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nirvana33 [79]
3 years ago
11

I need help with the following if possible

Mathematics
2 answers:
scoundrel [369]3 years ago
5 0
1. -1
2. 1
3. 5
4. -1
5.-1
6. -3
kenny6666 [7]3 years ago
3 0

Answer:

  1. 2+(-3)= 2-3= -1
  2. -3+4=1
  3. -4+9=5
  4. 7+(-8)= 7-8= -1
  5. -2+1= 1
  6. 6+ -9= -3
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Which coordinate grid shows Point A at (0.50, −1.25)? (1 point)
polet [3.4K]

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Answer:

  (d)  see below

Step-by-step explanation:

Since each grid square is 1/4 unit. the coordinates in terms of grid squares are ...

  A(0.50, -1.25) ⇒ (0.50/.25, -1.25/.25) = (2, -5) . . . grid squares

Point A is located 2 squares right and 5 squares down on the grid. This matches choice D.

5 0
3 years ago
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Use the graph to find the midpoint between O and H.
butalik [34]

Answer:

(-1, 4)

Step-by-step explanation:

-5+3/2 = -1

6+ 2/2 = 4

(-1, 4)

3 0
3 years ago
Fractions that have the same value
Lelechka [254]

Yes, they certainly do exist.

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6 0
2 years ago
A particle moves on a circle through points which been marked 0,1,2,3,4 (in a clockwise order). At each step it has a probabilit
Sedaia [141]

Answer:

Step-by-step explanation:

Given data:

SS={0,1,2,3,4}

Let probability of moving to the right be = P

Then probability of moving to the left is =1-P

The transition probability matrix is:

\left[\begin{array}{ccccc}1&P&0&0&0\\1-P&1&P&0&0\\0&1-P&1&P&0\\0&0&1-P&1&P\\0&0&0&1-P&1\end{array}\right]

Calculating the limiting probabilities:

π0=π0+Pπ1                 eq(1)

π1=(1-P)π0+π1+Pπ2     eq(2)

π2=(1-P)π1+π2+Pπ3    eq(3)

π3=(1-P)π2+π3+Pπ4    eq(4)

π4=(1-P)π3+π4             eq(5)

π0+π1+π2+π3+π4=1

π0-π0-Pπ1=0

→π1 = 0

substituting value of π1  in eq(2)

(1-P)π0+Pπ2=0

from

π2=(1-P)π1+π2+Pπ3  

we get

(1-P)π1+Pπ3 = 0

from

π3=(1-P)π2+π3+Pπ4

we get

(1-P)π2+Pπ4 =0

from π4=(1-P)π3+π4  

→π3=0

substituting values of π1 and π3 in eq(3)

→π2=0

Now

π0+π1+π2+π3+π4=0

π0+π4=1

π0=0.5

π4=0.5

So limiting probabilities are {0.5,0,0,0,0.5}

4 0
3 years ago
Consider f and c below. f(x, y, z) = yzexzi + exzj + xyexzk,
Jet001 [13]

Answer:

f(x,y,z)=ye^{xz}+C  

Step-by-step explanation:

We can write the given expression as :

\vec f(x,y,z)=yze^{xz}\,\vec\imath+e^{xz}\,\vec\jmath+xye^{xz}\,\vec k

As given,   f = ∇f.

∇f = \dfrac{\partial f}{\partial x}i  + \dfrac{\partial f}{\partial y}j  +\dfrac{\partial f}{\partial z}k 

We can write the partial derivative with respect to x, y and z.

\dfrac{\partial f}{\partial x}=yze^{xz}       ___(Equation 1)

\dfrac{\partial f}{\partial y}=e^{xz}               ______(Equation 2)

\dfrac{\partial f}{\partial z}=xye^{xz}             ______(Equation 3)

Take equation 2 and integrate with respect to y,

\dfrac{\partial f}{\partial y}=e^{xz}

f(x,y,z)=ye^{xz}+a(x,z)           ----------Equation 4

Derivate both sides w.r.t x , we get :

\frac{d}{dx}(yze^{xz})=yze^{xz}+\dfrac{\partial a}{\partial x}

or

\dfrac{\partial a}{\partial x}=0

integrate

a(x,z)=b(z)

put in equation 4 ,

we get :

f(x,y,z)=ye^{xz}+b(z)

take derivative wrt z

\frac{d}{dz} (ye^{xz}+b(z))\impliesxye^{xz}=xye^{xz}+\frac{db}{dz}

we can take here:

\frac{db}{dz} = 0

integrate:

\int\ {\frac{db}{dz} } \, =\int0

b(z) = C

The function can be written as :

from equation 4 :

f(x,y,z)=ye^{xz}+C  

Where C is a constant.

4 0
2 years ago
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