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3 years ago

Can someone tell me what x equals!

Mathematics

2 answers:

Zinaida [17]3 years ago

5 0

Answer:

Step-by-step explanation:

hope u find this helpfull

sertanlavr [38]3 years ago

5 0

Answer:

Step-by-step explanation:

x = 12

the triangle was made smaller (i forget what the term is)

it was made smaller by $1\frac{1}{3}$

to find that you divide 24 by its equivalent side on the other triangle which would be 18

24 / 18 = $1\frac{1}{3}$

to find what x is equal to you have to find the equivalent side on the other triangle. this would be 16. since the triangle with 16 is bigger the equation is:

16 / $1\frac{1}{3}$ = 12

so x = 12

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3 years ago

What is 5t+16+6-5t can u help me

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3 years ago

Read 2 more answers

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

3 years ago

Read 2 more answers

When 2(3/5 x + 2/3 y - 1/4 x -1 1/2 y + 3 )is simplified, what is the resulting expression?

vovangra [49]

Answer:

Simplifying the expression: $2(\frac{3}{5}x+\frac{2}{3}y-\frac{1}{4}x-1\frac{1}{2}y +3)$ we get $\mathbf{42x-100y-360}$

Step-by-step explanation:

We need to simplify the expression: $2(\frac{3}{5}x+\frac{2}{3}y-\frac{1}{4}x-1\frac{1}{2}y +3)$

First we will solve terms inside the bracket

$2(\frac{3}{5}x+\frac{2}{3}y-\frac{1}{4}x-1\frac{1}{2}y +3)$

Converting mixed fraction $1\frac{1}{2} y$ into improper fraction, we get: $\frac{3}{2}y$

Replacing the term:

$2(\frac{3}{5}x+\frac{2}{3}y-\frac{1}{4}x-\frac{3}{2}y +3)$

Now, taking LCM of: 5,3,4,2 we get 60

Now multiply 60 with each term inside the bracket

$2(\frac{3}{5}x\times60+\frac{2}{3}y\times60-\frac{1}{4}x\times60-\frac{3}{2}y\times60 +3\times60)\\2(3x\times12+2y\times20-1x\times15-3x\times30-3\times60)\\2(36x+40y-15x-90y-180)$

Now, combine like terms

$2(36x-15x-90y+40y-180)\\2(21x-50y-180)$

Now, multiply all terms with 2

$42x-100y-360$

So, Simplifying the expression: $2(\frac{3}{5}x+\frac{2}{3}y-\frac{1}{4}x-1\frac{1}{2}y +3)$ we get $\mathbf{42x-100y-360}$

3 0

3 years ago