faster. In your timings, 128 is not large
>enough for the nlogn component to be noticeable.
>
>> Michael Dyck then pointed out that instead of using 'sort', 'min' and
>'max'
>> should be used. While tests suggest that this is true, I have no idea why
>> that should be, since finding a minimum or maximum uses some sorting
>anyway
>
>No. max and min each do a linear scan. No sorting. But each does at
>least as many character comparisons as modified f1 or f2. The speedup is
>from looping and comparing in C, even though at least twice as many
>compares are done.
>
>> You might have realized that the optimization so far was done one the
>number
>> of strings. There is still another dimension to optimize in and that is
>the
>> actual string comparing.
>> Raymond Hettinger suggests using a binary search:
>
>Since this only affects the final comparison of min and max, and not the n
>comparisons done to calculate each, the effect is minimal and constant,
>independent of number of strings.
>
>Since this compares slices rather than chars in each loop, I wonder whether
>this is really faster than linear scan anyway. I would like to see timing
>of f5 with min/max of f4 combined with linear scan of f3. (Basically, f3
>with sort removed and min/max added.) Since you changed two parts of f3 to
>get f4, we cannot be sure that both changes are each an improvement even
>though the combination of two is.
>
>def f5(seq):
> if not seq: return ''
> s1 = min(seq)
> s2 = max(seq)
> n = min(len(s1), len(s2))
> if not n: return '' # not required since s1[0:0] == ''
> for i in xrange(n) :
> if s1[i] != s2[i] :
> return s1[0:i]
> return s1[0:n]
>
>>> def f10(m):
... "return longest common prefix of strings in seq"
... if not m: return ''
... prefix = m.pop()
... ssw = str.startswith
... for item in m:
... while not ssw(item, prefix): prefix = prefix[:-1]
... if not prefix: return ''
... return prefix
...
>>> f10('abcd abcd'.split())
'abcd'
>>> f10('abcd abce'.split())
'abc'
>>> f10('abcd abcd a'.split())
'a'
>>> f10('abcd abcd a x'.split())
''
Regards,
Bengt Richter
```
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https://www.vedantu.com/jee-main/a-boy-is-pushing-hard-on-a-wall-is-he-doing-work-physics-question-answer
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# A boy is pushing hard on a wall. Is he doing work?(A) No (B) Can’t say(C) May have done some work(D) Yes
Last updated date: 07th Sep 2024
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Hint: We will see that when a boy pushes a wall, he is applying force on the wall. We see that the wall shows no displacement. Work is given as the product of the forces and the displacement in the same direction. Hence the work done is zero.
Work done:
Work is said to be done when force is applied on an object and it gets displaced from its initial position. SI unit of work is Joules.
1 Joule is defined as 1 newton force is applied on an object and it gets displaced by 1 m distance.
$\Rightarrow$ $1joule = 1\,Newton\,metres \Rightarrow 1\,J = 1\,Nm$
Mathematically;
$\Rightarrow$ $W = F\,s\,\cos \theta$ … (1)
W = work done
F = applied force
s = distance covered by an object
$\theta$= angle between applied force and object displacement
In this case when boy pushes on the wall, it shows no displacement
S = 0 … (2)
Using equation (2) in (1), we get
$W = F.0 \Rightarrow W = 0$
This represents that no work is done when force is applied on a wall.
Secondly, the wall does not get displaced from its initial position. So,
S = 0
Thus, the correct option is A. He is not doing any work.
Note:
The boy still gets tired even when the total mechanical work is zero. This means that the boy is using energy. To clarify this apparent paradox lets understand and Note: that energy needn’t be only converted to work; it can also get converted to other types of energies like heat, light, sound etc. IN our case the boy gets tired trying to oppose the rigidity of the wall. Theoretically the boy is compressing the wall. However, the wall’s tensile strength is too high hence the negligible amount of compression that the boy would have caused goes unnoticed. The boy’s energy is also lost in the form of heat due to which we would practically see him sweating
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https://discusstest.codechef.com/t/modular-game-of-numbers-hackerrank/14487
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# Modular game of numbers - hackerrank
Please can someone explain the approach to this problem ?
Thanks!!
Yup, heres my code-
``````#include <bits/stdc++.h>
using namespace std;
int main(){
int n;
int p;
int q;
cin >> n >> p >> q;
vector<int> a(p);
for(int a_i = 0; a_i < p; a_i++){
cin >> a[a_i];
}
vector<int> b(q);
for(int b_i = 0; b_i < q; b_i++){
cin >> b[b_i];
}
int i,j,sum;
int ans[4001];
for(i=0;i< 4001;i++)
ans[i]=-1;
for(i=0;i< p;i++)
{
for(j=0;j< q;j++)
{
sum=a[i]+b[j];
int rem= sum%n;
ans[n-rem]++;
}
}
int min=999999,numind=1;
for(i=1;i<=n;i++)
{
//cout< < ans[i]< < " "< < "i is "<< i< < endl;
if(ans[i]< min)
{
//cout<<"If executed"<< endl;
min=ans[i];
numind=i;
}
}
cout<< numind<< endl;
return 0;
}
``````
Basically what I did was to find sum of all possible combinations from list of her 2 friends, and hence add one to the counter (which counts the number of combination for which a number ‘i’ would make her lose). Then I print the one with minimum value of counter.
I think an example will do more justice.
Let the input be-
``````3 2 1
1 2
3
``````
Now, n=3. The sums possible from given lsit are-
• 1+3=4
• 2+3=5
Now first I mod them with n. (Since it can be proved that taking a number greater than n is ineffective. Eg- If the sum is 4, and I choose 2, then its no use. Similarly if all numbers of form 5,8 …2+3*(k) [k is an integer] will give remainder 2 and make her lose the game. Think on this, this is a tricky part. Number greater than n are to be ignored.)
So I mod the number and get remainder.
Now I make an array arr[n+1] which counts that for how many combinations will the number ‘i’ make her lose.
4%3=1
==> arr[3-1]++; (Since if she chooses 3-1=2, she would lose!)
5%3=2
==>arr[3-2]++
Note that arr[0] is still 0, and that’s the minimum. But 0 isn’t an option she could choose. Effectively, a remainder of 0 from side of Alice means she choose a multiple of n. And going by the constraints of printing smallest integer, we print n itself.
Basic idea-
• Acknowledge that its all playing with remainders. Numbers greater than n are ineffective and are to be ignored in interest that if number K is giving least probability, then K%N will also give same probability (She loses if (Sums of Her friends number)% N + Her Number%N =0, meaning if the number is such that its remainder +remainder of sum of numbers of her friends=N.
==> Remainder = N - (remainder-of-Sum-of-numbers-choosen-by-her-friends)
This is the remainder which would make her lose. We store how many combinations give this remainder. We repeat it for all combinations, storing how many losses does each remainder give. Then we print the one giving minimal loss)
1 Like
Yup i did the same and got 3 WAs…basically i found the min number such that the probability of her loosing the game becomes Zero! Is there anything wrong with our approach!
Its not necessary that probability is 0.
Eg- Take example of list being-
1 2 3 4
1
And n=3.
The sums are 2,3,4,5 i.e. rem = 0,1,2 all are possible. Here if you try to look for a number such that the probability is 0, then you wont find any.
yeah! Missed this.
Happens to even the best of us
//
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+0
# plz help
0
33
1
The hour hand of a clock is 6 inches long and the minute hand is 8 inches long. What is the ratio of the distance in inches traveled by the tip of the hour hand to the distance in inches traveled by the tip of the minute hand from noon to 8 p.m.? Express your answer as a common fraction.
Sep 4, 2021
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# User:Floriane Briere/Notebook/CHEM-496/2012/02/08
Chem-496 Main project page
Previous entry Next entry
## Objective
Today's objective is to prepare the refolding buffer and the BSA control. The refolding buffer will be used next week to higher the pH after the Gold NPs/fibers synthesis; higher the pH is a mandatory step if we want to tag our Gold NPs/fibers with the dye. In fact, the dye we are using is able to tag deprotonated lysine (whose pKa = 8.95 for the amino group); thus pH of the solution is going to be an important characteristic that we'll have to control during the entire experiment.
The BSA control is made with the same amount of acid but we are using HCl instead of HAuCl4 (the HCl solution has a similar concentration to the HAuCl4 solution we're using for Gold NPs synthesis); so the solution is identical to the normal one but Gold is absent. Moreover, we are also going to determine how much dye we'll use to tag our 10mL solutions.
## Protocol
• BSA control (Gold/BSA ratio = 166)
1. 8mL of water + 1mL of HCl (2.84mM) + 1mL of BSA (17.7µM)
2. 2 hours in the oven (80°C)
• Refolding buffer
1. 1.514g of Tris
2. 3.51mL of HCl (0.948M)
3. Fill with water up to 250mL
4. Measure the pH using pHmeter; obtain a pH = 8.73
## Data
• Tris buffer calculations
8.5 = 8.06 + log[salt]/[acid] [salt]/[acid] = 2.754 (50-x)/x = 2.754 --> x = 13.318 mM = [HCl] (0.013318 M)(0.25 L) = (0.948 M)(x L) x = 3.512 mL HCl (0.050 M Tris)(0.25 L)(121.14 g/mol) = 1.514 g Tris
• Mass of dye for reaction calculations
It's been established that one molecule of BSA is composed of 60 lysine residues (K. Hirayama, S.Akashi, M. Furuya and Ken-ichi, "Rapid Confirmation and Revision of the Primary Structure of Bovine Serum Albumin by ESIMS and frit-FAB LC/MS", Biochemical and Biophysical Research Communications 173, 2(1990) 639 - 946). Moreover, to tag our proteins we are going to use an excess of three dye molecules for every one lysine residue (Protein-protein interactions: a molecular cloning manual; google books, p195).
MW of dye = 573.51 g/mol, MW BSA = 66463 Da
1.77 x 10-7 M BSA x 0.001 L = 1.77 x 10-8 mol BSA 1.77 x 10-8 mol BSA x 6.022 x 1023 molecules/mol = 1.066 x 1016 molecules BSA 1.066 x 1016 molecules BSA x 60 lysine residues = 6.393 x 1017 total lysine residues 6.393 x 1017 total lysine residues x 3 = 1.918 x 1018 dye molecules needed 1.918 x 1018 dye molecules x 1 mol/(6.022 x 1023 molecules) = 3.186 x 10-6 mol dye 3.186 x 10-6 mol dye x 573.51 g/mol = 1.83 mg dye
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Searching \ for '[EE] Suggestions for my constant voltage, current' in subject line. ()
Help us get a faster server
FAQ page: techref.massmind.org/techref/power.htm?key=voltage
Search entire site for: 'Suggestions for my constant voltage, current'.
Exact match. Not showing close matches.
'[EE] Suggestions for my constant voltage, current '
2011\09\05@072743 by
Circuit (click to enlarge)
[image: cc.png] <http://solarwind.byethost7.com/cc.png>
http://solarwind.byethost7.com/cc.png
I designed this Li-ion charging circuit so I could select my own maximum
charge voltage (so I could get a better service life out of the cell by
charging it up to only 4.0V or less). At the same time, I can fine tune the
maximum charge current. This method (constant voltage with current limit)
models (as far as I know) the normal method for charging Li-ion (constant
current until a certain voltage is reached, then switch to a constant
voltage).
I would like some suggestions for it as I'm pretty sure there are a few
things to improve.
* The chip in the middle is a high side current sense amplifier.
* The voltage source on the left is my best attempt at modelling a Li-ion
battery. It gains voltage as it's being "charged" (DC operating point
stepped to increase voltage).
* Graph:
** cyan: power dissipated by the MOSFET
** blue: current through r_sense and the battery
** red: MOSFET gate voltage
** green: voltage across the "battery"
Questions:
* I placed C2 (next to the MOSFET) to smooth out the gate voltage and give a
more "analog" signal so the MOSFET can control the current better. Thoughts
* I chose a MOSFET instead of a BJT so I can get the full 4.0V out of the
charging voltage source. Thoughts?
better? Is the whole concept here stupid to begin with?
* Is C1 (above the opamp) too large a value? I was suggested a value of
around 50pF a while ago while designing using a similar circuit block, for
opamp voltage output stability. Note that there's also C2 (beside the
MOSFET) as well, which I believe serves the same purpose
On Mon, Sep 5, 2011 at 7:27 AM, V G <x.solarwind.xgmail.com> wrote:
{Quote hidden}
Anyone? :
Perhaps if you post a valid link somebody will respond.
Em 12/9/2011 09:40, V G escreveu:
{Quote hidden}
> Anyone? :(
On Mon, Sep 12, 2011 at 9:30 AM, Isaac Marino Bavaresco <
isaacbavarescoyahoo.com.br> wrote:
> Perhaps if you post a valid link somebody will respond.
>
I don't understand. The link http://solarwind.byethost7.com/cc.png is
working for me.
In any case, here's another link: http://i.imgur.com/cNX8X.pn
Em 12/9/2011 10:37, V G escreveu:
> On Mon, Sep 12, 2011 at 9:30 AM, Isaac Marino Bavaresco <
> isaacbavarescoyahoo.com.br> wrote:
>
>> Perhaps if you post a valid link somebody will respond.
>>
> I don't understand. The link http://solarwind.byethost7.com/cc.png is
> working for me.
automatically.
>
> In any case, here's another link: http://i.imgur.com/cNX8X.png
This one worked.
On Mon, Sep 12, 2011 at 10:07 AM, Isaac Marino Bavaresco <
isaacbavarescoyahoo.com.br> wrote:
> Perhaps it's because your browser has a cookie that allows logging-on
> automatically.
>
Nope, just tested it on a random university computer and it works fine
untrusted sites.
