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3 years ago

Resolute academy algebra 1

Mathematics

1 answer:

Goryan [66]3 years ago

3 0

Answer:

Step-by-step explanation:

We can see in the expression that there are 2 x² blocks, 4 -x blocks, and 3 -1 blocks. Therefore, we can write this as

2 * x² + 4 * (-x) + 3 * (-1) = 2x²-4x-3

Comparing this with each answer, we have

A: x²-2x-4 + x² + 2x-1 = 2x²-5. This is not correct

B: 3x²-7x+1-(x²-3x+4) = 3x²-7x+1 -x²+3x-4 = 2x²-4x-3. This seems correct but we can check the other answers to be sure

C: (2x+1)(x-3) = 2x²-6x+x-3 = 2x²-5x-3. This is incorrect

D: $\frac{8x^{12} - 16x^7 - 12x^6}{4x^6} = 2x^6 - 4x - 3$ . This is incorrect

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An opaque bag contains 5 green marbles 3 blue marbles and 2 red marbles. What is the probability of randomly picking out a marbl

anastassius [24]

Answer= .2 probability

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2 years ago

Solve for p. 9(p-4) = -18 p=

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Step-by-step explanation:

9(p-4) = -18

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9p = -18 + 36

9p = 18

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Check:

9 ( 2 - 4 ) = - 18

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3 years ago

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A doctor ran 55 hours the mom ran 52 hours what was the total

Elan Coil [88]

This is I suppose basic addition. Addition is a mathematical process of combining or joining something to something else. The word "total" indicates that the addition will be used. If the doctor ran for 55 hours and the mom for 52 hours, by using the addition it will be:
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107 hours in total.

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3 years ago

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Work out the number of sides tile A has.<br> Both tile A and tile B are regular polygons

finlep [7]

Answer:

Step-by-step explanation:

since tile B is a regular polygon means it is an equilateral triangle. The angle inside B would be 60.

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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,

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

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

7 0

2 years ago

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