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3 years ago

Perform the indicated operation and simply the result.

2 answers:

5 0

I am assuming that the whole 3x - 6 in the fraction is the numerator

the this simplifies to 3x (3x - 6)

= 9x^2 - 18x (C)

Murrr4er [49]3 years ago

4 0

Answer:

9x^2 - 18x

Hope this helps! :)

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Please help solve!!! Thank you!!

Hatshy [7]

3 radical 11 ?? Maybe ??

5 0

3 years ago

What does x equal 3.25x-2.5= -9.75x+4

Andrews [41]

X would be 0.5. Hope this helped

5 0

3 years ago

Read 2 more answers

Use the disk method or the shell method to find the volumes of the solids generated by revolving the region bounded by the graph

diamong [38]

Answer:

a) 8π

b) 8/3 π

c) 32/5 π

d) 176/15 π

Step-by-step explanation:

Given lines : y = √x, y = 2, x = 0.

a) The x-axis

using the shell method

y = √x = , x = y^2

h = y^2 , p = y

vol = ( 2π ) $\int\limits^2_0 {ph} \, dy$

= $( 2\pi ) \int\limits^2_0 {y.y^2} \, dy$

∴ Vol = 8π

b) The line y = 2 ( using the shell method )

p = 2 - y

h = y^2

vol = ( 2π ) $\int\limits^2_0 {ph} \, dy$

= $( 2\pi ) \int\limits^2_0 {(2-y).y^2} \, dy$

= ( 2π ) * [ 2/3 * y^3 - y^4 / 4 ] ²₀

∴ Vol = 8/3 π

c) The y-axis ( using shell method )

h = 2-y = h = 2 - √x

p = x

vol = $(2\pi ) \int\limits^4_0 {ph} \, dx$

= $(2\pi ) \int\limits^4_0 {x(2-\sqrt{x} ) } \, dx$

= ( 2π ) [x^2 - 2/5*x^5/2 ]⁴₀

vol = ( 2π ) ( 16/5 ) = 32/5 π

d) The line x = -1 (using shell method )

p = 1 + x

h = 2√x

vol = $(2\pi ) \int\limits^4_0 {ph} \, dx$

Hence vol = 176/15 π

attached below is the graphical representation of P and h

3 0

3 years ago

Petrolyn motor oil is a combination of natural oil and synthetic oil. It contains

fenix001 [56]

So 5 liters of natural for every 7 synthetic, so 5 / 12 liters are natural oil.

so; 684 × (5/12) = 285 liters of natural oil

Hope this helps!

7 0

3 years ago

The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78

____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

$m-t_{a/2,n-1}\frac{s}{\sqrt{n} }$ ≤ μ ≤ $m+t_{a/2,n-1}\frac{s}{\sqrt{n} }$

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and $t_{a/2,n-1}$ is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and $t_{0.05,9}$ by 1.8331, we get that the 90% confidence interval for the mean speed is:

$74.2-(1.8331)\frac{5.3083}{\sqrt{10} }$ ≤ μ ≤ $74.2+(1.8331)\frac{5.3083}{\sqrt{10} }$

74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077

71.123 ≤ μ ≤ 77.277

8 0

3 years ago