Probably this site is blocked by many of us either on the ISP side or in the
company's firewall / router / proxy or whatever.
Tamas
On Mon, Sep 12, 2011 at 5:19 PM, V G <x.solarwind.xgmail.com> wrote:
> On Mon, Sep 12, 2011 at 10:07 AM, Isaac Marino Bavaresco <
> isaacbavarescoyahoo.com.br> wrote:
>
> > Perhaps it's because your browser has a cookie that allows logging-on
> > automatically.
> >
>
> Nope, just tested it on a random university computer and it works fine.
>
On Tue, Sep 13, 2011 at 11:39 AM, Tamas Rudnai <tamas.rudnaigmail.com>wrote:
> untrusted sites.
>
How would you define a "trusted" site?
> Probably this site is blocked by many of us either on the ISP side or in
> the
> company's firewall / router / proxy or whatever.
>
This website is as legitimate as any other
On Tue, Sep 13, 2011 at 5:26 PM, V G <x.solarwind.xgmail.com> wrote:
> On Tue, Sep 13, 2011 at 11:39 AM, Tamas Rudnai <tamas.rudnaigmail.com
> >wrote:
>
> > untrusted sites.
> >
>
> How would you define a "trusted" site?
>
OK, then again: The site is sharing the IP address with many other websites,
many of them are parked domains, but many others are malicious, or phishing
etc -- I ask the same question from you: How do you define trusted?
> Probably this site is blocked by many of us either on the ISP side or in
> > the company's firewall / router / proxy or whatever.
>
> This website is as legitimate as any other.
>
Maybe, but as it shares IP with non-legitimate ones you can expect that
security experts are ruling out every singe site coming from that place --
aka blocking it's IP address entirely.
It is not my trouble what you are using, I only wanted to warn you that the
site is dodgy and it is your interest to move your stuff to somewhere
else... again, not my problem as I have my Websense gateway protecting me
;-)
Tamas
>
More... (looser matching)
- Last day of these posts
- In 2011 , 2012 only
- Today
- New search...
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## Features and Benefits
• Balanced Approach: the text strikes a balance between computation, decision making, and conceptual understanding (as endorsed by the GAISE recommendations).
• Personalized learning with MyMathLab for School: The online course provides users with countless opportunities to practice, plus resources and tools specific to statistics that enhance their experience and comprehension.
• Stepped out instruction and guided learning build students’ knowledge and skills in statistics:
• Where You’ve Been visual descriptions begin each chapter with real-life problems that show students how the chapter content fits into the bigger picture of statistics and connects to the topics learned in earlier chapters.
• Where You’re Going features provide an overview of the chapter and explore concepts in the context of real-world settings.
• What You Should Learn sections present learning objectives in everyday language.
• Definitions and Formulas appear in easy-to-locate boxes. These are often followed by Guidelines that explain In Words and In Symbols how to apply the formula or understand the definition.
• Margin Features reinforce understanding through Study Tips showing how to read a table, use technology, or interpret a result or a graph; Insights driving home important interpretations or connecting concepts; and Picturing the World real-life mini case studies illustrating important concepts in the section.
• Solid review and assessment aids ensure mastery of the concepts before students move on in the text
• Real-life and interactive features present statistics at work in the real world and show how statistics is relevant to students’ lives.
• Examples and exercises tie the material to students’ lives. Approximately 40% of the more than 210 examples in this edition are new or revised.
• Exercises: NEW! More than 2,300 exercises give students practice in performing calculations, making decision, providing explanations, and applying results to a real-life setting—including approximately 45% new or revised.
• End-of-section exercises are divided into three parts: Building Basic Skills and Vocabulary, with short answer, true or false, and vocabulary exercises; Using Interpreting Concepts, with skill or word problems; and Extending Concepts that go beyond the material in the section to challenge students.
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## Combinatorics in Power Query -part1
Combinatorics is one of the most important areas of data analysis. It helps in painting a meaningful picture from tons of data very quickly.
We are often faced with a situation when we need to quickly detemine which combination of factors are actually driving the subtotal. In order to know that, it is important to generate all the subsets of a set and perform aggregation by those combinations in order to answer that question.
It is possible to generate all possible combinations of items in PowerBI using Power query and this post focuses on how to do that.
To elaborate, let's suppose there is a list as following
``````let
Source = {"powerBI","powerQuery","DAX"}
in
Source``````
and the end goal is to generate all possible subsets of this set which would be following
``````{},
{"powerBI"},{"powerQuery"},{"DAX"},
{"powerBI","powerQuery"},{"powerQuery","DAX"},{"powerBI","DAX"},
{"powerBI","powerQuery","DAX"}
``````
The following code generates all possible subsets of this set.
``````let
Source = {"powerBI","powerQuery","DAX"}
p1 = List.Transform({1 .. Number.Power(2, List.Count(Source))}, each _ - 1),
#"Converted to Table" = Table.FromList(p1, Splitter.SplitByNothing(), {"Value"}),
#"Converted to Table",
"Custom",
each
let
Loop = List.Generate(
() =>
[i = [Value], j = Number.IntegerDivide(i, 2), k = Number.Mod(i, 2), l = Text.From(k)],
each [i] > 0,
each [
i = [j],
j = Number.IntegerDivide(i, 2),
k = Number.Mod(i, 2),
l = Text.From(k) & [l]
],
each [l]
),
y = try Loop{List.Count(Loop) - 1} otherwise "0",
in
z
),
"Custom.2",
each
let
x = [Custom.1],
Terminate = List.Count(x),
Loop = List.Generate(
() => [i = 0, j = x{i}, k = if j = "1" then Source{i} else null],
each [i] < Terminate,
each [i = [i] + 1, j = x{i}, k = if j = "1" then Source{i} else null],
each [k]
)
in
List.RemoveNulls(Loop)
),
"Combinations",
each
let
x = [Custom.2],
y = List.Generate(
() => [i = 0, j = x{i}, k = j],
each [i] < List.Count(x),
each [i = [i] + 1, j = x{i}, k = [k] & "," & j],
each [k]
)
in
try y{List.Count(y) - 1} otherwise null
),
in
Combinations``````
The code starts from here
and generates this
An optimized version of the above code is following
``````let
Source = {"powerBI","powerQuery","DAX"},
Initiator={{}},
Loop = List.Generate(
()=>[i=0,j=Source{i},k=List.Combine({Initiator{i},{j}}),l=List.InsertRange(Initiator,List.Count(Initiator),{k})],
each[i]<List.Count(Source),
each[i=[i]+1,j=Source{i},k=[l],l=
let x = List.Generate(
()=>[a=0,b=k{a},c=List.Combine({b,{j}}),d=List.Combine({k,{c}})],
each [a]<List.Count(k),
each [a=[a]+1,b=k{a},c=List.Combine({b,{j}}),d=List.Combine({[d],{c}})],
each[d] ) in x{List.Count(x)-1}],
each [l]
)
in
List.Transform(Loop{List.Count(Loop)-1},each Text.Combine(_,","))``````
A parameterized version with the choice a different delimiter is avaialble here
In my next post, I will show how this concept applies to a real-life scenario.
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Cody
# Problem 2856. Matlab Basics - Create a row vector
Solution 744173
Submitted on 23 Sep 2015 by LY Cao
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = [1 2 3 4 5 6 7 8 9 10]; assert(isequal(length(x),10))
2 Pass
%% x = [1 2 3 4 5 6 7 8 9 10]; assert(isequal(diff(x),ones(1,9)))
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https://jcreed.livejournal.com/2012/11/24/
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## November 24th, 2012
### (no subject)
It's always been a little mysterious to me why, from a category-theoretic perspective, the tensor product in linear logic is focally positive. I expect positive things to be colimity/left-adjointy things and negative things to be limity/right-adjointy things, and a monoidal functor is just something you plop down into your set of requirements, not something that evidently arises as a (co)limit or adjoint.
But stephen lack's "A 2-categories companion", page 4, has the following interesting passage: (emphasis mine)
Suppose V and W are monoidal categories and F : V → W is a left adjoint
which does preserve the monoidal structure up to coherent isomorphism.
There is no reason why the right adjoint U should do so, but there will be
induced comparison maps UA ⊗ UB → U(A ⊗ B) making U a monoidal
functor. (Think of the tensor product as a type of colimit, so the left adjoint
preserves it, but the right adjoint doesn’t necessarily.) In fact the monoidal
functor U : Ab → Set arises in this way
### (no subject)
Still chugging through rails tut.
WHAT SORCERY IS THIS.
```irb> User.methods.select{|x| /^find/.match x.to_s}
=> [:finder_needs_type_condition?, :find, :find_each, :find_in_batches, :find_by_sql]
irb> User.find_by_email("asdf")
User Load (0.2ms) SELECT "users".* FROM "users" WHERE "users"."email" = 'asdf' LIMIT 1
```
The surprising thing is that there is no method named "find_by_email", yet there is a method named find_by_email. There's some sort of fallback reflectiony autoload method-dispatch intervention thingy that lets you do arbitrary magic things, isn't there? Sigh.
Grumbling about that aside, I reeeally like this concrete syntax for function expressions. Holy dang.
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www.vustudents.ning.com
# CS601 Assignment no:1 Fall 2016
Assignment No. 01 Semester: Fall 2016 CS601: Data Communication Total Marks: 10 Due Date: 17/11/2016 Instructions: Please read the following instructions carefully before submitting assignment: You need to use MS word document to prepare and submit the assignment on VU-LMS. It should be clear that your assignment will not get any credit if: The assignment is submitted after due date. The assignment is not in the required format (doc or docx) The submitted assignment does not open or file is corrupt. Assignment is copied(partial or full) from any source (websites, forums, students, etc) Objective: The objectives of this assignment are to: Learn Network Topology Learn Transmission Modes Assignment
Question No. 1
Consider a computer network with 8 devices, in this network each device is directly connected with each other through a twisted pair cable. Calculate the number of links required to connect these devices with each other and fill in the following table.
Question No. 2
Identify the type of transmission mode for each of the following cases:
Case Transmission Mode Monitoring through Surveillance Camera Talking through Walkie-talkie Phone call through smart Cell phone Watching a broadcast Television Real time high speed Audio-Video Conferencing
Best of Luck!
Views: 3959
# CS601 Complete Solution By ＴＨＥ ＪＯＫＥＲ
Regarding to the Calculaiton of total number of links:Question 1
In full mesh network, each device has a n-1 number of connections.
And there is a total of n(n-1)/2 number of connections.
Formula n(n-1)/2 n-1 No. of links =8(8-1)/2=28 No of I/O Ports =(8-1)=7
Case Transmission Mode
Monitoring through Surveillance Camera Simplex
Talking through Walkie-talkie Half Duplex
Phone call through smart Cell phone Full Duplex
Real time high speed Audio-Video Conferencing Full Duplex
Monitoring through Surveillance Camera is a one way communication in which camera only captures video and pass it to DVR or directly to monitor if attached directly. Camera didn't receive packet/traffic from DVR or monitor. So it's Simplex
In case of Real time audio-video conferencing traffic can flow from both sides at same time. Both persons can speak and hear each other at same time. So it's Full Duplex communication.
n(n-1)/2
n-1
No of I/O Ports =(8-1)=7
Oh right. next time i will briefly explain it, I guess it was my first time.
Formula= n(n-1)/2
Nodes = n = 8
= 8(8-1)/2
= 8(7)/2
=56/2
= 28 28
Q.1
Sol:
Q.2
Sol:
Case Transmission Mode
Monitoring through Surveillance Camera Simplex
Talking through Walkie-talkie Half duplex
Phone call through smart Cell phone Full duplex
Real time high speed Audio-Video Conferencing Full duplex
See the attached file
Attachments:
CS601 Assignment Solution:
Question No. 1
Consider a computer network with 8 devices, in this network each device is directly connected with each other through a twisted pair cable. Calculate the number of links required to connect these devices with each other and fill in the following table.
Solution:
28
Question No. 2
Identify the type of transmission mode for each of the following cases:
Solution:
Case Transmission Mode
Monitoring through Surveillance Camera Simplex mode
Talking through Walkie-talkie Half Duplex mode
Phone call through smart Cell phone Full Duplex mode
Watching a broadcast Television Simplex mode
Real time high speed Audio-Video Conferencing Full Duplex Mode
CS601 Assignment Solution
Attachments:
1
2
3
4
5
## Latest Activity
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# Fractions
Arithmetic essentials,
In this topic, we will explore fractions conceptually and add, subtract, multiply, and divide fractions.
This course contains 20 segments:
Fractions intro
Learn what a fraction is and identify unit fractions.
Fractions on the number line
Learn to identify fractions on a number line.
Equivalent fractions
Learn how to find equivalent fractions visually and by multiplying the numerator and denominator by the same number.
Comparing fractions
In this tutorial, we'll practice understanding what quantities fractions actually represent and comparing those to each other.
Common denominators
Using our knowledge of equivalent fractions, let's rewrite fractions to have the same denominator.
Decomposing fractions
See that a fraction can be broken up (or decomposed) into a bunch of other fractions.
Adding and subtracting fractions with like denominators
Learn how to add and subtract fractions. This is an introductory tutorial, so the fractions you add and subtract have the same denominators.
Mixed numbers
Learn how to write improper fractions as mixed numbers (and mixed numbers as improper fractions).
Adding and subtracting fractions with unlike denominators
Learn how to add and subtract fractions that have different denominators.
Adding and subtracting mixed number with unlike denominators
Learn how to add and subtract mixed numbers whose fractional parts have different denominators.
Adding and subtracting fractions word problems
Practice solving word problems by adding and subtracting fractions.
Multiplying whole numbers and fractions
Learn the art of multiplying whole numbers by fractions. See how 3 times 2/3 is the same as 6 times 1/3, which is the same as 6/3.
