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What is the product?( square root 3x+ square root 5) (square root 15x+ 2 square root 30)

hichkok12 [17]

$(\sqrt{3x}+\sqrt5)\cdot(\sqrt{15x}+2\sqrt{30})\\\\=(\sqrt{3x})(\sqrt{15x})+(\sqrt{3x})(2\sqrt{30})+(\sqrt5)(\sqrt{15x})+(\sqrt5)(2\sqrt{30})\\\\=\sqrt{(3x)(15x)}+2\sqrt{(3x)(30)}+\sqrt{(5)(15x)}+2\sqrt{(5)(30)}\\\\=\sqrt{45x^2}+2\sqrt{90x}+\sqrt{75x}+2\sqrt{150}\\\\=\sqrt{9x^2\cdot5}+2\sqrt{9\cdot10x}+\sqrt{25\cdot3x}+2\sqrt{25\cdot6}\\\\=\sqrt{9x^2}\cdot\sqrt5+2\sqrt9\cdot\sqrt{10x}+\sqrt{25}\cdot\sqrt{3x}+2\sqrt{25}\cdot\sqrt6$
$=3x\sqrt5+2\cdot3\sqrt{10x}+5\sqrt{3x}+2\cdot5\sqrt6\\\\=3x\sqrt5+6\sqrt{10}+5\sqrt{3x}+10\sqrt6$

3 0

3 years ago

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consider a dianmeter of 8 feet and height of h feet.which equation can be used to find v,the volumeof the cylinder in cubic feet

yarga [219]

Answer:

diameter times height times x = volume

Step-by-step explanation:

using the length times width times height equation for volume.

good luck :)

4 0

3 years ago

Find the standard deviation of the following data. Round your answer to one decimal place. x 0 1 2 3 4 5 P(X

lbvjy [14]

Answer:

$\sigma = 1.8$

Step-by-step explanation:

Given

$\begin{array}{ccccccc}x & {0} & {1} & {2} & {3} & {4}& {5} \ \\ P(x) & {0.2} & {0.1} & {0.1} & {0.2} & {0.2}& {0.2} \ \end{array}$

Required

The standard deviation

First, calculate the expected value E(x)

$E(x) = \sum x * P(x)$

So, we have:

$E(x) = 0 * 0.2 + 1 * 0.1 + 2 * 0.1 + 3 * 0.2 + 4 * 0.2 + 5 * 0.2$

$E(x) = 2.7$

Next, calculate E(x^2)

$E(x^2) = \sum x^2 * P(x)$

So, we have:

$E(x^2) = 0^2 * 0.2 + 1^2 * 0.1 + 2^2 * 0.1 + 3^2 * 0.2 + 4^2 * 0.2 + 5^2 * 0.2$

$E(x^2) = 10.5$

The standard deviation is:

$\sigma = \sqrt{E(x^2) - (E(x))^2}$

$\sigma = \sqrt{10.5 - 2.7^2}$

$\sigma = \sqrt{10.5 - 7.29}$

$\sigma = \sqrt{3.21}$

$\sigma = 1.8$ --- approximated

5 0

3 years ago

How does the value of log Subscript 2 Baseline 100 compare with the value of Log Subscript 6 Baseline 20?

SCORPION-xisa [38]

Answer:

The value of $\log_2 100$ is about 4 times the value of $\log_6 20$

Explanation:

1. Change of base rule

$\log_ab=\dfrac{\log_{10}b}{\log_{10}a}=\dfrac{\log b}{\log a}$

2. Apply change of base rule to each of the given logarithm

$\log_2 100=\dfrac{\log 100}{\log 2}\approx 6.64$

$\log_6 20=\dfrac{\log 20}{\log 6}\approx 1.67$

3.Compare

$\dfrac{6.64}{1.67}\approx 3.97\approx 4$

4. Conclusion

The value $\log_2 100$ is about 4 times the value of $\log_6 20$

3 0

4 years ago

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On January 1, 2010, Joline deposits $1000 in a bank account that does not earn interest. For the next 12 months she deposits $25

hjlf

Answer:

$1000 + $25m and $1300 - 25m ;

1100 + 2.75m

2010 (4th month)

2011 ( 7th month)

Step-by-step explanation:

Joline's account :

Initial deposit = $1000

Monthly deposit = $25

Monthly withdrawal = $25

Number of months = 12

Let m = number of month at any point in time

Initial deposit + total deposit

Amount in Joline's account ; first 12 months

$1000 + $25m - - - (1)

Next 12 months :

$1000 + $25 *12 - 25m

1300 - 25m

Sofia's account :

Deposit amount = $1100

Interest per month = 0.25% = 0.0025

1100 + (1100 * 0.0025 * m)

1100 + 2.75m - - (2)

To find the month:

Equate equation 1 and 2, then find m

$1000 + $25m = 1100 + (1100*0.0025*m)

1000 + 25m = 1100 + 2.75m

25m - 2.75m = 1100 - 1000

22.25m = 100

m = 100 / 22.25

m = 4.494

In year 2010, they have approximately the same amount in the fourth month

1300 - 25m = 1100 + 2.75m

1300 - 1100 = 2.75m + 25m

200 = 27.75m

m = 200 / 27.75

m = 7.207

In 2011 ; they have approximately the se amount in the 7th month

4 0

3 years ago