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2 years ago

Solve |8y+4|=2|y-1| please show the work!

Mathematics

1 answer:

Snowcat [4.5K]2 years ago

3 0

8y+4 = 2(y-1)
8y+4 = 2y-2
8y-2y+4 = -2
6y+4 = -2
6y = -2-4
6y = -6
y = -1

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2 years ago

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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

What is the product of (X4/3)(X2/3) ?

katrin [286]

Hi, the answer to this would be x6/9. I'm assuming the x4/3 and x2/3 are fractions and the x's aren't exponents. Now how I got x6/9 is shown here.

1st Step: Started off by regrouping the terms
1/3x3 x^4x^2

2nd Step: we can easily simplify 3x3 to just 9. And now we're left with 1/9x^4x^2

3rd Step: Now we can simplify the 1/9 to just x^4x^2/9

4th Step: Now we can use the product rule which is simple. So We add the exponents and simplify it to just one exponent. So x4+2=6 that simplifies to just x^6.

Final Answer: x^6/9.
Hope this helped you :)

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3 years ago

If x=5 and y=2<br> ...........................

azamat

${ \bold{ - 5}}$