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Sliva [168]
3 years ago
15

Solve the Equation. 14x+28=14

Mathematics
1 answer:
Korvikt [17]3 years ago
8 0

Answer:

x= -1

Step-by-step explanation:

1. First, you need to combine like terms. Constants with constants and variables with variables. Since there is only one variable, there is nothing to combine it with so nothing happens to it. The constants, on the other hand, can combine.

2. You have to subtract 28 from 28 and 14 so that +28 can cancel out on the left side and combine on the right.

3. That will leave you with 14x= -14.

4. Now you need to get x on its own so you need to divide 14 (opposite of multiplication to cancel out) on both sides.

5. That will leave you with x= -1. That is your final answer.

Hope this helps! Please vote me as Brainliest. Let me know if you need more of an explanation. Remember, cheating is never the answer. Try your best before you seek answers. Hope it's not too late. Good luck with whatever you are doing! Have an amazing day!

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OLEGan [10]

Answer:

3.636 hour. dividing 200miles /55m/h =3.636h

3 0
2 years ago
Read 2 more answers
Thirty students are eating lunch at 5 tables. Each table has the same number of students. How many students are sitting at each
kaheart [24]
30 ÷ 5
6 students at each table

Hope this helps!
5 0
3 years ago
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What is the value of x in the following equation? 5x + 12 = 6(x + 1) <br> A. 6 <br> B. 5 C. 24 D. 7
Maurinko [17]

Answer:

A: 6

Step-by-step explanation:

Our equation is 5x+12=6(x+1)

Our possible numbers are 6, 5, 24, or 7.

To figure this out you start putting the possible numbers in place of x so we'll start off using 6:

5*6+12=6(6+1).

5*6=30. 30+12=42. 6+1=7, 6*7=42.

This equals the same on both sides.

Next we'll go to 5:

5*5+12=6(5+1)

5*5=25, 25+12=37. 5+1=6, 6*6=36.

This doesn't equal the same amount on both sides.

Next is 24:

5*24+12=6(24+1)

5*24=120, 120+12=132. 24+1=25, 6*25=150

This doesn't equal the same amount.

Next is 7:

5*7+12=6(7+1)

5*7=35, 35+12=47. 7+1=8, 6*8=48.

This doesn't equal.

The only number that equals the same amount on both sides is A:6

4 0
3 years ago
Three times each day, a quality engineer samples a component from a recently manufactured batch and tests it. Each part is class
Tresset [83]

Answer:

Step-by-step explanation:

Hello!

Three samples of components manufactured are taken per day. They are classified as:

D: "Conforming (suitable for its use)"

E: "Downgraded (unsuitable for the intended purpose but usable for another purpose)"

F: "Scrap (not usable)"

This classification includes the three events that may occur in your sample space S. The experiment consists in recording the categories of the three parts tested in a day.

a. List the 27 outcomes in the sample space.

The possible outcomes in the space sample are the combinations of the three events. To avoid using the same letters as in the following questions I've named the evets as D, E, and F

S={DDD, DED, DFD, DEF, DFE, DEE, DFF, DDE, DDF , EDE, EEE, EFE, EED, EEF, EDF, EFD, EDD, EFF , FDF, FEF, FFF, FFE, FFD, FDE, FED, FDD, FEE}

b. Let A be the event that all the parts fall into the same category. List the outcomes in A.

A: "All the parts fall into the same category"

You have three possible outcomes for this event, that the three compounds are conforming, "DDD", that the three are unconforming, "EEE", or that the three compounds are scrap, "FFF". There are only three possible outcomes for this event.

S={DDD, EEE, FFF}

c. Let B be the event that there is one part in each category. List the outcomes in B.

B: "There is a part in each category"

This means, for example, The first one is conforming "D", the second one is unconforming "E" and the third one is scrap "F", then the first one may be unconforming "E", the second one is conforming "D" and the thirds one is scrap "F", and so on, you have 6 possible outcomes for this event:

S={DEF, DFE, EDF, EFD, FDE, FED}

d. Let C be the event that at least two parts are conforming. List the outcomes in C.

C: "At least two parts are conforming"

For this event, you can have two of the compounds to be considered conforming or the three of them.

S={DDD, DED, DFD, DDE, DDF , EDD, FDD}

A total of 7 combinations fit this event.

I hope you have a SUPER day!

6 0
2 years ago
I’m timed !!! Please help
Doss [256]
1,450 because you would plug 75 in for x
f(x)= 20(75)-50
4 0
3 years ago
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