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serious [3.7K]
3 years ago
5

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}

or

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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Isaac goes to an amusement park where tickets for the rides cost $10 per sheet and tickets for the shows cost $15 each.
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In a café, I order a cup of tea and a piece of cake and it costs £1.10. The next time I order 2 cups of tea and one piece of cak
defon

Answer:

£0.50

Step-by-step explanation:

t = one cup of tea

c = one piece of cake

t + c = £1.10

2t + c = £1.70

the cost increases by £0.60 (£1.70 - £1.10) when you order one more cup of tea which means that one cup of tea costs £0.60

substitute £0.60 into t + c = £1.10

£0.60 + c = £1.10

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3 years ago
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Do these two statements contradict
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2 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
2 years ago
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