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Rus_ich [418]
3 years ago
8

3(2x + 7) = 6(x - 4)

Mathematics
2 answers:
Lesechka [4]3 years ago
3 0

let's solve for x :

  • 3(2x + 7) = 6(x - 4)

  • 6x + 21 = 6x - 24

here, 6x is present on both sides in the equation, and they cancel out so, there's actually no solution to this equation.

lesya [120]3 years ago
3 0

Answer:

<h2>6x+21=6x-24</h2>

Step-by-step explanation:

<h2>or,6x-6x=-24-21. or,0 = -45</h2>
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Simplify the equation 3x-4+2=-2x-5+5x Down to its simplest form do not solve the equation
lana66690 [7]

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3x-4+2-2x-5+5x

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7 0
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Calculate the slope of the line that contains the points (5,-2) and (-3,-7).
photoshop1234 [79]

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4 0
3 years ago
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
3 years ago
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