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Sergio039 [100]
3 years ago
12

Heccin hurry

Mathematics
2 answers:
kipiarov [429]3 years ago
8 0

Question:

The model below represents 4x + (-4) = -2x + 8.

4 long green tiles and 4 square red tiles = 2 long red tiles and 8 square green tiles.

What is the value of x when solving the equation 4x + (-4) = -2x + 8 using the algebra tiles?

x = -4

x = -2

x = 2

x = 4

Answer:

x = 2

Step-by-step explanation:

Given

4x + (-4) = -2x + 8 which represents a model of coloured tiles of various lengths (shapes)

Required

Find x

To find x, we'll solve the expression 4x + (-4) = -2x + 8 using the knowledge of algebra.

4x + (-4) = -2x + 8

Open bracket

4x - 4 = -2x + 8

Collect like terms

4x + 2x = 4 + 8

Perform addition arithmetic operation on both sides of the equation

6x = 12

Multiply both sides by ⅙

⅙ * 6x = ⅙ * 12

x = 2

Hence, the value of x that satisfies the expression 4x + (-4) = -2x + 8 is 2

Semenov [28]3 years ago
6 0

Answer:

<h2>x = 2</h2>

Step-by-step explanation:

Given the equation model 4x+(-4) = -2x+8

To find the value of x, the following steps must be followed

4x+(-4) = -2x+8\\4x-4 = -2x+8\\subtracting\ 8\ from\ both\ sides\\4x-4-8=-2x+8-8\\4x-12=-2x\\4x+2x=12\\6x=12\\x=\frac{12}{6}\\ x = 2

The value of x is 2

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Find the area of the region enclosed by f(x) and the x axis for the given function over the specified u reevaluate f(x)=x^2+3x+4
lord [1]

Answer:

A = 68 unit^2

Step-by-step explanation:

Given:-

The piece-wise function f(x) is defined over an interval as follows:

                       f(x) =  { x^2+3x+4    , x < 3

                       f(x) =  { x^2+3x+4     , x≥3

                        Domain : [ -3 , 4 ]

Find:-

Find the area of the region enclosed by f(x) and the x axis

Solution:-

- The best way to tackle problems relating to piece-wise functions is to solve for each part individually and then combine the results.

- The first portion of function is valid over the interval [ -3 , 3 ]:      

                       f(x) = x^2+3x+4

- The area "A1" bounded by f(x) is given as:

                      A1 = \int\limits^a_b {f(x)} \, dx

Where,  The interval of the function { -3 , 3 ] = [ a , b ]:

                     A1 = \int\limits^a_b {x^2+3x+4} \, dx\\\\A1 = \frac{x^3}{3} + \frac{3x^2}{2} + 4x |\limits_-_3^3 \\\\A1 = \frac{3^3}{3} + \frac{3*3^2}{2} + 4*3 - \frac{-3^3}{3} - \frac{3(-3)^2}{2} - 4(-3)\\\\A1 = 9 + 13.5 +12 + 9-13.5+12\\\\A1 =42 unit^2

- Similarly for the other portion of piece-wise function covering the interval [3 , 4] :

                     f(x) = 4x+10

- The area "A2" bounded by f(x) is given as:

                      A2 = \int\limits^a_b {f(x)} \, dx

Where,  The interval of the function { 3 , 4 ] = [ a , b ]:

                     A2 = \int\limits^a_b {4x+10} \, dx\\\\A2 = 2x^2 + 10x |\limits_3^4 \\\\A2 = 2*(4)^2 + 10*4 - 2*(3)^2 - 10*3\\\\A2 = 32 + 40 - 18-30\\\\A2 =26 unit^2

- The total area "A" bounded by the piece-wise function over the entire domain [ -3 , 4 ] is given:

                     A = A1 + A2

                     A = 42 + 26

                     A = 68 unit^2

8 0
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