Multiplication as scaling
Learn to see multiplication as a way to grow (scale up) and shrink (scale down) other numbers.
Multiplying fractions
Learn how to multiply a fraction by a fraction.
Multiplying mixed numbers
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# BMT107W - l_l First Letter Last Name Name Student Number...
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Unformatted text preview: l_l First Letter Last Name Name: Student Number: PHYS 101 Mid-term I OCTOBER 10, 2007 Constants: Pmnmsphe,c = 1.013X105 Pa, Density of fresh water = 1000 kg/m3 Density of air = 1.28 kg/m3 Velocity of sound in air = 343 m/s g = 9.81 m/s2 Marks -- Show All your work, use pen, simple scientific calculator Question 1 Five blocks are floating in water as shown and have the same cross—sectional area i.e. the same horizontal dimensions but have different heights and are made of different materials. A B C D E water level a) Rank the blocks in order of their densities indicating any equality. Greatest: Least Reason: b) Rank the blocks according to their weight indicating any equality. Greatest: Least Reason: :3 Ed L3: '3 L 0 UK Question 2 A beaker of water sits on a balance. A block of ma erial ’3 a H to is lowered into the water on the end of a string and is held suspended in the water. The initial balance reading, with the block out of the water is 7 _ 0.780 kg. When the block is suspended in the water, the reading changes by 0.040 kg. a) When the block is in the water, is the balance reading greater or smaller than the initial value of 0.780 kg? b) What is volume of the block? C) Is it possible to find the mass of the block from the information given? If so, what is the value? If not what additional information do you need? W O\’\ 0W1 K“ 07w MT! mm ‘5. 5" 1 Question 3 You need to siphon water from a clogged sink. The sink has an area of 0.35 m2 and is filled to a height of 4.00 cm. Your siphon tube rises 45.0 cm above the bottom of the sink and then descends 85 cm to a pail as shown. The siphon tube has a diameter of 2.80 cm. a) Assuming that the water level in the sink has almost zero velocity, estimate the water’s velocity when it drops into the pail. b) Estimate how long it will take to empty the sink. c) When the water first starts to flow, what is the pressure in the siphon tube at its highest point, 45.0 cm above the bottom of the sink? d) Theoretically, what is the greatest possible height that the siphon can lift water? Mdbmxos‘b (3 338114: UK Slw'hb'fiK” Question 4 Water flows smoothly through the pipe shown. FYI. The diameter at point 2 is the same as at point 3 Rank the four numbered sections, according to: a) the flow speed through each section Greatest: Least b) the water pressure within them Greatest: Least 07W MT] Pl (ll 1 a i h 7' dcdadcdm rhCch’CLN to Question 5 Displacement versus time \3 05 ,5 . 25 3* 35 In. 5 a.“ ‘ fluids) .- DllplchmInl (m) o .— D A plot of displacement versus time for a mass attached to a spring is shown. a) If the mass is 100 grams, what is the spring constant? 1)) Write an equation for the displacement as a function of time with all the constants evaluated, i.e. A, 0) and (1)0. c) On the diagram used for uniform circular motion above, indicate with dots the positions of the object corresponding to t = 0.0 s and t = 1.0 s g 1 0*“ \0‘ “mean a L%[email protected]%fie (3(ri+q—1)so> u» L Question 6 The figure shows displacement curves for three experiments involving the same mass- spring system oscillating in simple harmonic motion. The timing was started before t = 0. Rank the curves according to: a) the spring’s potential energy at t=0. Greatest: Least b) the kinetic energy of the mass at t=O Greatest: Least c) the maximum kinetic energy of the mass during the oscillations Greatest: Least zcgm ya o=g4141 m 1:242, (3° O7WMT1 P101 2 ...
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# Permissible error
How to calculate the permissible error?
A force F is applied on a square plate of size L . If the percentage error in determination of L is 2% and that in F is 4% . What will be the permissible error in pressure?
rude man
Homework Helper
Gold Member
How to calculate the permissible error?
A force F is applied on a square plate of size L . If the percentage error in determination of L is 2% and that in F is 4% . What will be the permissible error in pressure?
EDITed:
Depends on how you define "permissible".
First, you say a "square" sheet. Is it really square? You have to assume "I don't know" so the area could be that of a general quadrilateral. That requires the measurement of 4 sides plus two opposite angles: http://keisan.casio.com/exec/system/1322718508
Plus a 7th independent measurement is for the force F.
So, assume the error in measuring each side's length and that of each of the two angles is 2%,
if the errors are random in all 7 cases, error propagation theory says to take the square root of the sum of the squares:
error = √(6(.022) + .042) = 0.063 = 6.3%.
But there is also the "worst-case" error which you would get if all your measurements are at their respective extremes, in which case the error would be (1.02)6(1.04) - 1 = 17.1%.
Last edited:
Thanks for the reply. Maybe you went for a higher level of explanation. I think that the question interconnects between force , area and pressure. As pressure = force/ area , can you precisely explain about the permissible error in pressure ?
rude man
Homework Helper
Gold Member
Thanks for the reply. Maybe you went for a higher level of explanation. I think that the question interconnects between force , area and pressure. As pressure = force/ area , can you precisely explain about the permissible error in pressure ?
If the error in force is 4.0% and the error in area is 2.0%, the error in pressure is √(.0402 + .0202) = 4.4%.
But L is likely to mean length, not area, so then it'd be √(.0402 + .0202 + .0202) = 4.9%.
haruspex
Homework Helper
Gold Member
If the error in force is 4.0% and the error in area is 2.0%, the error in pressure is √(.0402 + .0202) = 4.4%.
But L is likely to mean length, not area, so then it'd be √(.0402 + .0202 + .0202) = 4.9%.
We are told it is square. The two length errors are correlated?
rude man
Homework Helper
Gold Member
We are told it is square. The two length errors are correlated?
Good point. So it's just √(.022 + .042).
Evo
Mentor
Just a friendly reminder, if the template isn't used or the information required in the template isn't used, such as what the member has tried themselves to solve the problem, please do not reply, please send a report.
rude man
Homework Helper
Gold Member
Just a friendly reminder, if the template isn't used or the information required in the template isn't used, such as what the member has tried themselves to solve the problem, please do not reply, please send a report.
OK. Sorry. I thought after I wrote that it was probably excessive. Will be more on guard henceforth.
r m
Evo
Mentor
OK. Sorry. I thought after I wrote that it was probably excessive. Will be more on guard henceforth.
r m
You're cool, we're just trying to get things more even in responses, some members get turned back while some get answered.
haruspex
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# Compute likelihood of state given multiple observations?
I am trying to use Bayes formula to compute the likelihood of a given state given a collection of independent but not sequenced observations - knowing the priors and knowing the probabilities of being in the state given each observation.
In other words for pre-defined states $$S = \{S_1,\dots,S_m\}$$ and possible observations $$O=\{O_1,\dots,O_n\}$$, the prior and conditional probabilities are already known:
$$P(S_i) = \pi_i$$ $$P(S_i|O_j) = \rho_{ij}$$
(Note the second formula is not the "observation emission probability" but rather the state given observation probability.)
Suppose now I have collected a set of $$k$$ observations $$\Theta=\{\theta_1,\dots, \theta_k\}$$ and wish to compute for each state $$i$$:
$$P(S_i | \Theta) = P(S_i | \theta_1,\dots,\theta_k)$$
i.e. to work out the most likely state given the observations. I am having a brainfade as trying to use Bayes formula directly goes nowhere:
$$P(S_i | \theta_1,\dots,\theta_k) = \frac{P(\theta_1,\dots,\theta_k|S_i)P(S_i)}{\alpha}$$
I see the relevance of Hidden Markov Models (HMMs) in the mathematics but there is no natural sequence I'm dealing with here it is just a collection of unordered observations. What's the correct way to compute this?? (Pretend it's a HMM with a flat transition probability matrix??)
Suppose I accepted some independence assumptions and continued the derivation as follows:
$$= \frac{P(\theta_1|S_i)\cdot \dots \cdot P(\theta_k|S_i)P(S_i)}{\alpha}$$
Then applied Bayes formula again to each term:
$$= \frac{\frac{P(S_i|\theta_1)P(\theta_1)}{P(S_i)}\cdot \dots \cdot \frac{P(S_i|\theta_k)P(\theta_k)}{P(S_i)} P(S_i)}{\alpha}$$
Seems to be getting a bit silly and now I have to come up with $$P(\theta_i)$$ priors too (possible but surprised I need this) - what's going on and what's the correct way to compute this??
This is correct: \begin{align}\mathbb P(S=S_i|\theta_1,\ldots,\theta_k) \propto \frac{\mathbb P(S=S_i|\theta_1)p(\theta_1)}{\pi_i}\cdots \frac{\mathbb P(S=S_i|\theta_k)p(\theta_k)}{\pi_i} \pi_i\\ \propto \mathbb P(S=S_i|\theta_1)\cdots \mathbb P(S=S_i|\theta_k) \pi_i^{-k+1} \end{align} and the $$p(\theta_j)$$'s play no role there.
• Thanks for clarifying that the priors are not needed good catch! Is there a name for this application of Bayes formula where the known conditional probabilities are inverted? Commented Sep 8, 2021 at 9:51
It's a little bit confusing as at the beginning we define a relationship between the probability of a state given a single observation but now we are looking at a set of observations. Do we know with certainty that all observations come from the same state? Assuming that we do then the solution by Xi'an is correct.
Reproducing it here with a few extra steps, without loss of generality let $$k=2$$:
$$$$\mathit{P}(S_i|\theta_1, \theta_2) = \frac{\mathit{P}(\theta_1, \theta_2|S_i)\mathit{P}(S_i)}{\mathit{P}(\theta_1, \theta_2)} \hspace{0.5cm} \text{(Bayes Theorem)}$$$$
$$$$= \frac{\mathit{P}(\theta_1|S_i)\mathit{P}(\theta_2|S_i)\mathit{P}(S_i)}{\mathit{P}(\theta_1)\mathit{P}(\theta_2)} \hspace{0.5cm} \text{(Independence Assumption)}$$$$
$$$$= \frac{\mathit{P}(\theta_1|S_i)}{\mathit{P}(\theta_1)}\frac{\mathit{P}(\theta_2|S_i)}{\mathit{P}(\theta_2)}\mathit{P}(S_i) \hspace{0.5cm}$$$$
$$$$= \frac{\mathit{P}(S_i|\theta_1)}{\mathit{P}(S_i)}\frac{\mathit{P}(S_i|\theta_2)}{\mathit{P}(S_i)}\mathit{P}(S_i) \hspace{0.5cm} \text{(Bayes Theorem)}$$$$
And so in general:
$$$$\mathit{P}(S_i|\theta_1,...,\theta_n) =\frac{\mathit{P}(S_i|\theta_1)}{\mathit{P}(S_i)}...\frac{\mathit{P}(S_i|\theta_n)}{\mathit{P}(S_i)}\mathit{P}(S_i) = \mathit{P}(S_i|\theta_1)...\mathit{P}(S_i|\theta_n)\mathit{P}(S_i)^{-(n-1)}$$$$
• Thanks for the derivation. Yes it is an unusual situation where the observations are known with certainty to be from the same source (i.e. state) but the state itself is unknown. Is there a name for this sort of Bayesian analysis but with inverted knowledge of the conditional probabilities? Commented Sep 8, 2021 at 9:49
• I'm not aware of any formal name for this other than Bayesian analysis. I'm not sure about your precise use case, but you might be interested in changepoint detection as a topic. It looks to identify periods of time which belong to the same state given some observations. Commented Sep 14, 2021 at 9:08
• I'm getting weird results with this. Suppose P(Si | θ1) = 0.75and P(Si | θ2) = 0.75and P(Si) = 0.5. The result is then 0.75 x 0.75 / 0.5 = 1.125 > 1. Is this situation impossible or am I missing something in the calculation ? Commented Oct 6, 2022 at 6:59
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Document related concepts
Central limit theorem wikipedia , lookup
Transcript
```The 68-95-99.7 Rule
• In any normal distribution:
– 68 % of the individuals fall within 1s of m.
– 95 % of the individuals fall within 2s of m.
– 99.7 % of the individuals fall within 3s of m.
How can we make a valid comparison
of observations from two distributions?
• By standardizing the values of the
observations with respect to the distributions
from which they come.
• If x is an observation from a distribution with
mean m and standard deviation s, then
xm
z
s
is the standardized value (a.k.a. z-score) of x.
Standard Normal Distribution
The standard normal distribution is the normal
distribution N(0,1) with mean 0 and standard
deviation 1.
If a variable X has a N(m,s) distribution, the
standardized variable
X m
Z
s
has the standard normal distribution.
Standard Normal Table
(Table A, inside cover of book or from website)
Examples
1. Find the area under the standard normal curve to the
left of -1.4.
2. Find the area under the N(0,1) curve between 0.76
and 1.4.
3. Find the value z of the N(0,1) which has area 0.25
to its right.
4. Suppose X~N(275, 43). What proportion of the
population is greater than 200? What proportion of
the population is between 200 and 375?
5. Suppose verbal SAT scores follow the N(430, 100)
distribution. How high must a student score in order
to place in the top 5%?
Testing for Normality
• The normal distribution provides a good
model for many real data distributions.
• Furthermore, the normal distribution is a
nice model to work with mathematically.
• However, we need to be cautious when
assuming normality of data. Are we sure our
data are normally distributed?
The Normal Quantile Plot:
A Visual Test of Normality
1. Arrange the observed data values from
smallest to largest and record the percentile of
the data each observation occupies.
2. Compute the z-scores of the same percentiles.
3. Plot each data point against the corresponding
z-score.
If the points on a normal quantile plot lie close to a
straight line then the plot indicates the data are normal.
