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3 years ago

13 – 6x = 2(8 + 3x) x= ???????

Mathematics

2 answers:

ruslelena [56]3 years ago

8 0

Answer:

$13-6x=2\left(8+3x\right)$

$13-6x=16+6x$

$13-6x-13=16+6x-13$

$-6x=6x+3$

$-6x-6x=6x+3-6x$

$-12x=3$

$\frac{-12x}{-12}=\frac{3}{-12}$

$x=-\frac{1}{4}$

$x=-0.25$

Alika [10]3 years ago

6 0

Answer:

-1/4 = x

Step-by-step explanation:

first, we distribute the 2 into the parenthesis

13 - 6x = 2(8 + 3x)

13 - 6x = 16 + 6x

then, we add 6x to both sides to move all of the x terms onto one side

13 - 6x = 16 + 6x

13 = 16 + 12x

next, we subtract 16 on both sides

13 = 16 + 12x

-3 = 12x

next, we divide both sides by 12 to isolate x

-3 = 12x

-3/12 = x

finally, we simplify the fraction

-3/12 = x

-1/4 = x

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2 years ago

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VLD [36.1K]

Answer:

The bounded area is 5 + 5/6 square units. (or 35/6 square units)

Step-by-step explanation:

Suppose we want to find the area bounded by two functions f(x) and g(x) in a given interval (x1, x2)

Such that f(x) > g(x) in the given interval.

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$\int\limits^{}_{} {x} \, dx = \frac{x^2}{2}$

$\int\limits^{}_{} {1} \, dx = x$

$\int\limits^{}_{} {x^2} \, dx = \frac{x^3}{3}$

Then our integral is:

$\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx = (\frac{2^3}{2} + 2 - \frac{2^2}{2}) - (\frac{(-1)^3}{3} + (-1) - \frac{(-1)^2}{2} )$

The right side is equal to:

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2 years ago