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KiRa [710]
3 years ago
6

What is -6a + a + 28a

Mathematics
2 answers:
densk [106]3 years ago
7 0

These are all like terms, meaning that they involve the same literal expression (in this case, "a").

So, we can factor a out of all the terms, and we have

-6a+a+28a=a(-6+1+28) = 23a

Alisiya [41]3 years ago
4 0

-6a+a+28a=-6a+29a=23a

Hope this helps.

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IM IN A HURRY PLEASE HELP ME QUESTION IS DOWN BELOW WORTH 15 POINTS each
Blizzard [7]

The value of x is 2 and the length of JK is 4

<h3>How to solve the unknown variables?</h3>

The given parameters from the circle are:

  • Center = Point S
  • Segment JK = 8
  • Segment LK = 2x + 4
  • Congruent SN = SP = 7

The lines SR and SQ are the radii of the circle P

This means that lines JK and JL are congruent

So, we have:

JK = KL

Substitute LK = 2x + 4 and JK = 4

4 = 2x + 4

Rewrite the above equation as:

2x + 4 = 8

Subtract 4 from both sides

2x + 4 - 4 = 8 - 4

Evaluate the difference

2x = 4

Divide both sides by 2

2x = 4/2

This gives

x = 2

Substitute x = 2 in LK = 2x + 4

LK = 2*2 + 4

Evaluate the product of 2 and 2

LK = 4 + 4

This gives

LK = 8

The point N divides JK into 2 equal segments

So, we have

JN = JK/2

JN= 8/2

JN = 4

Hence, the value of x is 2 and the length of JK is 4

Read more about circles at:

brainly.com/question/11833983

#SPJ1

5 0
2 years ago
Layla wants to build a wooden box with a volume of 450 cubic centimeters. She started with a width of 10cm and a height of 3cm.
puteri [66]

Answer:

Stop cheating and do your own work....

Mr E

Step-by-step explanation:

7 0
3 years ago
In Exercises 1-4, tell whether the ordered pair is a solution of
Wittaler [7]

Answer: 1. No     2. yes    3. yes    4. no

Step-by-step explanation:

If you locate the points on the graph, 1 and 4 are not in the shaded and 2 and 3 are, which means they are not solutions.

7 0
3 years ago
Which statements are true for solutions when there is a system of two linear equations in two variables? Select all that apply.
monitta

Answer: A system of linear equations consists of two or more linear equations made up of two or more variables, such that all equations in the system are considered simultaneously.

To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time.

In order for a linear system to have a unique solution, there must be at least as many equations as there are variables.

The solution to a system of linear equations in two variables is any ordered pair  

(

x

,

y

)

that satisfies each equation independently. Graphically, solutions are points at which the lines intersect.

Key Terms

system of linear equations: A set of two or more equations made up of two or more variables that are considered simultaneously.

dependent system: A system of linear equations in which the two equations represent the

same line; there are an infinite number of solutions to a dependent system.

inconsistent system: A system of linear equations with no common solution because they

represent parallel lines, which have no point or line in common.

independent system: A system of linear equations with exactly one solution pair  

(

x

,

y

)

.

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Need help pleaseI was bad at math in school so lwant to learn
aleksklad [387]

The probability of an event is expressed as

Pr(\text{event) =}\frac{Total\text{ number of favourable/desired outcome}}{Tota\text{l number of possible outcome}}

Given:

\begin{gathered} \text{Red}\Rightarrow2 \\ \text{Green}\Rightarrow3 \\ \text{Blue}\Rightarrow2 \\ \Rightarrow Total\text{ number of balls = 2+3+2=7 balls} \end{gathered}

The probability of drwing two blue balls one after the other is expressed as

Pr(\text{blue)}\times Pr(blue)

For the first draw:

\begin{gathered} Pr(\text{blue) = }\frac{number\text{ of blue balls}}{total\text{ number of balls}} \\ =\frac{2}{7} \end{gathered}

For the second draw, we have only 1 blue ball left out of a total of 6 balls (since a blue ball with drawn earlier).

Thus,

\begin{gathered} Pr(\text{blue)}=\frac{number\text{ of blue balls left}}{total\text{ number of balls left}} \\ =\frac{1}{6} \end{gathered}

The probability of drawing two blue balls one after the other is evaluted as

\begin{gathered} \frac{1}{6}\times\frac{2}{7} \\ =\frac{1}{21} \end{gathered}

The probablity that none of the balls drawn is blue is evaluted as

\begin{gathered} 1-\frac{1}{21} \\ =\frac{20}{21} \end{gathered}

Hence, the probablity that none of the balls drawn is blue is evaluted as

\frac{20}{21}

8 0
1 year ago
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