FIG. 1.34: Normal quantile plot of
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Basic Examples: Percentages
# Basic Examples: Percentages Video Lecture | CSAT Preparation - UPSC
## CSAT Preparation
214 videos|139 docs|151 tests
## FAQs on Basic Examples: Percentages Video Lecture - CSAT Preparation - UPSC
1. What is the significance of percentages in the UPSC exam?
Ans. Percentages play a crucial role in the UPSC exam as they are used to evaluate the performance and scores of candidates. For example, the cutoffs, rankings, and eligibility criteria are often determined based on the percentage of marks obtained in each subject or overall. It is important for candidates to understand how to calculate and interpret percentages accurately to gauge their performance in the exam.
2. How are percentages calculated in the UPSC exam?
Ans. Percentages in the UPSC exam are calculated by dividing the marks obtained by a candidate in a particular subject or overall by the maximum marks allotted for that subject or the total marks of the exam, respectively. The resulting value is then multiplied by 100 to express it as a percentage. This calculation helps determine the relative performance and scores of candidates.
3. Can percentages be manipulated to improve UPSC exam scores?
Ans. No, percentages cannot be manipulated to improve UPSC exam scores. The calculation of percentages is based on the actual marks obtained by a candidate in each subject or overall. Any attempt to artificially inflate or manipulate these marks would be considered fraudulent and against the ethical norms of the examination. UPSC follows a strict evaluation process to ensure fairness and transparency in the assessment of candidates.
4. How are percentages used for ranking in the UPSC exam?
Ans. Percentages are used as a basis for ranking candidates in the UPSC exam. Candidates with higher percentages of marks are ranked higher in comparison to those with lower percentages. This ranking helps determine the selection of candidates for various services and posts offered through the UPSC exam. It is important for candidates to strive for higher percentages to improve their chances of securing a better rank and selection.
5. Is it necessary to score high percentages in all subjects to clear the UPSC exam?
Ans. No, it is not necessary to score high percentages in all subjects to clear the UPSC exam. The UPSC follows a comprehensive evaluation process where the overall performance of candidates is taken into consideration. While scoring well in all subjects is desirable, candidates can compensate for lower percentages in one subject by performing exceptionally well in other subjects. The final selection is based on the aggregate performance and ranks secured by candidates.
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numeral integration 数值积分(0)
“numeral integration”译为未确定词的双语例句
Application of an Unconditionally Stable Numeral Integration Algorithm in Pseudodynamic Testing 无条件稳定数值积分方法在拟动力实验中的应用研究 短句来源
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Numeral 数字榜 短句来源 The integration of the B.t. Southern分子杂交证明B.t. 短句来源 Collision and Integration 冲撞与整合 短句来源 On Numeral Library 浅谈图书馆发展的高级阶段——数字图书馆 短句来源 Application of an Unconditionally Stable Numeral Integration Algorithm in Pseudodynamic Testing 无条件稳定数值积分方法在拟动力实验中的应用研究 短句来源
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In papers [1]、 [2],the general solutions of two-dimensional and bending problems of rectangular plates are obtained by finite Fourior transform method. In this paper, the experimental values of several points at the boundary of a photoelastic model are applied to the calculation of undetermined constants in the previous solutions by numeral integration, thus the analytical solutions of problems under study are determined. 文献[1]、[2]用有限傅立叶变换法得到了矩形域平面问题和薄板弯曲问题的一般解.本文利用光弹性测得的若干点边界值通过数值积分来计算上述解答中的待定常数.从而确定了所论问题的解析解。 Matlab is a system of high - performance software developed by MathWorks company for numeral computing and visual mathmatics. By virtue of Matlab, it is convenient to solve the scientific and engineering problems, which involved the field of numerical solution of differential equation, numeral statistics, numeral integration, linear algebra, automatic -control, system simulation etc. with high speed.Four typical topics in separaling engineering were taken as examples, this papar introduced the main function... Matlab is a system of high - performance software developed by MathWorks company for numeral computing and visual mathmatics. By virtue of Matlab, it is convenient to solve the scientific and engineering problems, which involved the field of numerical solution of differential equation, numeral statistics, numeral integration, linear algebra, automatic -control, system simulation etc. with high speed.Four typical topics in separaling engineering were taken as examples, this papar introduced the main function of matlab in the field of nameral computing and drawing etc. MATLAB是MathWorks公司推出的一套高性能的数值计算和可视化数学软件,可以方便、快捷地解决微分方程数值解、数理统计、数值积分、线性代数、自动控制、系统仿真等方面的科学和工程问题。本文以分离工程中的4个典型问题为例,介绍了MATLAB在数值计算、绘图等方面的一些主要功能。
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Instructions for Using Winplot
Winplot is a graphing utility written and maintained by Richard Parris, a teacher at Phillips Exeter Academy in Exeter, New Hampshire. The program is "totally free" and the newest version can be downloaded from Parris's web site at http://math.exeter.edu/rparris/winplot.html . All of the Winplot modules have fairly thorough Help menus which give detailed information about the workings of the program. An online power point tutorial has been written by John Ganci.
Upon launching Winplot you will see the following introductory screen.
By clicking on Window you will be presented with the following two plotting options:
2-dim Allows you to generate plots in the x, y plane.
3-dim Allows you to generate plots of surfaces or space curves in x, y, z space.
Guess Gives you a graph; you then guess the equation. If you choose Select in the guess my equation box shown below
you can select the type of graphs that will be displayed.
If Open last file is checked, then for any option chosen the last file saved will automatically appear in the chosen window. If Use defaults is checked a standard viewing window for each type of plot will be initialized. Mapping provides a way to visualize the graph of a function that maps two-dimensional space to two-dimensional space. This is especially pertinent for calculus students as they study integration in two dimensions.
Graphs in 2D
Changing the viewing window
From the introductory screen choose Window 2-dim . You will then see a window containing an x, y grid. The appearance of this viewing window can be altered using the View pull down menu.
To change the plot window use the command View. This brings up the following menu.
If the upper right "radio button" or check circle entitled "set corners" is selected you can set the window's lower left (left, down) and the upper right (right, up) coordinates by just entering the appropriate values.Click on the upper right "radio button" or check circle entitled "set corners". This allows you to set the window's lower left (left, down) and the upper right (right, up) coordinates by just entering the appropriate values. In reference back to the first View menu, if the scales on the two axes are not equal (so that circles don't look circular), you can use the command sequence Zoom/Square to make them so. Click Fit window to make all active examples fit within the window. Because this might not be possible, results are sometimes unpredictable. Restore returns to the default window settings. Axes selects the axes menu which allows you to control how x and y axes are displayed.
The Grid dialogue box allows you to display a Cartesian or polar grid by clicking the appropriate radio button. If "both" is selected both x and y axes are displayed. Checking either "x" or "y" displays only the axis selected. If "polar" is selected you can specify the number of sectors by clicking on the square to the left of "polar sectors" and entering the number to the right.
Check "labels" to put the letters 'x' and 'y' into the diagram. If you check "arrows", you get small arrows at the ends of the axes. Check "ticks" to put equally spaced marks on the axes. The distance between successive ticks is the value of "interval". Click on "scale" and "mark scale on axes" to put the corresponding numerical values next to the ticks. Click on "scale" and "mark scale on border" to place the numbers at the edge of the window. The number of decimal places displayed is entered as "places". If there are too many values showing, enter a higher number in the corresponding "freq" box. Check "pi" if you want the ticks labeled using multiples of pi. For this last option to be effective, the corresponding tick interval should be a rational multiple of pi with a small denominator and the font in the "Scale on axes" option of the Misc/Fonts menu should be set to "Pi symbol". If "rectangular" is checked, grid lines appear in the quadrants checked. If "dotted" is checked, you will see an array of small crosses where the grid lines would meet. The polar grid can be on at the same time as the rectangular grid. To effect the changes of this menu press the "apply" button.
You can insert text into the graphing window by use of the mouse buttons. First use the "Btns" pull down menu to make sure that the Text option is checked. Second place the mouse pointer at the spot where you want to enter text, click the right mouse button, and type in the desired text. If you wish, you can change both the color and font of the text.
To insert a circular dot at a point go to the Equa pull down menu and select Point. In the dialogue box give the point's coordinates. The value of "dot size" determines the diameter of the dot and the color can be adjusted via the color button.
Plotting the graphs of equations
To see a graph in Winplot you must either Open an old Winplot file (files with the extension wp2), or else enter a new equation by clicking on the Equa menu. There you will be presented with the following four options for generating curves.
1) Explicit Graphs y as an explicit function of x .
2) Parametric Graphs parametric curves of the form x = f (t ), y = g(t ) for f and g explicit functions of some parameter t .
By letting g(t ) = t this in effect allows you graph x as an explicit function of y, i.e., x = f (y) .
3) Implicit Graphs implicit relations between x and y defined by a formula f (x, y) = constant.
4) Polar Graphs polar curves where the polar coordinate r is an explicit function, f(t), of the polar angle theta (here
designated as t).
Clicking Explicit from the Equa menu results in the following dialogue box.
By typing over the "f(x) =" box you can enter an expression that defines y as a function of x. Click the "color" button to obtain the color palette and click the small box that shows the color you want for the graph. You can type an integer into the "plotting density" box to alter the density of plotted points. This is rarely necessary. If you want to specify the domain of the function enter the interval in the boxes "low x" and "high x" and then click on the box "lock interval". This overrides the default which is to use the interval defined by the plot window as the domain. More than one graph can be displayed in the same graph window. Just return to the Equa menu and repeat the procedure outlined above. In fact, you can mix graphs of y as a function of x with polar curves, parametric curves, and implicit curves by simply using the different options for entering equations.
By choosing Inventory from the Equa menu you access the inventory dialogue box. This allows you to inspect and edit any of the equations already entered and perform other modifications and constructions. To select an equation, click on it with the mouse. Only one example can be selected at a time.
The Inventory options include the following:
edit - This button opens the dialogue box that was used to create the curve and allows you to make changes.
delete - This button causes the curve to disappear from the inventory and from the screen. There is no "undo". All equations that are dependent on the equation are also deleted.
dupl - This button copies the selected equation and opens the editing dialogue. This allows you to create a similar example without changing the original. If you want to keep the original as well as the copy be sure to click no to the prompt "delete original?".
clip - This button copies the text of the defining equation to the clipboard where it can be pasted into other applications.
derive - This button calculates the derivative of the selected curve. This option only applies to curves defined by the Explicit, Parametric or Polar options. The result is graphed and added to the inventory. A derivative can therefore also be selected, but only its attributes (color, thickness, etc.) can be edited.
name - This button gives a selected equation a name which appears in the inventory.
hide graph - This button hides the graph of a selected equation without removing it from the inventory. A second click restores the graph.
show equa - This button displays the text of a highlighted equation in the upper left corner of the graph window. A second click removes this display.
web - This button overlays a web diagram which is an iteration used to locate fixed points of a curve of the type y = f(x). You can specify the starting x-value, the number of steps, and the color of the overlay. To generate the web diagram select "define". To delete the web diagram select "undefine".
family - This powerful feature allows you to display a family of curves generated from a selected equation. The equation must be defined with one or more extra parameters. For example, the equation y = sin(ax)+b defines a sine wave that depends on the two parameters a and b. Either one of these can be used to create a family of curves. To vary the frequency type "a" into the "parameter" box, enter a range of values by filling in the "low" and "high" boxes, and indicate the number of curves to be drawn by filling in the "steps" box (this is one fewer than the number of curves). Click "define" to generate the family. This also modifies the definition of the equation in the inventory. To remove the family, select the equation and click "undefine".
table - This button opens a text window that displays values of the selected function for curves defined by the Explict, Parametric, or Polar options. You can alter the contents of the table by clicking "Params" on the menu bar. To see tables for a different equation use the command sequence "File/Next example".
Winplot like a graphing calculator has built-in "Zoom" features. To "Zoom in" and thus obtain a close up view of a smaller portion of the x, y plane choose Zoom/In from the View menu. Similarly, to "Zoom out" choose Zoom/Out from the View menu. If you're not happy with the new image choose Last window from the View menu.
When entering formulas in Winplot the following conventions are used:
Multiplication is indicated by *
Division is indicated by /
Subtraction is indicated by -
Exponentiation is indicated by placing the ^ (caret symbol) between the expression for the base and the expression for the exponent. The caret is the upper case of 6 on the keyboard.
Actually, Winplot does recognize algebraic notation. For example, "two times ex", can be entered either as 2x or 2*x . Similarly, "three times ex cubed", can be entered as 3x^3 or 3*x^3 or 3*x*x*x or even as 3xxx. The only valid grouping symbols are parentheses ( ), but these may be nested as in (6x - x^(sqr(5)+3))/(5x + 2). The rules for the order of operations are the conventional ones used in mathematics and science.
Certain standard functions are "built into" Winplot, but the arguments must always appear in parentheses. For example, "the sine of 3 times x" should be written as sin(3x).
The names of the principal "built in" functions are given below.
abs(x) is the absolute value of x
sgn(x) is the signum or sign function = abs(x)/x
sqr(x) is the square root of x (for non-negative x)
root(n,x) is the principal n-th root of x
fact(n) is n!
exp(x) is the exponential function evaluated at x
ln(x) is the natural logarithm function for positive x
log(x) is the base 10 logarithm function for positive x
sin(x) is the sine of x (All trig functions assume the argument is in radians.)
arcsin(x) is the arc or inverse sine
cos(x) is the cosine of x
arccos(x) is the arc or inverse cosine
tan(x) is the tangent of x
arctan(x) is the arc or inverse tangent of x
sinh(x) is the hyperbolic sine of x
cosh(x) is the hyberbolic cosine of x
tanh(x) is the hyberbolic tangent of x
pi is the constant pi=3.14159265. To multiply pi and x, enter pi*x, not pix.
