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4vir4ik [10]
4 years ago
10

4(-8x + 5) – (-33x – 26) ​

Mathematics
2 answers:
STALIN [3.7K]4 years ago
6 0

Answer:

x+46

Step-by-step explanation:

4(-8x + 5) – (-33x – 26) ​

The first step is to distribute and simplify.

4 distributed to -8x will be -32x, and same thing goes with 5, which is 20.

-

The first part of the equation is -32x+20

Now, this is where it gets confusing.

If there is a sign before a pair of parentheses, you always put 1. In this case, it's -1 distributed to -33x and -26.

So, the second part of the expression will be +33x+26

The whole expression is

-32x + 20 + 33x + 26

add like terms, -32x+33x= x.

20+26=46.

The answer is x+46.

Keith_Richards [23]4 years ago
5 0

Answer:

1x+46 is the answer

Step-by-step explanation:

=4(-8x + 5) – (-33x – 26)

opening brackets to simplify

=-32x+20+33x+26

=-32x+33x+20+26

=1x+46 is the answer

hope it will help :)

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\rule{200}4

Answer : <em>The</em><em> </em><em>required</em><em> </em><em>ratio</em><em> </em><em>is</em><em> </em><em>(</em><em>1</em><em>4</em><em>m</em><em>-</em><em>6</em><em>)</em><em>:</em><em>(</em><em>8</em><em>m</em><em>+</em><em>2</em><em>3</em><em>)</em><em> </em><em>.</em>

\rule{200}4

Here we are given that the ratio of sum of first n terms of two AP's is (7n + 1):(4n + 27) .

That is.

\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{7n +1}{4n +27}  \\

As , we know that the sum of n terms of an AP is given by ,

\small\sf\longrightarrow \pink{ S_n =\dfrac{n}{2}[2a +(n-1)d]} \\

Assume that ,

  • First term of 1st AP = a
  • First term of 2nd AP = a'
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Using this we have ,

\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{\dfrac{n}{2}[2a + (n-1)d]}{\dfrac{n}{2}[2a' +(n-1)d'] } \\

\small\sf\longrightarrow \dfrac{7n+1}{4n+27}=\dfrac{2a + (n -1)d}{2a' + (n -1)d' } . . . . . (i) \\

Now also we know that the nth term of an AP is given by ,

\longrightarrow\sf\small \pink{ T_n = a + (n-1)d}\\

Therefore,

\longrightarrow\sf\small \dfrac{T_{m_1}}{T_{m_2}}= \dfrac{ a + (n-1)d }{a'+(n-1)d'}. . . . . (ii)\\

\longrightarrow\sf\small \dfrac{T_1}{T_2}=\dfrac{2a + (2n-2)d}{2a'+(2n-2)d'} . . . . . (iii)\\

From equation (i) and (iii) ,

\longrightarrow\sf\small n-1 = 2m-2\\

\longrightarrow\sf\small n = 2m -2+1 \\

\longrightarrow\sf\small n = 2m -1 \\

Substitute this value in equation (i) ,

\longrightarrow \sf\small \dfrac{2a+ (2m-1-1)d}{2a' +(2m-1-1)d'}=\dfrac{7(2m-1)+1}{4(2m-1) +27}\\

Simplify,

\longrightarrow\sf\small \dfrac{ 2a + (2m-2)d}{2a' +(2m-2)d'}=\dfrac{14m-7+1}{8m-4+27}\\

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\longrightarrow\sf\small \dfrac{[a + (m-1)d]}{[a' + (m-1)d']}=\dfrac{ 14m-6}{8m+23}\\

From equation (ii) ,

\longrightarrow\sf\small \underline{\underline{\blue{ \dfrac{T_{m_1}}{T_{m_2}}=\dfrac{ 14m-6}{8m+23}}}}\\

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