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Pani-rosa [81]
2 years ago
6

Where is the midpoint of AB if A(6,4) and B(5,-2)?

Mathematics
1 answer:
Nikitich [7]2 years ago
3 0

Answer:

Midpoint={5.5,1}

Step-by-step explanation:

M={x1+x2/2,y1+y2/2}

M={6+5/2,-2+4/2}

M={11/2,2/2}

M={5.5,1}

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Compute the probability of the event E1 that Bob wins in a duel against Eve alone, assuming he shoots first. (Hint: Let x be the
skad [1K]

Answer: Hello your question is incomplete attached below is the missing

n ( 1 + n )

Step-by-step explanation:

P( Bob hits target ) = 1/3

P( Eve hits target ) = 2/3

P( Carol hits target ) = 1

<u>Compute the P that Bob wins in a duel against Eve alone</u>

P(Bob hits the target in first shot ) = n = 1/3

P(Bob hits the target in second shot ) = n^2 = ( 1/3 * 1/3 ) = 1/9

hence the probability of Bob winning( i.e. P( Bob wins Event E1  ) = n + n^2 = n ( 1 + n )

6 0
2 years ago
What is the answer for (2x/y4)2
Soloha48 [4]

Answer:

4x/y4

Step-by-step explanation:

2(2x/y4)

2 x 2x/ y4

=4x/y4

8 0
2 years ago
Which diagram represents the net of a cube? Circle all that apply.
Zigmanuir [339]

Answer:

B 100%

Step-by-step explanation:

3 0
1 year ago
This extreme value problem has a solution with both a maximum value and a minimum value. use lagrange multipliers to find the ex
noname [10]

L(x,y,z,\lambda)=10x+10y+2z+\lambda(5x^2+5y^2+2z^2-42)

L_x=10+10\lambda x=0\implies1+\lambda x=0

L_y=10+10\lambda y=0\implies1+\lambda y=0

L_z=2+4\lambda z=0\implies1+2\lambda z=0

L_\lambda=5x^2+5y^2+2z^2-42=0

\begin{cases}L_x=0\\L_y=0\\L_z=0\end{cases}\implies1+\lambda x=1+\lambda y=1+2\lambda z=0\implies x=y=2z

5x^2+5y^2+2z^2=5(2z)^2+5(2z)^2+2z^2=42z^2=42\implies z^2=1

z^2=1\implies z=\pm1\implies x=y=\pm2

There are two critical points, at which we have

f\left(2,2,1\right)=42\text{ (a maximum value)}

f\left(-2,-2,-1\right)=-42\text{ (a minimum value)}

3 0
3 years ago
Help! I’ll give extra points:)
Fofino [41]

Answer:

if I'm not wrong I think it's B

3 0
2 years ago
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