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A DDR3 DIMM has three notches - one on each end (which matches the retaining springs on the motherboard socket) - and one offset notch on the row of connector pins - to ensure you insert it the correct way into the socket.
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What is 1 plus m?
1 + m is the simplest expression you can make for this problem.
What is larger 1 m or 1 cm?
1 m
0.5 cm is equal to how many meters?
1 m = 100cm 1/100 m = 1 cm 1/100 / 2 m = 0.5cm 1/100 x 1/2 m = 0.5cm 1/200 m = 0.5cm 5x10-3 m = 0.5cm
How many 1 m is to cm?
1 m = 100 cm 1 cm = 0.01 m
What is 1mm x 2 m?
1 mm = 1/1000 m → 1 mm × 2 m = 1 × 1/1000 m × 2 m = 0.002 m² Or 1 m = 1000 mm → 1 mm × 2 m = 1 mm × 2 × 1000 mm = 2000 mm²
What guitar chords make the key of A?
Here's the 1-7 I. A ii. Bm iii. C#m IV. D V. E vi F#m Vii G#dim
Change a cognos report to a new different catalog?
You can do it with a macro. Dim ImpAppObject as Object Dim ImpRepObject as Object Dim strPath as String Dim strExtension as String Dim strFolder as String Dim PromptVal As String Dim strPath1 as String Sub Main() strPath = "M:\Algemeen\Werkmap\In" strFolder = dir("M:\Algemeen\Werkmap\In\*.IMR") strPath1 = "M:\Algemeen\Werkmap\Uit" Set ImpAppObject = CreateObject("CognosImpromptu.Application") ImpAppObject.Visible 0 ImpAppObject.OpenCatalog "L:\BESTMATE 3.014PRPI.CAT","Creator" Do While StrFolder <>"" Set ImpRepObject = ImpAppObject.OpenReportNoExecute(strpath & strfolder) ImpRepObject.SaveAs strpath1+strfolder ImpRepObject.CloseReport strFolder = Dir loop ImpAppObject.Quit Set ImpAppObject = Nothing End Sub
How to compute temperature?
The temperature is calculated by the formula KE = dim/2 N k T, where KE = total kinetic energy of the group of atoms (sum of 1/2 m v^2), dim = 2 or 3 = dimensionality of the simulation, N = number of atoms in the group, k = Boltzmann constant, and T = temperature.
What words can be made using the letters P Z E i H i D M?
him, dim, zip
Words that start with d and end with m?
dim dims diagram diaphragm denim deem dream
3 letter words ending with m?
3 letter word ending in m are aim arm gym jam ham mom hum gum dim him
What rhymes with m?
Tim limb dim It is much easier to answer a question about rhyming if you ask for words that rhyme with a word, not a letter.
How do you find equation for a pyramid of oranges stacked 1 2x2 3x3 4x4?
The equation for n layers is S(n) = n(n+1)(2n+1)/6It is simplest to prove it by induction.When n = 1,S(1) = 1*(1+1)(2*1+1)/6 = 1*2*3/6 = 1.Thus the formula is true for n = 1.Suppose it is true for n = m. That is, for a pyramid of m levels,S(m) = m*(m+1)*(2m+1)/6Then the (m+1)th level has (m+1)*(m+1) oranges and soS(m+1) = S(m) + (m+1)*(m+1)= m*(m+1)*(2m+1)/6 + (m+1)*(m+1)= (m+1)/6*[m*(2m+1) + 6(m+1)]= (m+1)/6*[2m^2 + m + 6m + 6]= (m+1)/6*[2m^2 + 7m + 6]= (m+1)/6*(m+2)*(2m+3)= (m+1)*(m+2)*(2m+3)/6= [(m+1)]*[(m+1)+1)]*[2*(m+1)+1]/6Thus, if the formula is true for n = m, then it is true for n = m+1.Therefore, since it is true for n =1 it is true for all positive integers.
What is the c plus plus program for magic square?
