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nignag [31]
3 years ago
8

What are the values of x in the equation x(x+6)=4(x+6)

Mathematics
2 answers:
stiks02 [169]3 years ago
7 0
       x(x + 6) = 4(x + 6)
   x(x) + x(6) = 4(x) + 4(6)
       x² + 6x = 4x + 24
       <u>     - 4x  - 4x        </u>
       x² + 2x = 24
x² + 2x - 24 = 24 - 24
 x² + 2x - 24 = 0
x = <u>-(2) ± √((2)² - 4(1)(-24))</u>
                       2(1)
x = <u>-2 ± √(4 + 96)</u>
               2
x = -<u>2 ± √(100)
</u>             2<u>
</u>x = <u>-2 ± 10
</u>          2
x = -1 ± 5
x = -1 + 5    U    x = -1 - 5
x = 4    U    x = -6
<u />
Gwar [14]3 years ago
5 0
Since both sides has same factor (x+6), we can set (x+6) equal to 0.

So we would have one of solution x = -6. (because this would have us x·0 = 4·0 )

Now when x is NOT equal to -6, we can divide both sides by (x+6), leaving us 

x = 4

So answer is x = -6 or 4.
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Final Answer: 175.5
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y varies directly as x, y=14 when x=18. Determine x when y=31. Estimate your answer to two places after the decimal, if necessar
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Y : x
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She shared 2/3 of her cupcakes


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the moon is about 240000 miles from earth what is this distance written as a whole number multiplied by a power of ten
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4 0
3 years ago
The region bounded by y=x^2+1, y=x, x=-1, x=2 with square cross sections perpendicular to the x-axis.
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.

This area then can be calculated as the integral between x1 and x2 for f(x) - g(x).

We want to find the area bounded by:

f(x) = y = x^2 + 1

g(x) = y = x

x = -1

x = 2

To find this area, we need to f(x) - g(x) between x = -1 and x = 2

This is:

\int\limits^2_{-1} {(f(x) - g(x))} \, dx

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

We know that:

\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:

(4 + 2 - 2) - ( -1/3 - 1 - 1/2) = 4 + 1/3 + 1 + 1/2 = 5 + 2/6 + 3/6 = 5 + 5/6

The bounded area is 5 + 5/6 square units.

3 0
3 years ago
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