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romanna [79]
3 years ago
14

How do I use a Pascal Triangle

Mathematics
1 answer:
sergeinik [125]3 years ago
5 0
Okay every row is built from thr row above it. this gives us the coefficients for an expanded binomial of the form (a + bn) , where n is row of triangle.

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A man runs 200 m in 25 s. What is his average speed?
Romashka [77]

Answer:

his average speed is 8

Step-by-step explanation:

3 0
3 years ago
The steps below show the incomplete solution to find the value of x for the equation
mafiozo [28]
Step 1: 6x - 2x - 5 = -6 + 21 
Step 2: 6x - 2x - 5 = 15 
Step 3: 4x - 5 = 15 

The next step would be to add 5 to both sides of the equation. 
so Step 3: 4x - 5 = 15 .......add 5 to both sides would make 
Step 4: 4x = 20 
 
So after the next step you would have 4x = 20 

Hope this helps! :) 

    
7 0
4 years ago
Kathy starts with a number. She adds 3 to her number then doubles the result. Finally she subtract 7. If she ended with 9 what n
Ivahew [28]

if she has 9, add 7. you have 16. divide it by half, 8. subtract 3 and you get 5.

8 0
3 years ago
Read 2 more answers
) Consider point p = (0, 0, 0) of paraboloid z = x 2 + ky2 , k > 0. (a) Show that the unit vectors of x-axis and y-axis are e
lapo4ka [179]

Answer:

Answer: K = 4k and H=<u>(2+2k) </u>= k+1

                                          2

Step-by-step explanation:

Part(a) :Let's parametrize the given surface-

X (u, v) = ( u²,v²,u²+kv² )

Now : Xu = (1,0,2u) and Xv = (0,1,2kv)

using below formula:

N (u, v) = <u>Xu x X v</u>

               [ Xu x Xv]

N(u,v) = <u>(-2u,-2kv , 1)</u>

          √ {4u²+4k²v²+1}

Now value of normal vector at p=(0,0,0) is N(p)

N(p) = N(0,0 )= (0,0,1)

so finally at p=(0,0,0) : Xu = (1,0,0) , Xv = (0,1,0), Np = (0,0,1)

Clearly Tp(S) is spanned by (1,0,0),(0,1,0)

 dN.Xu = d/dt N(t,0)/ t=0 = (-2,0,0)

 dN.Xv = d/dr N(0,r)/r=0=(0,-2k,0)

since Xu ,Xv ∈Tp(S)

aXu + bXv ∈Tp(S)

-dNp (aXu + bXv) =(2a,2kb,0)

and represent above in matrix form:

   2    0              a

   0    2k           b

-dNp  is symmetric matrix with eigen values (LAMDA) ∧1=2 and ∧2 =2k

Part (b):

 

Product of Eigen Values is called Gaussian Curvature(K).

Mean of Eigen Values is called Mean Curvature(H).

Therefore K = 4k and H=<u>(2+2k) </u>= k+1

                                             2

5 0
4 years ago
The coordinates of the midpoint of segment GH are M (4,3) and the coordinates of one endpoint are G (5,-6)
Mariulka [41]

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ G(\stackrel{x_1}{5}~,~\stackrel{y_1}{-6})\qquad H(\stackrel{x_2}{x}~,~\stackrel{y_2}{y}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{x+5}{2}~~,~~\cfrac{y-6}{2} \right)~~=~~\stackrel{\stackrel{midpoint}{M}}{(4,3)}\implies \begin{cases} \cfrac{x+5}{2}=4\\[1em] x+5=8\\ \boxed{x=3}\\ \cline{1-1} \cfrac{y-6}{2}=3\\[1em] y-6=6\\ \boxed{y=12} \end{cases}

5 0
3 years ago
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