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Alexeev081 [22]
3 years ago
13

Solve: 3х + y = 5x-7 when , y = 5 Show your work

Mathematics
1 answer:
Juliette [100K]3 years ago
6 0

Answer: x=6

Step-by-step explanation:

Hope this helps

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(Literal Equations) Solve each equation or formula for the variable indicated
Serggg [28]

Answer:

y = 3/2( x-v)

Step-by-step explanation:

2/3y + v = x

Subtract v from each side

2/3y + v-v = x-v

2/3y = x-v

Multiply each side by 3/2

3/2 *2/3y  =3/2( x-v)

y = 3/2( x-v)

4 0
3 years ago
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Allyson loves to eat cheese sticks. A small order comes with 3 cheese sticks, and a large order comes with 6 cheese sticks. Last
alukav5142 [94]
She would have gotten 5 large orders of cheese sticks.

Hope this helps!
3 0
3 years ago
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Please Help!<br> Show how to solve 8x^2+14x+5=0 by completing the square.
Kisachek [45]
X=12                          cause i looked it up online
3 0
3 years ago
I NEED HELP PLEASE !!!
Igoryamba

Answer:

64

180 - 65 - 51 = 64

6 0
3 years ago
A coin is tossed four times What is the probability of getting four heads? (1 mark) a) b) What is the probability of getting exa
BlackZzzverrR [31]
<h2>Answer:</h2>

1. A coin is tossed four times. What is the probability of getting four heads?

Each toss has a 1/2 chance of getting a head.

So the chance of getting all four heads can be calculated as :

1/2\times1/2\times1/2\times1/2=1/16

2. A coin is tossed four times. What is the probability of getting two heads?

Each toss can have 2 results, so 4 flips will have 2^{4} or 16 results.

Getting two heads means getting two tails also. So, we can get the number of times two heads come = \frac{4!}{2!2!} = 6

We can write the groups like - {HHTT,HTHT,HTTH,THHT,THTH,TTHH}

So, the required probability is : 6/16 or 3/8.

3. Not getting two heads means getting 3 tails and 1 head or all tails.

Probability of having all tails = 1/2\times1/2\times1/2\times1/2=1/16

Probability of one head(1st trial) and three tails = 1/16

Probability of one head (2nd trial) and three tails (the 1st, 3rd and 4th trials) = 1/16

Probability of one head (3rd trial) and three tails (the 1st, 2nd and 4th trials) = 1/16

Probability of one head (4th trial) and three tails (the 1st, 2nd and 3rd trials) = 1/16

So, the total probability showing only one or none head and at least three tails = 1/16+1/16+1/16+1/16+1/16=5/16

6 0
3 years ago
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