#include<iostream> #include<vector> #include<cassert> class magic_square { friend std::ostream& operator<< (std::ostream&, const magic_square&); private: size_t dim; std::vector<std::vector<size_t>> elem; public: magic_square (const size_t); magic_square (const magic_square& m): dim {m.dim}, elem {m.elem} {} magic_square (magic_square&& m): dim {m.dim}, elem {std::move (m.elem)} {} size_t size () const { return dim*dim; } size_t order () const { return dim; } size_t* operator[] (const size_t); #ifndef NDEBUG bool assert_valid(); #endif DEBUG }; magic_square::magic_square (const size_t order): dim {order}, elem {} { if (dim < 3) throw std::range_error ("Magic square's must have an order greater than 2"); elem.resize (dim); for (size_t row=0; row!=dim; ++row) elem[row].resize(dim); switch (dim % 4) { case (0): { size_t box = dim / 4; size_t lower = 1; size_t upper = dim * dim; for (size_t x=0; x!=dim; ++x) { size_t row = x+1; for (size_t y=0; y!=dim; ++y) { size_t col = y+1; if (((row<=box row>dim-box) && (col<=box col>dim-box)) ((row>box && row<=dim-box) && (col>box && col<=dim-box))) elem[x][y] = lower; else elem[x][y] = upper; ++lower; --upper; } } } break; case (1): case (3): { size_t num=0; size_t row=0; size_t col=(dim-1)/2; while (num<dim*dim) { elem[row][col] = ++num; size_t new_row = !row?dim:row-1; size_t new_col = col+1; if (new_row==dim) { if (new_col==dim) { new_col=col; --new_row; } else new_row=dim-1; } else if (new_col==dim) new_col=0; if (elem[new_row][new_col]!=0) { new_row=row+1; new_col=col; } row = new_row; col = new_col; } } break; case (2): { size_t num = 0; size_t quarter = dim / 2; magic_square temp (quarter); for (size_t loop=0; loop!=4; ++loop) { num = loop * quarter * quarter; switch (loop) { case (0): for (size_t row=0; row!=quarter; ++row) for (size_t col=0; col!=quarter; ++col) { elem[row][col] = temp.elem[row][col] + num; } break; case (1): for (size_t row=quarter; row!=dim; ++row) for (size_t col=quarter; col!=dim; ++col) { size_t x = row-quarter; size_t y = col-quarter; elem[row][col] = temp.elem[x][y] + num; } break; case (2): for (size_t row=0; row!=quarter; ++row) for (size_t col=quarter; col!=dim; ++col) { size_t x = row; size_t y = col-quarter; elem[row][col] = temp.elem[x][y] + num; } break; case (3): for (size_t row=quarter; row!=dim; ++row) for (size_t col=0; col!=quarter; ++col) { size_t x = row-quarter; size_t y = col; elem[row][col] = temp.elem[x][y] + num; } break; } } size_t div = (quarter-1) / 2; for (size_t row=0; row!=quarter; ++row) { size_t min_col=row==div?1:0; size_t max_col=row==div?div+1:div; for (size_t col=min_col; col!=max_col; ++col) std::swap (elem[row][col], elem[row+quarter][col]); } if (div=div-1) { for (size_t row=0; row!=quarter; ++row) { for (size_t col=dim-div; col!=dim; ++col) { std::swap (elem[row][col], elem[row+quarter][col]); } } } } break; } #ifndef NDEBUG assert_valid(); #endif NDEBUG } #ifndef NDEBUG bool magic_square::assert_valid() { size_t old_sum = 0; size_t sum = 0; for (size_t row=0; row!=dim; ++row) { sum = 0; for (size_t col=0; col!=dim; ++col) sum += elem[row][col]; if (!old_sum) old_sum = sum; assert (old_sum sum); return true; } #endif NDEBUG std::ostream& operator<< (std::ostream& os, const magic_square& s) { for (size_t x=0; x!=s.dim; ++x) { for (size_t y=0; y!=s.dim; ++y) { os << s.elem[x][y] << '\t'; } os << '\n'; } return os; } int main() { std::cout << "Magic square of order 3\n" << magic_square (3) << std::endl; std::cout << "Magic square of order 4\n" << magic_square (4) << std::endl; std::cout << "Magic square of order 6\n" << magic_square (6) << std::endl; }
Words that begin with d and end with m?
Dam, darkroom, decorum, deem, deform, denim, diagram, dim, disarm, doom, dorm, dream and drum begin with the letter d. They end with the letter m.
What are all the words you make out of letters l r i y c u d m?
mild, mud, lid, rim, dim, duly
What is m over m to the -4 power?
It depends on the interpretation of the question: Trivially, (m/m)-4 = 1-4 = 1. More interestingly, m-a = 1/ma or ma = 1/m-a So, m/m-4 = m*(1/m-4) = m*(m4) = m5