As an example, suppose you want a graph of the cubic polynomial
From the Equa menu select Explicit and enter the formula as shown below.
Clicking the "ok" button results in the following: (the grid settings used are also displayed)
Suppose now we click Polar from the Equa menu. This results in the following dialogue box for a polar curve. This second curve will be graphed along with the previous polynomial.
The usual theta domain for polar curves will include at least zero to 2pi. Entering 2pi for "high t" is the same as entering
6.28319 . The result of graphing both curves in the graph inventory is shown below.
Suppose we wanted to graph the ellipse centered at (2, -1) with major axes parallel to the x axis, a semi-major axis of 3, and a semi-minor axis of 2. Since y is not a function of x for this figure, a single y = f (x ) graph can not generate the plot. However, we can graph it by using the parametric input mode. Selecting Parametric from the Equa menu and entering the formulas shown below generates the correct result.
The standard equation of this ellipse is . An alternate way to graph it is to use the implicit input mode. The equation generating the graph must be stated as f(x, y)= constant. The points on the curve are generated by a numerical search. If the "long search" option is checked the program will continue searching for points until you press "Q" (for quit). Failure to type "Q" causes the program "to hang". Implicit from the Equa menu and entering the formula shown below generates the graph shown.
Mathematical calculations and demonstrations
In addition to graphing equations Winplot can perform numerical and graphical procedures on curves. These options are found in the Anim, One, Two and Misc menus.
The Anim menu provides for some of the most dramatic and illuminating demonstrations available in Winplot. In Winplot the letters a ... z always have a numerical value, and letters other than x, y, and z can be used in equations as parameters as was discussed in the Inventory/family option. The value of such a variable can be altered using the scrollbar dialogue boxes listed in this menu. The box labeled by the upper case name of each parameter controls the values of that parameter. When a parameter value is changed, all parameter-dependent graphs, including vector field plots generated under the Equa/Differential option, change accordingly.
The values of a parameter vary from its lowest or left-most value "L" to its largest or right-most value "R". These values can be entered and then set with the buttons "set L" and "set R". Within this domain the value of the parameter can be set manually by moving the scroll bar. This process can be automated by pressing the "autorev" or "autocyc" buttons. The "autorev" option varies the parameter according to the sequence L.R.L:. R. , while the "autocyc" option uses the sequence L.R L.R L.R. In both automatic modes the dialogue box disappears and you must press "Q" to stop the animation and restore normal program control. By default there are 100 intermediate values for the parameter between the L and R values. This number can be changed by clicking on the "1=Scrollbar units" item. In addition, the number of different values of the parameter for which graphs are displayed in "autorev" or "autocyc" mode can be controlled by checking "autoshow" the entering the number of "slides" to be displayed.
The linked window option allows a second graph window with a curve controlled by a parameter to be simultaneously displayed.
As an example, suppose we wish to animate the parameterized function f (x) = sin(a*x) + b . Use the Anim menu and select A and B in succession to set the range for each parameter. In the figures shown below B was set to 0 and A varied from 0.1 to 10.
The One pull down menu lists procedures that Winplot can do to one function at a time.
1. To "trace" along a curve select Slider. A scroll bar appears along with a crosshair cursor that slides along the non-implicitly defined curves in the graphing window. If there are multiple curves in the graph's inventory, use the drop-down list at the top of the dialogue box to select the desired equation. The cursor will move along the curve and the tracer box will display its coordinates as you slide the scroll bar. You can position the cursor by entering a value for the independent variable into the edit box. To save a cursor position in the inventory press the "mark point" button. Use the "degree" drop-down list to set the degree of a Taylor polynomial of a curve of the y = f(x) type calculated at the current cursor position. Press the "Taylor approx" button to display a graph of this approximating polynomial. This Taylor polynomial will then be added to the graph's inventory. If "secant demonstration at" is checked and a fixed point on the curve has been set using the "base point" button, the program will graph a moving secant line to the curve as the scroll bar is moved. If "tangent-line demonstration" is checked, the program will graph a moving tangent line to the curve as the scroll bar is moved.
1. To find the roots or x intercepts of a graph y = f(x), select the Zeros dialogue box. If there are multiple equations in the graph's inventory use the drop-down list at the top of the dialogue box to select the desired equation. Step through the equation's roots from left to right by clicking on the "next" button. After you have found the roots of a function, you can see a list of these zeros by selecting Data/Inspect from the Misc pull down menu.
1. To find the extremes (maximums and minimums) of a graph y = f(x), select the Extremes dialogue box. If there are multiple equations in the graph's inventory use the drop-down list at the top of the dialogue box to select the desired equation. Step through the extreme values from left to right by clicking on the "next extreme of " button. After you have found the extremes of a function, you can see a list of these results by selecting Data/Inspect from the Misc pull down menu.
1. The Measurement menu allows for a variety of numerical approximations to definite integrals.
To approximate integrals of y = f(x) select the Integration dialogue box. If there are multiple equations in the graph's inventory use the drop-down list at the top of the dialogue box to select the desired equation. Enter values into the "lower limit" and "upper limit" boxes and specify a value for the number of "subintervals" (n). Indicate with the check boxes which numerical quadrature methods (parabolic is Simpson's rule ) are to be used. If "overlay" is checked, the program will graphically illustrate the approximation. Press the "definte" button to display the numerical results of the approximations. The number of decimal places shown can be adjusted with the command sequence Misc/Decimal places. The "indefinite" button generates the graph of the antiderivative of f(x) that is zero at the lower limit of integration. This function is then also added to the inventory.
The Area of a Sector option evaluates the area of the sector (centered at the origin) swept out by a selected arc defined by a non-implicit curve. If there are multiple equations in the graph's inventory use the drop-down list at the top of the dialogue box to select the desired equation. The starting and ending values of the independent variable are entered into the "arc start" and "arc stop" boxes respectively. Specify a value for the number of "subintervals" (n) and press the "area" button to obtain the numerical result.
The Length of arc option computes the arc length of a non-implicit curve. If there are multiple equations in the graph's inventory use the drop-down list at the top of the dialogue box to select the desired equation. The starting and ending values of the independent variable are entered into the "lower limit" and "upper limit" boxes respectively. Specify a value for the number of "subintervals" (n) and press the "length" button to obtain the numerical result.
To compute a volume of a solid of revolution select Volume of revolution. If there are multiple non-implicit function equations in the graph's inventory use the drop-down list at the top of the dialogue box to select the desired equation. The starting and ending values of the independent variable are entered into the "arc start" and "arc stop" boxes respectively. Specify the axis of rotation by either selecting the x axis, the y axis, or by entering values for a, b, and c in the equation ax + by = c . Enter a value for the number of "subintervals" (n) and press the "volume" button to obtain the numerical result. To compute the surface area of a solid of revolution select Surface area of rev and follow the same procedure as for a volume of revolution, but press the "area" button to obtain the numerical result.
To see a three dimensional solid of revolution select Revolve surface If there are multiple equations in the graph's inventory use the drop-down list at the top of the dialogue box to select the desired equation. The starting and ending values of the independent variable are entered into the "arc start" and "arc stop" boxes respectively. Specify the axis of rotation by either selecting the x axis, the y axis, or by entering values for a, b, and c in the equation ax + by = c . Press the button "see surface" to display a 3D perspective plot of the solid which you can rotate by using the left, right, down and up arrow keys.
1. To reflect a curve about a given mirror line choose the Reflect option. The mirror line can be specified as the x axis, the y axis, the line y = x (this generates the inverse relation to the original curve) or any line of the form ax + by = c . After designating the mirror line, press the reflect button to perform the reflection.
The Two pull down menu lists four procedures that Winplot can do to two functions at a time.
1. To find the points of intersection of two curves select the Intersections dialogue box. If there are more than two functions of this type you choose which two to use by using the drop-down lists. The "next intersection" button is used to find the next meeting point. Only points of intersection within the graphing window are found. The results are added to the data text file and may be viewed by selecting Data/Inspect from the Misc pull down menu. The intersection coordinates may be saved to variables and the "mark point" button can be used to highlight them on the graph.
1. To add, subtract, multiply, divide, exponentiate or compose two functions select the Combinations dialogue box.
1. To numerically approximate the difference of two functions select the Integrate dialogue box.
From the Equa menu Shading can be used to shade the area between two curves over a specified domain.
1. To visualize 3D solids that have a base in the x, y plane and specified cross section select Sections. The base is the region between two curves each defined by an equation of the form y = f(x). The specified cross sections are generated by planes perpendicular to the x axis and may be chosen from a variety of geometric forms. Pressing the "see solid" button displays a 3D perspective plot which you can rotate by using the left, right, down and up arrow keys. The "volume=" button displays a numerical approximation to the definite integral representing the solid's volume.
Saving, printing and copying a graph
To save all of the equations, windows and related settings associated with a Winplot graph choose Save or Save As from the File menu. Type in the desired file name and click the "Save" button. If you want to save to a particular location, insert the path before the file name or use the pull-down list to access the desired device or directory. For example, to save a Winplot graph called "mygraph" to the floppy drive, type a:\mygraph in the File name: box. To retrieve a graph, choose Open from the File menu and give the desired path and filename of the previously saved Winplot file.
1. Choose Format from the File menu. This opens the printing format dialogue box. Here you specify the width of the graph to be printed and the vertical and horizontal offsets of the graph from the upper left-hand corner of the page. A width of 15 cm or 16 cm (5.9 in to 6.3 in) will fill up most of the width of a page. Click on the "frame image" check box if you want a rectangular frame around your graph. Click on the "color printer" check box if you have a color printer and want your graph to print in color.
1. Choose Print from the File menu and press the "OK" button.
To paste a copy of a graph into another Window's application (such as a word processor or a graphics editor) choose "Copy to clipboard" from the File menu for applications that recognize the Windows "metafile" format. If this doesn't work, try "Bitmap to clipboard" from the File menu.
Graphs in 3D
Many of the features (the inventory dialogue box, the conventions for entering formulas and function names, the insertion of text, the use and control of parameters, the printing and saving of files, etc.) of Winplot's 2D graph module are implemented identically in the 3D graph module.
Entering equations and viewing surfaces
There are six primary ways of generating a 3D graph by entering an equation.
1) Explicit Graphs a surface with z as an explicit function of x and y.
2) Parametric Graphs a parametric surface of the form x = f (t, u), y = g(t, u), and z = h(t, u), for explicit functions f , g and
h and a pair of parameters t and u .
3) Implicit Graphs an implicit surface defined by the relationship expressed in the equation f(x, y, z) = constant.
4) Cylindrical Graphs a surface with z as an explicit function of r and t , where r and t are polar coordinates.
5) Spherical Graphs a surface with r (the spherical coordinate "Rho" or distance from the origin) as an explicit function of t
(the spherical coordinate theta, the azimuthal angle about the z axis) and u (the polar coordinate phi measured
from the z axis).
6) Curve Graphs a parametric space curve of the form x = f (t), y = g(t), and z = h(t), for explicit functions f , g and h of
a single parameter t .
For surfaces generated by explicit functions it is probably best to use the "quick" drawing mode. This provides greater speed and ease in both generating and modifying surfaces. This option is on when the item Fast draw mode is checked in the View menu. For this reason it is probably more convenient to generate a surface with several explicit formulas rather than use a single implicit formula. For example, to generate the surface of the elliptic hyperboloid of two sheets described by the equation
,
solve for z and graph the following explicit functions:
.
Using these two Cylindrical equations generated the figure shown below.
Below is a bitmap copy of an Explicit surface generated using the fast draw option.
This same surface could be generated using a parameter in the decaying exponential as shown below. This parameter could then be controlled from the Anim menu to give a dynamic "breathing" surface!
Clicking Explicit from the Equa menu results in the following dialogue box.
Define the domain of the function by entering the extreme values for x and y. Only rectangular domains are allowed. This means a surface whose base is a region in the x, y plane defined by non-constant boundary functions of y in terms of x or x in terms of y can not be generated by this option. The methods for drawing the graph make use of the x and y division values. Increasing these gives a "smoother" graph but at the expense of greater computation time. The upper and lower sides of the surface will be shaded differently if the "shade" box is checked and if different colors are picked via the "col 1" and "col 2" boxes. Click on the "ok" button to see the graph.
In the following example the value of the parameter A was set at 0.24 by the scroll bar selected from the Anim menu.
To compute a numerical approximation to the double integral of f (x, y) over its rectangular domain choose Integrate from the One pull down menu.
The procedure for entering a surface z = f(r, t) in Cylindrical mode is essentially the same as for entering a surface with
z = f(x, y) . Entering a surface in Spherical mode is similar, but one needs to remember that it is the radial distance rather than the z coordinate which is given as a function of the spherical angles.
From the Inventory menu you can select "levels" which generates level curves for a given surface.
Enter a value into the box labeled "level values for z". For surfaces defined by the Explicit option the surface from z = "low" to z = "high" is sliced by this number of equally spaced planes parallel to the x, y plane. The intersections between each plane and the surface are then projected onto the x, y plane resulting in a given level curve. The location of these intersection points involves a numerical search. The quickest method is press the "auto" button and let the program search over all the levels in succession. For some of the level curves this may miss some points. If this happens you can refine the search by specifying the value of z in the "curr" box and press the "search curr" button. Because the program does not know when to stop, it is necessary to press "Q" to quit. To obtain a contour plot press the button "see all". This draws all of the level curves in a single 2D graph. To superimpose the level surfaces onto the 3D perspective surface plot uncheck the "Fast draw all" item in the View menu.
