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8090 [49]
3 years ago
10

For what value of k will these pairs of simultaneous equations have no solution?

Mathematics
1 answer:
yKpoI14uk [10]3 years ago
4 0
Kx-3y=k
y=4x+1 ( we will put it in 1st equation )
kx - 3 (4x+1)=k
kx - 12 x - 3 =k
kx - 12 x = k+3
x(k-12) = k+3
x=\frac{k+3}{k-12}
Also y=\frac{5k}{k-12}
We can not divide with zero.
k-12=0
k=12
For k= 12 equations have no solutions.
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A 25 kg rock , is raised to the top of a hill 10 m high. What is the PE of the rock ?
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Potential  \:  \: Energy =(mass) \times (gravity \: acceleration) \times (height) \\

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If gravity acceleration is 10 :

PE = 25 \times 10 \times 10

PE = 2500

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If gravity acceleration is 9.8 :

PE = 25 \times 9.8 \times 10

PE = 25 \times 98

PE = 2450

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2 years ago
A product that is a chemically bonded combination of reactants will form in which type of reaction? Select three options.
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Answer: synthesis

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Step-by-step explanation:

3 0
2 years ago
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A stadium has 49000 seats. Seats sell for $25 in section A, $20 in section B, and $15 in section C. The number of seats in secti
eimsori [14]

Answer:

A stadium has 49000 seats.  

Seats sell for $25 in Section A, z

$20 in Section B,-------------------x seats

$15 in Section C. ------------------y seats  

(x+y)=z  

25(x+y)+20x+15y=1052000

25x+25y+20x+15y=1052000

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9x+8y=210400------------------1

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Step-by-step explanation:

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3 years ago
PLZZZ HELPPP THIS IS A TIMED TESTTT *look at the picture*
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Answer:

x=14

Step-by-step explanation:

7 0
3 years ago
You roll a pair of honest dice. If you roll a total of 7, you win $18; if you roll a total of 11, you win $54; if you roll any o
amm1812
Answer: -1
The negative value indicates a loss

============================================================

Explanation:

Define the three events
A = rolling a 7
B = rolling an 11
C = roll any other total (don't roll 7, don't roll 11)

There are 6 ways to roll a 7. They are
1+6 = 7
2+5 = 7
3+4 = 7
4+3 = 7
5+2 = 7
6+1 = 7
Use this to compute the probability of rolling a 7
P(A) = (number of ways to roll 7)/(number total rolls) = 6/36 = 1/6
Note: the 36 comes from 6*6 = 36 since there are 6 sides per die

There are only 2 ways to roll an 11. Those 2 ways are:
5+6 = 11
6+5 = 11
The probability for event B is P(B) = 2/36 = 1/18

Since there are 6 ways to roll a "7" and 2 ways to roll "11", there are 6+2 = 8 ways to roll either event. 
This leaves 36-8 = 28 ways to roll anything else
P(C) = 28/36 = 7/9

-----------------------------

In summary so far,
P(A) = 1/6
P(B) = 1/18
P(C) = 7/9

The winnings for each event, let's call it W(X), represents the prize amounts.
Any losses are negative values
W(A) = amount of winnings if event A happens 
W(B) = amount of winnings if event B happens
W(C) = amount of winnings if event C happens
W(A) = 18
W(B) = 54
W(C) = -9

Multiply the probability P(X) values with the corresponding W(X) values
P(A)*W(A) = (1/6)*(18) = 3
P(B)*W(B) = (1/18)*(54) = 3
P(C)*W(C) = (7/9)*(-9) = -7

Add up those results
3+3+(-7) = -1

The expected value for this game is -1.
The player is expected to lose on average 1 dollar per game played.


Note: because the expected value is not 0, this is not a fair game.


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3 years ago
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