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Semenov [28]
3 years ago
12

Two cylindrical containers, C and D, with different capacities are shown below. What is the total volume of liquid held in these

containers if half of container C is full and one third of container D is full?

Mathematics
1 answer:
Zielflug [23.3K]3 years ago
6 0

Answer:

Option D. (x + 4)(x + 1)

Step-by-step explanation:

From the question given above, the following data were obtained:

C = (6x + 2) L

D = (3x² + 6x + 9) L

Also, we were told that half of container C is full and one third of container D is full. Thus the volume of liquid in each container can be obtained as follow:

Volume in C = ½C

Volume in C = ½(6x + 2)

Volume in C = (3x + 1) L

Volume in D = ⅓D

Volume in D = ⅓(3x² + 6x + 9)

Volume in D = (x² + 2x + 3) L

Finally, we shall determine the total volume of liquid in the two containers. This can be obtained as follow:

Volume in C = (3x + 1) L

Volume in D = (x² + 2x + 3) L

Total volume =?

Total volume = Volume in C + Volume in D

Total volume = (3x + 1) + (x² + 2x + 3)

= 3x + 1 + x² + 2x + 3

= x² + 5x + 4

Factorise

x² + 5x + 4

x² + x + 4x + 4

x(x + 1) + 4(x + 1)

(x + 4)(x + 1)

Thus, the total volume of liquid in the two containers is (x + 4)(x + 1) L.

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Answer:

The point estimate for population mean is 8.129 Mpa.

Step-by-step explanation:

We are given the following in the question:

Data on flexural strength(MPa) for concrete beams of a certain type:

11.8, 7.7, 6.5, 6.8, 9.7, 6.8, 7.3, 7.9, 9.7, 8.7, 8.1, 8.5, 6.3, 7.0, 7.3, 7.4, 5.3, 9.0, 8.1, 11.3, 6.3, 7.2, 7.7, 7.8, 11.6, 10.7, 7.0

a) Point estimate of the mean value of strength for the conceptual population of all beams manufactured

We use the sample mean, \bar{x} as the point estimate for population mean.

Formula:

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

\bar{x} = \dfrac{\sum x_i}{n} = \dfrac{219.5}{27} = 8.129

Thus, the point estimate for population mean is 8.129 Mpa.

4 0
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Given the slope m=4 what is the perpendicular slope
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Derivative of the denominator:
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Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx

and then split the fraction,

\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx

Based on our previous test, we know that a simple substitution will work for the first integral: \rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx

So the first integral changes,

\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx

integrating to a log,

\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2

So we have this going on,

\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx

Bringing all that stuff together as a single square,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx

Making the substitution: \rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx

giving us,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du

simplying a lil bit,

\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du

and hopefully from this point you recognize your arctangent integral,

\rm ln|x^2+2x+5|+\frac52arctan(u)

undo your substitution as a final step,
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Hope that helps!
Lemme know if any steps were too confusing.

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