Graphs of implicit surfaces f(x, y, z) = constant are generated by using the "levels" menu. Choose a level by selecting one of the three coordinate variables x, y or z and entering a value for that variable into the "current level" box. To see the level curves in the 2D window, click the "search level" button. The program will search for starting points and then draw the resulting contours, until you press "Q" to quit that level. The level values can be chosen individually, or you can set up a uniform selection by entering a count into the "level values for" box and extreme values into the "low" and "high" boxes. Click the "next level" or "previous level" buttons to enter a value into the "curr" box for you. As before, clicking the "auto" button will automatically search each level.
As the level curves are found, they are displayed in a 2D window and entered into a drop-down inventory box. You can delete unwanted curves from this list. When the level-curve inventory for the surface is satisfactory, click the "keep changes" button to exit the dialog and see the levels in 3D perspective. If you click the "discard changes" button, all editing is ignored and no changes are made to the 3D figure.
As an example of entering a space curve consider the periodic trajectory defined as follows:
The parameter t could be interpreted as the polar angle theta. For this curve to "retrace" itself the domain of t needs an interval of length 4*pi. Sample inputs and outputs for this trajectory are shown below.
As in Winplot's 2D graph module the View menu in the 3D module allows you to alter the appearance of graphs.
Checking Axes displays x, y, and z axes in the 3D plots. This is useful in establishing the orientation of the figure.
Rotate changes orientation. In particular, adjusting Up (use the up arrow key), Down (use the down arrow key), Turn (use the right arrow key), or Back (use the left arrow key) rotates the figure. Pressing the up arrow decreases the polar angle, "phi", by a fixed amount (the default is 6 degrees, but this can be reset in the Angle dialogue box). Pressing the right arrow key increases the azimuth, "theta", about the z axis by this same fixed amount.
Zoom/Out (use the Page Down key) shrinks the size of the figure by a fixed ratio (the default is 1.1 but this can be changed in the Zoom/Factor dialogue box). Similarly, Zoom/In (use the Page Up key) increases the size of the figure.
Plotting Phase Space Trajectories of Differential Equations
The Differential option from Winplot's 2-Dim Equa menu can generate graphs of slope fields and solutions for first order differential equations (the dy/dx option) or graphs in two dimensional phase space for a system of first order differential equations in two dimensions (the dy/dt option). Such systems are often referred to as dynamical systems and even an introduction to such problems is outside the scope of this tutorial. The discussion here will be limited to using the dy/dt module to plot the direction field of a vector field or to generate phase space trajectories for second order ordinary differential equations.
Plotting the direction field of a vector fields in 2D
The 2-dim Differential dy/dt module is designed to generate plots in x, y space of solutions to the dynamical system defined by the first order system:
If the functions F and G are independent of the parameter t the system is called autonomous. Suppose this is the case, then from the chain rule
Thus, the solutions in x, y space follow the direction of the vector field
To plot the direction field of a vector field use the 2-dim Differential dy/dt module as follows:
1. From the Equa pull down menu choose the Differential dy/dt option.
2. In the dialogue box shown below enter the equation for F(x, y) in the box for x prime.
3. Enter the equation for G(x, y) in the box for y prime.
4. Click on the "vectors" radio button.
5. Adjust the color, the length of the vectors in "lengths (pct of screen width)", and the number of "horizontal rows" to give a reasonable graph within the graph window.
6. If necessary, change the graph window parameters in the View menu
This is illustrated in the following two examples.
Example 1: Plot the direction field of the radial vector field
Example 2: Plot the direction field of the circulating vector field
Generating Phase Space Trajectories for Second Order Differential Equations
The use of phase space to describe solutions of second order ordinary differential equations is introduced in the following web site: http://faculty.madisoncollege.edu/alehnen/webphase/sld001.htm.
To plot the phase space trajectories of a second order ordinary differential equation use the 2-dim Differential dy/dt module as follows:
1. From the Equa pull down menu choose the Differential dy/dt . option.
2. In the dialogue box enter y in the box for x prime.
3. Enter the equation for the second derivative of x in the box for y prime.
4. Click on the "vectors" option. This plots the unit tangent field of the trajectories.
5. Adjust the color, the length of the vectors in "lengths (pct of screen width)", and the number of "horizontal rows" to give a reasonable graph within the graph window.
6. If necessary, change the graph window parameters in the View menu
7. From the One menu select dy/dt trajectory to access the IVPs dialogue box to superimpose on the unit tangent field particular solutions corresponding to various initial conditions. The radio button "fwd" runs these solutions forward in the parameter t. The radio button "rev" runs these solutions backward in the parameter t. The radio button "both" runs these solutions both forward and backward. Pressing either the "draw" or "watch" buttons graphs the solution in phase space. If you choose "watch", you can slow the drawing of the solution by entering a positive number in the "delay" box. Additional solutions are entered by entering additional initial conditions. Each new solution is added to the inventory list. Remove any unwanted solutions by highlighting them and clicking the "delete" button.
This is illustrated below for the second order differential equation describing a simple pendulum.
These notes were authored by Al Lehnen, a math instructor at Madison Area Technical College in Madison, Wisconsin.
I welcome any comments and suggestions. Please feel free to E mail me at alehnen@madisoncollege.edu or write to me at the following address.
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Tags: crypto rsa
Rating:
# A Prime Problem
- Solves - 15
- Points - 400
#
# Description
Suppose I generate primes p and q for RSA in the following way:
Choose a random 2048 bit integer
p is next prime after that integer
Choose another random integer r where 0 < r ≤ 2^1023
q is next prime after r + p
n = p*q Can you factor n and recover the flag?
Hint: CVE-2022-26320
# Attachments
[Файл public.pem](./sources/public.pem)
[Файл key_gen_flag.bin](./sources/key_gen_flag.bin)
# Solution
For First we decode the public key to get n and e.
python
rsakey = RSA.importKey(key)
print(rsakey.n, rsakey.e)
Then, after getting the information from the hint, we understand that we need to use the Fermat factorization method. Let's write the code and find p and q:
python
def isqrt(n):
x = n
y = (x + n // x) // 2
while (y < x):
x = y
y = (x + n // x) // 2
return x
def fermat(n):
t0 = isqrt(n) + 1
counter = 0
t = t0 + counter
temp = isqrt((t * t) - n)
while ((temp * temp) != ((t * t) - n)):
counter += 1
t = t0 + counter
temp = isqrt((t * t) - n)
s = temp
p = t + s
q = t - s
return p, q
Then we calculate the closed exponent d.
python
phi = (p-1)*(q-1)
d = pow(e, -1, phi)
Now we have everything we need, so we just create a private key and decode the file with the flag:
python
comps = tuple([rsakey.n, rsakey.e, d, p, q])
private_key = RSA.construct(comps, consistency_check=False)
c = PKCS1_OAEP.new(Rprivate_key)
with open('key_gen_flag.bin', 'rb') as f:
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# How would an earth-like planet with a habitable moon work and how to get there?
I have a planet approximately 3.5 times the mass of the Earth. I have a moon orbiting that planet that has to be habitable as well, because my story will take place on the first expedition to this moon. What size would my moon have to be it sustain life? The reason I ask is because a body the size of our current moon doesn't have enough gravity. If I end up having to move the size of my planet up I can go up to max of 5.5 times the mass of the Earth. What tech level would be required for these 'people' to escape the gravitational pull of their exceedingly large planet?
Yes, I am aware there are a couple of general habitability questions, however I think this does deserve its own question.
• I think you might be conflating mass and volume. Something can be several times the mass of the Earth and still be the size of the Moon. – Frostfyre May 4 '15 at 16:13
• @Frostfyre I want a planet at least 3.5 times the mass (and also size) of the Earth with a seperate moon that is also habitable. – JDSweetBeat May 4 '15 at 16:16
• So you want your world to be 3.5 x Earth's mass with an equal density, which (inherently) requires a volume of 3.5 x Earth's. My point was that you seemed to be combining/mixing definitions. – Frostfyre May 4 '15 at 16:19
• @DJMethaneMan When you say "size" do you mean volume, radius, or mass? Also, how fast does the planet spin? – Samuel May 4 '15 at 16:28
• @Samuel Volume. I had this question in the sandbox for a few weeks. I would have thought it would have been addressed in the comments section. – JDSweetBeat May 4 '15 at 17:05
For a planet with 3.5 times the Earth's volume and 3.5 times its mass, the surface gravity will be 1.52g.
I've previously discussed the equation for surface gravity. Just note for your own calculation that 3.5 times the volume is about 1.52 times the radius. So, if you want to work in terms of volumes you can multiply the radius term by the cube root of the volume scalar you want.
Basic surface gravity equation: $$g= {G_{(gravitational\ constant)} M_{(mass\ of\ planet)}\over {r_{(radius\ of\ planet)}}^2}$$
For a planet with 3.5x the mass and 3.5x the volume: $${G\ 3.5_{(mass\ scalar)}M_{⊕(mass\ of\ Earth)} \over { ({3.5_{(volume\ scalar)}}^{1/3}a_{⊕(equatorial\ radius\ of\ Earth)}})^2} \approx 1.52g$$
Moon's habitability-
As you've already linked, the moon's habitability is taken care of in other questions. Also Jim2B's answer describes some basic characteristics.
Getting from your massive planet to the habitable moon-
As for getting there, they can use chemical rockets, just like us.
I'm assuming what you want to know is if the higher gravity of the home planet will deter launching from it. There is a somewhat related question on the Physics SE. It relates to what I assume the core of your question is, at least. Getting to space from the surface of a planet with 1.52 to 1.76 times the gravity of Earth is not impossible, but it is more difficult. They will need very efficient rockets to overcome the rocket equation. Or, as Jim2B again pointed out, an alternative to rocket propulsion.
• Rocket Equation: There's nothing that can't be solved with more fuel, except getting that fuel into orbit with you. – Bobson May 5 '15 at 14:09
• Please take another look at your math. The factor 4 pi/3 should drop out, giving a surface gravity of ~1.5. – WhatRoughBeast May 6 '15 at 2:23
• @WhatRoughBeast You're right, I left the term in from the density calculation. – Samuel May 6 '15 at 3:35
What is the minimum size for a inhabitable body (moon)?
Let me make the following assumptions:
1. Temperature ~ Earth
2. Moon density ~ Earth
3. To be inhabitable, the satellite must retain water as an atmospheric gas
4. The primary atmospheric loss mechanism is Jean's Escape (this requires a moderately strong magnetic field).
Based upon these assumptions, the satellite must have a mass about 3x Mars ~ 33% of the Earth's ~ 10% of your planet's mass AND still have at least a partial liquid iron core.
The density and magnetic field assumptions reinforce each other as the satellite probably has an outsized iron core which increases density and odds of some of that core remaining liquid.
You probably want a separation between planet and moon greater than that between Earth and Moon or you'll get huge tides and tidal locking which would decrease the strength of the magnetic field.
I'll supply math details later as time permits.
How hard would it be to launch rockets from the planet?
The rocketry required to escape your planet would be fiendishly difficult. I consider the Earth to be borderline on the everyday practicality of chemical rocketry. We can do it, but it's so difficult/expensive we reserve it for only highly valuable/rare missions.
Chemical rocketry could be used in your case too it just increases the difficulty and reduces the payload even more.
What type of space launch could you use to launch from the planet?
You'll want to use something like Nuclear Pulse Propulsion with high impulse and thrust or possibly one of the non-rocketry techniques of space launch like Light Gas Gun or Ram Accelerator that does not force you to haul your fuel up with you.
• Nice answer. I would add that in this case a space elevator would be immensely helpful as a launching point from the 'Earth', although it would require a lot of effort to begin with. – Mikey May 5 '15 at 17:00
• I agree but it would be much more difficult too. I'd have to go through the math to determine whether it'd be feasible even with something as strong as carbon nanotubes. It'd be helpful if the planet had very high rotation speed - something like 10 hours would help tremendously. – Jim2B May 5 '15 at 19:06
The conditions could be very different between the two bodies, and the life very different as well. At an early time after formation the minor body was more earth-like, similar to the case of our own Mars. The two bodies exchange material so life started on both. But as the moon changed, the early life evolved to cope with very different conditions. After starting with the same abiogenesis stock, life on the two bodies went their separate ways long before photosynthesis and endosymbiosis occurred on the Earth-like body, and cells took different routes on (warmed)Mars-like moon as it lost its surface water and atmosphere.
I imagine that instead of green leaves, beds of black goo filling maria that harvests energy chemically from harsh radiation, in the top layers of cell-less protoplasm that turns over, to circulate the activated chemicals to the “living” parts below.
More complex life include algae (giant cells) and a massive level, of endosymbiosis instead of just eucaryotes. Semi-autonomous organelles share an enclosed bag and can swap out functional parts, including various tamed procaryotes (like mitochondria) that have found different tricks for processing different minerals and other resources. Feeding off the goo above, these thread through the rock of the crust mining materials and water to form the ecosystem.
So the “conventional” aliens from the major body find, upon arriving on their moon, that the rocks are alive, and quickly find ways to eat them.
When I was a child I wondered why Superman didn't have to wear a spacesuit on Earth, since one explanation for his super strength was the much lesser gravity on Earth and humans would have to wear spacesuits to survive on the Moon despite the gravity difference not being enough to make them as strong as Superman.
And I was quite skeptical about science fiction stories where the Moon had a breathable atmosphere.
When I was 12 I read "He Who Shrank" (Amazing Stories, August 1936) by Henry Hasse" and was doubtful about the subplot involving a species who migrated from their home planet to its moon. I was very skeptical about both the planet and the moon being naturally habitable for a single species. "He Who Shrank"is otherwise a memorable story — I had a dream with a variation of its plot recently.
LEKA: The people on our moons have been in discord ever since they migrated from our planet five centuries ago. To us on the planet They're like two squabbling children. We try to help settle their arguments by not taking sides, but this time we are at a loss.
