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3 years ago

5x-25expresión racional

Mathematics

1 answer:

Nikitich [7]3 years ago

5 0

Answer:

expresión 5

Step-by-step explanation:

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Plz help me I don’t know how to do Number 1 and 2

Rainbow [258]

1. This shape is called a parallelogram and it’s area formula is A=W*H, so you would plug in 12 1/3 for H and 17 1/5 for W. The answer would be 215yd.

2. This is a triangle and it’s area formula is A=1/2bh. Plug in 14 as H and 17 for b. The answer would be 119ft.

6 0

3 years ago

Two supplementary angles are in the ratio of 31:5 find angles

ad-work [718]

Answer:

The angles are 155° and 25°.

Step-by-step explanation:

Given:

Two supplementary angles are in the ratio of 31:5.

Now, to find the angles.

The sum of two supplementary angles = 180°

Let the ratio of the angles be $31x\ and\ 5x$ .

So, according to question:

$31x+5x=180\°$

$36x=180\°$

Dividing both sides by 36 we get:

$x=5$

So, $31x=31\times 5=155\°.$

And, $5x=5\times 5=25\°.$

Therefore, the angles are 155° and 25°.

7 0

3 years ago

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,

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

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

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Solve the equation N/-2=-15

otez555 [7]

Answer:

Step-by-step explanation:

N/-2=-15

N= -15 times -2

N= 30

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Solve 6-2x=3 (bensjdkmenskkw)

Kamila [148]

Answer:

answer: 4.5

Step-by-step explanation:

2x=3+6

2x=9

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5x-25expresión racional ​

5x-25expresión racional