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Paraphin [41]
3 years ago
7

How can you analyze connections between linear equations and use them to solve problems

Mathematics
1 answer:
Alexeev081 [22]3 years ago
6 0

Answer:

Step-by-step explanation:

One of the more obvious "connections" between linear equations is the presence of the same two variables (e. g., x and y) in these equations.

Assuming that your two equations are distinct (neither is merely a multiple of the other), we can use the "elimination by addition and subtraction" method to eliminate one variable, leaving us with an equation in one variable, solve this 1-variable (e. g., in x) equation, and then use the resulting value in the other equation to find the value of the other variable (e. g., y).  By doing this we find a unique solution (a, b) that satisfies both original equations.  Not only that, but also this solution (a, b) will also satisfy both of the original linear equations.

I urge you to think about what you mean by "analyze connections."  

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vichka [17]
33=17x-18 add 18 both sides
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3 years ago
Help pls, will mark brainliest
BaLLatris [955]

Here , we are provided with a table which shows 5 consecutive terms of an arithmetic sequence . But before solving further , let's recall that ;

The n'th term of a Arithmetic Sequence let's say it be {\sf T_n} is given by ;

  • {\boxed{\bf T_{n}=T_{1}+(n-1)d}}

Where , <u>d</u> is the common difference

Now , here we are given with ;

{\quad \qquad \sf \blacktriangleright \blacktriangleright \blacktriangleright T_{1}=8 \: and \: T_{5}=-4}

We have to find the 2nd , 3rd and 4th term respectively ,

Now , by using the above formula , 5th term can be written as ;

{: \implies \quad \sf T_{1}+(5-1)d=T_{5}}

Putting the values and transposing 1st term to RHS , we have ;

{: \implies \quad \sf 4d = -4-8}

{: \implies \quad \sf d=-\dfrac{12}{4}}

{: \implies \quad \sf d=-3}

Now , as we got the common difference , so we can find out the missing terms now ;

{: \implies \quad \sf T_{2}= T_{1}+(2-1)d}

{: \implies \quad \sf T_{2}= 8 +d}

{: \implies \quad \sf T_{2}= 8-3}

{: \implies \quad \bf \therefore \:  T_{2}= 5}

Now

{: \implies \quad \sf T_{3}= T_{1}+(3-1)d}

{: \implies \quad \sf T_{3}= 8 +2d}

{: \implies \quad \sf T_{3}= 8-6}

{: \implies \quad \bf \therefore \:  T_{3}= 2}

Also ,

{: \implies \quad \sf T_{4}= T_{1}+(4-1)d}

{: \implies \quad \sf T_{4}= 8 +3d}

{: \implies \quad \sf T_{4}= 8-9}

{: \implies \quad \bf \therefore \:  T_{4}= -1}

Now , The given table can be written as ;

{\begin{array}{|c|c|c|c|c|c|}\cline{1-6} \bf n & \sf 1 & 2 & 3 & 4 & 5 \\ \cline{1-6} \bf T_{n} & \sf 8 & 5 & 2 & -1 & -4 \end{array}}

Note :- Kindly view the answer from web , if you're not able to see the full answer from here ;

brainly.com/question/26750175

8 0
2 years ago
Simplify the expression 2x+3(x-6)
Bingel [31]

Answer:

5x-18

Step-by-step explanation:

2x+3x-18

5x-18

if ur in college how do u not know this :P

8 0
2 years ago
alicia's mom drove 265 miles on thursday and 478 miles on friday. she has 143 miles more to drive on saturday. about how many mi
prohojiy [21]

Answer:

she will drive a total of 886 miles

Step-by-step explanation:


8 0
2 years ago
Find the area of the regular octagon if the apothem is 4.2 and a side is 3.4 In. Round to the nearest whole number. A. 114 B. 57
Elena-2011 [213]
The area of the regular octagon is calculated as half of the product of the perimeter and the apothem (ap), using the formula of the area of the regular polygon.
 We have then:
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 Where,
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 ap: apotema
 Substituting values:
 A = ((8 * 3.4) * (4.2)) / 2
 A = 57.12 in ^ 2
 Answer:
 
the area of the regular octagon is:
 
B. 57
6 0
3 years ago
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