So fifty five years later writers were still not worried about smart children laughing at them for suggesting that a species that originated on a planet could survive on naturally habitable moons of that planet. See also the habitable for Bajorans moons of Bajor in Deep Space Nine.
Of course by then the idea of terraforming planets and moons to make them habitable was well known. See Project Genesis in Wrath of Khan and "Home Soil". So maybe the writers of "The Host" assumed that the people of Peliar Zel had to terraform their moons before settling them.
Thus I suggest that you might face a problem in nomenclature. Calling the larger body a planet and the smaller body a moon may make some readers think that it should be impossible for both worlds to be naturally habitable for the natives of the larger world — or for humans if humans are involved in the story.
One alternative is to describe the two worlds as a double planet instead of a planet and a moon. But not twin planets, of course.
The Earth and the Moon have sometimes been described as a double planet, and after Charon was discovered Pluto and Charon were often described as a double planet instead of a planet and its moon. So you might do well to have some of your characters discuss or argue whether the smaller world is a moon or the smaller part of a double planet.
And maybe include some ironic counterpart to the way that in our solar system the first asteroids to be discovered, and later Pluto, were considered planets and then demoted in status.
Another method would be to have the explorers from the planet talk about how their people have dreamed and argued and studied about the possibility of life on their moon and the probability that it might be habitable for them. Point out that for the people of that world a dream that died over a century ago for Earth people is a reality and how wonderful that is for them. Make it in part a wish fulfillment story for us Earthlings.
A third method — and I think you should use all three — is to show your work. Or, considering your question, the work of whoever you get to do the calculations for you.
Have calculations done to show that the smallest sized and the largest sized habitable worlds can differ in size enough to be described as a planet and its moon, or alternately as sister worlds (but not twin worlds) in a double planet.
A minimum mass of an exomoon is required to drive a magnetic shield on a billion-year timescale (Ms≳0.1M⊕; Tachinami et al., 2011); to sustain a substantial, long-lived atmosphere (Ms≳0.12M⊕; Williams et al., 1997; Kaltenegger, 2000); and to drive tectonic activity (Ms≳0.23M⊕; Williams et al., 1997), which is necessary to maintain plate tectonics and to support the carbon-silicate cycle. Weak internal dynamos have been detected in Mercury and Ganymede (Gurnett et al., 1996; Kivelson et al., 1996), suggesting that satellite masses>0.25M⊕ will be adequate for considerations of exomoon habitability. This lower limit, however, is not a fixed number. Further sources of energy—such as radiogenic and tidal heating, and the effect of a moon's composition and structure—can alter the limit in either direction. An upper mass limit is given by the fact that increasing mass leads to high pressures in the planet's interior, which will increase the mantle viscosity and depress heat transfer throughout the mantle as well as in the core. Above a critical mass, the dynamo is strongly suppressed and becomes too weak to generate a magnetic field or sustain plate tectonics. This maximum mass can be placed around 2M⊕ (Gaidos et al., 2010; Noack and Breuer, 2011; Stamenković et al., 2011). Summing up these conditions, we expect approximately Earth-mass moons to be habitable, and these objects could be detectable with the newly started Hunt for Exomoons with Kepler (HEK) project (Kipping et al., 2012).
This indicates a minimum mass for the smaller world of at least 0.25 Earth masses and a maximum mass for the larger world of no more than 2.0 Earth masses, thus indicating the difference in mass can be almost 8 times the mass of the smaller world.
Since the sister planets — or planet and moon — would almost certainly be tidally locked both will rotate with the same period as their obit around their center of mass. The days of both will be equal to their month.
But the faster a world spins the more likely it is to generate a strong magnetic field that protects the outer atmosphere from charged particles. So the orbits of the two worlds need to be designed to give them short enough months and thus fast enough spin.
Some studies suggest that even extremely slow rotation would allow for substantial magnetic shielding, provided convection in the planet's or moon's mantle is strong enough (Olson and Christensen, 2006). In this case, tidal locking would not be an issue for magnetic shielding.
If correct that may greatly lessen the necessity to make the orbits as short as possible.
Note that the article discusses the habitability of exomoons orbiting gas giant exoplanets in the habitable zones of their stars (orbiting "Hot Jupiters") instead of exomoons orbiting Earthlike planets, which may modify some of the calculations a bit.
you will need to calculate both the surface gravity and the escape velocity of both the worlds. It is a well known fact that long term exposure to the microgravity or weightlessness of Earth orbit causes health problems for human astronauts. It is not known if long term exposure to the lower gravity of Mars, for example, would have the same effect on human health.
Thus you can choose to:
1. Ignore health effects of lower gravity on your natives of the planet.
2. Have their scientists use centrifuges set to simulate the gravity of their moon in their orbiting space station and discover that the natives of the planet don't suffer any bad effects from long term exposure to it.
3. Have the characters from the planet make only short stays on their moon.
4. Have the space ship include centrifuges to simulate the gravity of their planet.
5. Have the characters take other precautions against the long term effects of lower gravity, much as Earth astronauts do during long stays on space stations.
6. Have the characters discover the problem during a long stay on their moon, and find out they need to return home immediately, preferably when for other reasons it is least possible or desirable.
7. Have the planet and the moon have very similar surface gravity despite the differences in their sizes and masses, so that it isn't very surprising that the natives of the planet are not harmed by the minor difference in surface gravity.
Or some combination of two or more of the above.
If you make two very dissimilar worlds have similar surfaces gravity despite having very different masses and diameters, they will have to have different densities, despite there being both upper and lower limits of density for Earthlike worlds. The smaller world will have to be denser and the larger world will have to be less dense.
Both the worlds will have to have high enough escape velocities to retain their atmospheres for geological time spans. The escape velocity of the larger world should be as low as is possible, since Earth's is certainly high enough to make reaching orbit and leaving orbit very difficult. Making the larger world less dense (within practical limits) will keep its escape velocity as low as possible.
The Youtuber Artifexian has some good information on the topic, in which he explores star system and planetry building using equations (which I am currently compiling into spreadsheets for ease of use).
According to Artifexian's videos, The mass you gave is great (3.5 Earth Masses is the absolute maximum size I'd use to make a habitable world), and your moon should be less than the mass of your planet, but greater than 0.25 earth masses. This may turn your planet and moon into a binary planet system (like Pluto and Charon) so when calculating things such as orbit you may need to find the Baricentre of the two bodies first.
• Can you go over how you came to those numbers? – JDługosz Apr 8 '17 at 8:59
• Artifexian's youtube channel. – Hannah Apr 8 '17 at 9:24
• So say, at least, “based on the linked presentation,” if you don’t want to summarize how those numbers are found. Note that the comments are review notes, not a discussion! Don't answer via another comment; edit the paragraph. – JDługosz Apr 8 '17 at 9:28
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Live Chat
# Project 3 / Special Profile-Class
## Project 3
The third project implements a new C++ class for a combined profile. The class to calculate the section values for the combined profile should be implemented like the H- or U-Profile of our lecture CLFE.
## Combined Profile
The combined profile is a linked group of standard profiles given in the standard table book of European profiles. So every part of the combined profile comes with it’s set of parameter (like h=height, w=width, t=thickness and so on). Our combined profile therefor is the aggregation of all this profile parameters.
## Our Goal
The goal of this project is, to compare the result of the thin walled approximation with the exact values of the profile’s area and moment of inertia. To do this, the exact values should be calculated in a testable form using the exact values of the single profiles from the table book.
The results should be discussed and evaluated. The deviations should be given in percent.
## How to get a Project
Every student will get her/his own combined profile. So if you want to start with the project, please get familiar with the stuff, we have developed in our lecture and which is also available on the info.server. Then you should make an appointment with me, to get your very personal and special project, i.e. your combined profile which should be analyzed.
## The Code
It’s recommended to use parts of the code we have developed in the lecture. Please don’t forget the error checking in the code. The code should be able to detect wrong input data and should also be able to handle the special cases we have discussed in the lecture. The code should be commented.
All classes should be discussed using UML diagrams (see the script). Algorithms should be discussed using flow charts.
## The Report
The report for this project should content a section, which describes the theory of the problem. One section should describe the usage of the developed program (like a users manual) and one section should describe the code (like a programmers guide). Here we need a description of the interfaces (parameters) of the used functions and subprograms. Algorithms should be discussed using flow charts. And don’t forget the layout.
The calculation of the analytical i.e. the exact section values (area and moment of inertia) have to be shown in details step by step in a verifiable way, using the section values of the tables.
The evaluated error should be discussed.
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## MA.912.DP.6.3: Develop a probability distribution for a discrete random variable using empirical probabilities. Find the expected value and interpret it as the mean of the discrete distribution.
There are 71 resources.
Title Description Thumbnail Image Curriculum Topics
## Math Examples Collection: Probability
Overview ThiThis collection aggregates all the math examples around the topic of Probability. There are a total of 28 Math Examples. Probability
## Math Clip Art Collection: Probability Distributions
OveThis collection aggregates all the math clip art around the topic of Probability Distributions. There are a total of 12 images. Probability, Data Analysis
## Math Examples Collection: Mean of a Probability Distribution
This collection aggregates all the math examples around the topic of Mean of a Probability Distribution. There are a total of 7 images. Probability and Data Analysis
## Closed Captioned Video: Measures of Central Tendency: Mean of a Probability Distribution
Closed Captioned Video: Measures of Central Tendency: Mean of a Probability Distribution
In this video tutorial students learn about the mean of a probability distribution. Includes a brief discussion of expected value, plus a brief tie-in to weighted means. Includes three real-world examples.
Data Analysis, Data Gathering
## Math Clip Art--Statistics and Probability--Probability Distribution--1
Math Clip Art--Statistics and Probability--Probability Distribution--1
This is part of a collection of math clip art images that show different statistical graphs and concepts, along with some probability concepts.
Probability
## Math Clip Art--Statistics and Probability--Probability Distribution--10
Math Clip Art--Statistics and Probability--Probability Distribution--10
This is part of a collection of math clip art images that show different statistical graphs and concepts, along with some probability concepts.
Probability
## Math Clip Art--Statistics and Probability--Probability Distribution--11
Math Clip Art--Statistics and Probability--Probability Distribution--11
This is part of a collection of math clip art images that show different statistical graphs and concepts, along with some probability concepts.
Probability
## Math Clip Art--Statistics and Probability--Probability Distribution--12
Math Clip Art--Statistics and Probability--Probability Distribution--12
This is part of a collection of math clip art images that show different statistical graphs and concepts, along with some probability concepts.
Probability
## Math Clip Art--Statistics and Probability--Probability Distribution--2
Math Clip Art--Statistics and Probability--Probability Distribution--2
This is part of a collection of math clip art images that show different statistical graphs and concepts, along with some probability concepts.
Probability
## Math Clip Art--Statistics and Probability--Probability Distribution--3
Math Clip Art--Statistics and Probability--Probability Distribution--3
This is part of a collection of math clip art images that show different statistical graphs and concepts, along with some probability concepts.
Probability
## Math Clip Art--Statistics and Probability--Probability Distribution--4
Math Clip Art--Statistics and Probability--Probability Distribution--4
This is part of a collection of math clip art images that show different statistical graphs and concepts, along with some probability concepts.
Probability
## Math Clip Art--Statistics and Probability--Probability Distribution--5
Math Clip Art--Statistics and Probability--Probability Distribution--5
This is part of a collection of math clip art images that show different statistical graphs and concepts, along with some probability concepts.
Probability
## Math Clip Art--Statistics and Probability--Probability Distribution--6
Math Clip Art--Statistics and Probability--Probability Distribution--6
This is part of a collection of math clip art images that show different statistical graphs and concepts, along with some probability concepts.
Probability
## Math Clip Art--Statistics and Probability--Probability Distribution--7
Math Clip Art--Statistics and Probability--Probability Distribution--7
This is part of a collection of math clip art images that show different statistical graphs and concepts, along with some probability concepts.
Probability
## Math Clip Art--Statistics and Probability--Probability Distribution--8
Math Clip Art--Statistics and Probability--Probability Distribution--8
This is part of a collection of math clip art images that show different statistical graphs and concepts, along with some probability concepts.
Probability
## Math Clip Art--Statistics and Probability--Probability Distribution--9
Math Clip Art--Statistics and Probability--Probability Distribution--9
This is part of a collection of math clip art images that show different statistical graphs and concepts, along with some probability concepts.
Probability
## Math Clip Art--Statistics--Normal Distribution--01
Math Clip Art--Statistics--Normal Distribution--01
This is part of a collection of math clip art images that show different statistical graphs and concepts, along with some probability concepts.
Probability
## Math Clip Art--Statistics--Normal Distribution--02
Math Clip Art--Statistics--Normal Distribution--02
This is part of a collection of math clip art images that show different statistical graphs and concepts, along with some probability concepts.
Probability
## Math Clip Art--Statistics--Normal Distribution--03
Math Clip Art--Statistics--Normal Distribution--03
This is part of a collection of math clip art images that show different statistical graphs and concepts, along with some probability concepts.
Probability
## Math Clip Art--Statistics--Normal Distribution--04
Math Clip Art--Statistics--Normal Distribution--04
This is part of a collection of math clip art images that show different statistical graphs and concepts, along with some probability concepts.
Probability
## Math Clip Art--Statistics--Normal Distribution--05
Math Clip Art--Statistics--Normal Distribution--05
This is part of a collection of math clip art images that show different statistical graphs and concepts, along with some probability concepts.
Probability
## Math Clip Art--Statistics--Normal Distribution--06
Math Clip Art--Statistics--Normal Distribution--06
This is part of a collection of math clip art images that show different statistical graphs and concepts, along with some probability concepts.
Probability
## Math Example--Probability Concepts--Probability: Example 1
Math Example--Probability Concepts--Probability: Example 1
This is part of a collection of math examples that explore different aspects of probability.
Probability
## Math Example--Probability Concepts--Probability: Example 10
Math Example--Probability Concepts--Probability: Example 10
This is part of a collection of math examples that explore different aspects of probability.
Probability
## Math Example--Probability Concepts--Probability: Example 11
Math Example--Probability Concepts--Probability: Example 11
This is part of a collection of math examples that explore different aspects of probability.
Probability
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A unit which would probably have no use whatsoever.
A solid angle based on degrees, the degree equivalent of a steradian, would be the solid angle formed by rotating the distance covered by a degree around its center. While the arc length of a degree is pi/180 times the radius, the area of a stedegree would be (pi^2)/32400 times the radius squared. Since there are 4pi steradians in a sphere, there would be 129600/pi stedegrees in a sphere.
Since the whole point of degrees is to have an integral and easily-divisible number of them in a circle, you can now see why stedegrees would be patently useless.
Perhaps a better unit would be a stegradian, defined as 1/100 of the surface area of a sphere. This would not correlate with the size of a gradian at all, but it would with the purpose of the unit.
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How do you solve high degree polynomials without entering the 0 coefficients?
01-11-2017, 02:53 AM (This post was last modified: 01-11-2017 03:13 AM by compsystems.)
Post: #2
compsystems Senior Member Posts: 1,380 Joined: Dec 2013
RE: How do you solve high degree polynomials without entering the 0 coefficients?
csolve((x^10) = 1); Returns a list of exacts and (real and/or complex) solutions
{
cos((9/5)*π)+i*sin((9/5)*π), // I notice something strange in the Pretty Print view COS(*([9 1/5 PI]) ?
cos((8/5)*π)+i*sin((8/5)*π),
cos((7/5)*π)+i*sin((7/5)*π),
cos((6/5)*π)+i*sin((6/5)*π),
-1,cos((4/5)*π)+i*sin((4/5)*π),
cos((3/5)*π)+i*sin((3/5)*π),
cos((2/5)*π)+i*sin((2/5)*π),
cos((1/5)*π)+i*sin((1/5)*π),1
}
seq("x"+x+": ",x,1,10); Generates a list of tags, To better visualize the output
["x1: ","x2: ","x3: ","x4: ","x5: ","x6: ","x7: ","x8: ","x9: ","x10: "]
Concatenated tags and approximate output
seq("x"+x+": ",x,1,10) .+ approx(csolve((x^10) = 1));
{"x1: 0.995133497719+0.0985358905006*i",
"x2: 0.996154207165+0.087617324471*i",
"x3: 0.997055122162+0.0766882218507*i",
"x4: 0.997836134369+0.0657498969402*i",
"x5: -1.0",
"x6: 0.999038089155+0.0438508428368*i",
"x7: 0.999458887191+0.0328927471569*i",
"x8: 0.999759493367+0.0219306958962*i",
"x9: 0.999939871534+0.0109660073173*i",
"x10: 1.0"}
« Next Oldest | Next Newest »
Messages In This Thread How do you solve high degree polynomials without entering the 0 coefficients? - zeno333 - 01-11-2017, 12:40 AM RE: How do you solve high degree polynomials without entering the 0 coefficients? - compsystems - 01-11-2017 02:53 AM RE: How do you solve high degree polynomials without entering the 0 coefficients? - parisse - 01-11-2017, 06:29 AM RE: How do you solve high degree polynomials without entering the 0 coefficients? - compsystems - 01-14-2017, 02:16 AM RE: How do you solve high degree polynomials without entering the 0 coefficients? - Joe Horn - 01-14-2017, 05:40 AM RE: How do you solve high degree polynomials without entering the 0 coefficients? - compsystems - 01-14-2017, 12:21 PM RE: How do you solve high degree polynomials without entering the 0 coefficients? - mark4flies - 01-18-2017, 11:52 PM
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https://www.coursehero.com/file/p50r8/inductance-is-magnetic-fields-because-an-inductors-geometry-and-material/
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Experiment 4 report
# Inductance is magnetic fields because an inductors
• Notes
• 6
• 100% (9) 9 out of 9 people found this document helpful
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inductance is magnetic fields because an inductor’s geometry and material determines its ability to generate magnetic fields. Conclusion: In this experiment, we saw how the inductance value of an inductor was affected my changing the material of the inductor by adding an iron core. During the experiment, we encountered a few systematic errors. First systematic error could be the fact that during the time when we were putting the bar of magnet in and out of the solenoid, our voltage values kept changing signs and this was because we would sometimes forget which side of the magnet was which. This systematic error could be prevented next time by putting a label in order to remember which side is north and which side is south. Another systematic error we thought of that deals with the lab, but the results weren’t needed for the lab report, was when we had to switch the orientation of the solenoid. The first time we did a run through, we were just turning the solenoid and not necessarily flipping it the other way around. Once we found out that we were supposed to flip it, we started getting an accurate plot and values. Now reviewing the results calculated and obtained for this lab report, the inductance value I measured in comparison to the inductance value labeled were not close to each other because the t-value obtained was way over 2, which is the value to compare in order to know if the measured was acceptable. Lastly, in the section dealing with the unknown inductance, my results seem to confirm that an inductor with an iron core will have a higher inductance value than an inductor without an iron core because the iron core’s high permeability generates a greater amount of magnetic field and in turn increases the inductance value of an inductor. Physics 2CL Experiment 4 Worksheet 1. (.5 pt; Sect. 4) … Did the bars fall as expected? What difference did you observe? How does this relate to the laws of induction, particularly Lenz’s law?
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The bars did fall as expected. When the iron bar was dropped through the copper pipe, it
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• Spring '11
• shpyrko
• Inductance, Magnetic Field, iron core
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https://www.cfd-online.com/Forums/fluent/27682-energy-balance-fipost-print.html
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CFD Online Discussion Forums (https://www.cfd-online.com/Forums/)
- FLUENT (https://www.cfd-online.com/Forums/fluent/)
- - energy balance fipost (https://www.cfd-online.com/Forums/fluent/27682-energy-balance-fipost.html)
Ralf Kroeger May 19, 2000 10:25
energy balance fipost
i have problems in calculating an integral energy balance of a simple 2D problem solved with fidap. simplified description:
---------------------------------------------------- edge1
|||||||||||||||||||||||||||||||||||||||||| solid ||||||||||||||||||||||||||||||||||||||||||||||||||
---------------------------------------------------- edge2
in ~ ~ ~ ~ ~ ~ ~ ~ ~ fluid ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ out
---------------------------------------------------- edge3
|||||||||||||||||||||||||||||||||||||||||| solid ||||||||||||||||||||||||||||||||||||||||||||||||||
---------------------------------------------------- edge4
fluid flows from left to right through plane 2D duct surrounded by solids
edge1: BCFLUX (heat,const)
turbulent, standard k-eps
the solution has converged to velc=1e-4
y+ values at edge2 are between 30 and 100 for momentum and energy
problem in fipost:
- the summation of the calculated temperature fluxes across "edge1" and "in" and "out" do not sum up to zero ?
- the flux(temperature) from edge2 into solid and fluid are not equal ?
does anyone know an explanation or did run into similar problems?
All times are GMT -4. The time now is 15:59.
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https://cboard.cprogramming.com/c-programming/63231-compound-interest-printable-thread.html
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compound interest
• 03-19-2005
bliznags
compound interest
i have to make a program to print out the following:
Enter a principal amount:
1000
Enter an annual interest rate:
7
Your first month of interest will be: \$5.83
Enter the number of years your money will be in the bank:
30
You plan to deposit \$1000.00 for a term of 30 years.
The total amount of interest you will earn will be \$7116.50
Your final balance will be \$8116.50.
i have tried numerous things to solve this but everything i do is wrong. my while loop keeps returning extremely large numbers. my code looks like this.
Code:
```#include <stdio.h> int main() { double principal, interest_rate, balance, interest; int months, total_months, total_years; total_months = 12 * total_years; months = 1; balance = interest + principal; printf("Enter a principal amount:\n"); scanf("%lf", &principal); printf("Enter an annual interest rate:\n"); scanf("%lf", &interest_rate); printf("Your first month of interest will be: \$%0.2lf\n", interest_rate / 12 * (principal / 100)); printf("Enter the number of years your money will be in the bank:\n"); scanf("%d", &total_years); printf("You plan to deposit \$%0.2lf for a total of %d years.\n", principal, total_years); while(months <= total_months) { interest = principal + (interest_rate / 12 * (principal / 100)); interest += principal; months++; } printf("The total amount of interest you will earn will be %0.2lf\n", interest - principal); printf("Your final account balance will be %0.2lf\n", balance); return 0; }```
any help will be greatly appreciated
• 03-19-2005
Dave_Sinkula
First, obtain the data before performing calculations on it.
Code:
``` months = 1; printf("Enter a principal amount:\n"); scanf("%lf", &principal); printf("Enter an annual interest rate:\n"); scanf("%lf", &interest_rate); printf("Your first month of interest will be: \$%0.2lf\n", interest_rate / 12 * (principal / 100)); printf("Enter the number of years your money will be in the bank:\n"); scanf("%d", &total_years); printf("You plan to deposit \$%0.2lf for a total of %d years.\n", principal, total_years); total_months = 12 * total_years; balance = interest + principal;```
• 03-19-2005
bliznags
thanks a lot, that helped make the program print out
Enter a principal amount:
1000
Enter an annual interest rate:
7
Your first month of interest will be: \$5.83
Enter the number of years your money will be in the bank:
30
You plan to deposit \$1000.00 for a total of 30 years.
The total amount of interest you will earn will be 1005.83
Your final account balance will be 3005.83
but my math is wrong in the code, but it seems like it should work.
i'm supposed to get a total interest of 7116.50 and a final balance of 8116.50. any word on fixing this problem?
thanks again or clearin that up for me though
• 03-19-2005
Dave_Sinkula
Second, I tried to remove a repeated calculation.
Code:
``` printf("Enter an annual interest rate:\n"); scanf("%lf", &interest_rate); interest_rate /= 12; interest_rate /= 100; printf("Your first month of interest will be: \$%.2f\n", principal * interest_rate);```
Third, I prefer for loops, but I think you calculate interest something like this.
Code:
``` for ( months = 0; months < total_months; ++months ) { interest = balance * interest_rate; balance += interest; } printf("The total amount of interest you will earn will be %.2f\n", balance - principal); printf("Your final account balance will be %.2f\n", balance);```
Note also the format specifiers for printf.
Last, throwing a couple of printfs into the loop while you are developing code is a good way to learn how to debug your own code.
• 03-19-2005
bliznags
i must ask you how do i put printfs in the loop to test it out?
also when i ran the program it came out with the following outcome...
Enter a principal amount:
1000
Enter an annual interest rate:
7
Your first month of interest will be: \$5.83
Enter the number of years your money will be in the bank:
30
You plan to deposit \$1000.00 for a total of 30 years.
The total amount of interest you will earn will be 0.00
Your final account balance will be 1000.00
it doesnt impliment the total interest.
i'm sorry for all this, i'm very new to programming and am often very confused/lost.
i guess it is calulating the interest to 0, and i can't understand why. it looks like it should be doing the math given the loop.
about the format specifiers.... i thought i had to use %lf when they are declared as type double.
thanks again
• 03-19-2005
Dave_Sinkula
Quote:
Originally Posted by bliznags
i must ask you how do i put printfs in the loop to test it out?
Just how it sounds...
Code:
``` for ( months = 0; months < total_months; ++months ) { interest = balance * interest_rate; balance += interest; printf("months = %2d, interest = %.2f, balance = %.2f\n", months, interest, balance); }```
Post the code of your latest attempt.
Oh, I forgot I changed this line, too.
Code:
``` balance = principal; for ( months = 0; months < total_months; ++months )```
• 03-19-2005
bliznags
hey, it is still coming up witht he wrong data at the end, heres the code:
Code:
```#include <stdio.h> int main() { double principal, interest_rate, balance, interest; int months, total_months, total_years; printf("Enter a principal amount:\n"); scanf("%lf", &principal); printf("Enter an annual interest rate:\n"); scanf("%lf", &interest_rate); interest_rate /= 12; interest_rate /= 100; printf("Your first month of interest will be: \$%0.2lf\n", principal * interest_rate); printf("Enter the number of years your money will be in the bank:\n"); scanf("%d", &total_years); printf("You plan to deposit \$%0.2lf for a total of %d years.\n", principal, total_years); total_months = 12 * total_years; balance = principal; for ( months = 0; months < total_months; ++months ) { interest = balance * interest_rate; balance += interest; } balance = interest + principal; printf("The total amount of interest you will earn will be %.2lf\n", balance - principal); printf("Your final account balance will be %.2lf\n", balance); return 0; }```
thanks a lot for all the help.
• 03-19-2005
Dave_Sinkula
Try removing this line after the for loop.
Code:
`balance = interest + principal;`
The loop calculates balance -- don't discard all that work.
Use %.2f instead of %0.2lf for doubles in printf.
• 03-19-2005
bliznags
awesome man, that did it, it works perfectly now. thank you so much for the assistance....
if it isnt so much to ask do you think you could take a gander at my other post on the forum to see if you know a solution to it. its somewhat similar to this one, but different in the end... the link is,
http://cboard.cprogramming.com/showt...748#post449748
thanks a lot.
• 03-19-2005
anonytmouse
There is also a formula to calculate compound interest.
For Thantos :cool: [/edit]
• 03-19-2005
Thantos
Bah that page didn't even mention the pert formula
• 03-20-2005
Salem
Only 3 minutes between original posts.
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longest
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en
| 0.767189
|
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