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NNADVOKAT [17]
3 years ago
8

Simplify root 2 divided by root 5 + root 3 by Rationalising in the denominator​

Mathematics
1 answer:
Inessa05 [86]3 years ago
7 0

Answer:

root4 by 2

Step-by-step explanation:

multiply the whole no with root5-root3 up and down also

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Refer to Table 4-4. If these are the only four sellers in the market, then the market quantity supplied at a price of $4 is
NeTakaya

All things being equal; demand decreases, as price increases.

The quantity supplied when price is $4 is 28

From table 4.4 (see attachment).

When price = $4, we have the following supplies:

  • <em>Firm A = 6</em>
  • <em>Firm B = 6</em>
  • <em>Firm C = 8</em>
  • <em>Firm D = 10</em>

<em />

So, the total supply at $4 is:

Total = 4 + 6 + 8 + 10

Total = 28

Hence, the quantity supplied when price is $4 is 28

Read more about demand and supply at:

brainly.com/question/13353440

5 0
2 years ago
How do you multiply mixed fraction like this one?<br><br> 2 3/4 x 3 1/2<br> Somoene help me please
In-s [12.5K]

Answer:

Hope it will help you a lot.

7 0
2 years ago
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Of the people who fished at Clearwater Park today,
Vladimir79 [104]
The probability is 2/25 or an 8% chance. By taking the probability of the fishers from Clearwater (56/70) and multiplying that by the probability of the non fishers from Mountain view (8/80) we can find the total probability of the situation. 
8 0
4 years ago
The number of texts per day by students in a class is normally distributed with a 
kobusy [5.1K]

Answer:

1, 2, 6

Step-by-step explanation:

The z score shows by how many standard deviations the raw score is above or below the mean. The z score is given by:

z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean, \ \sigma=standard\ deviation

Given that mean (μ) = 130 texts, standard deviation (σ) = 20 texts

1) For x < 90:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{90-130}{20} =-2

From the normal distribution table, P(x < 90) = P(z < -2) = 0.0228 = 2.28%

Option 1 is correct

2) For x > 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0

From the normal distribution table, P(x > 130) = P(z > 0) = 1 - P(z < 0) = 1 - 0.5 = 50%

Option 2 is correct

3) For x > 190:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{190-130}{20} =3

From the normal distribution table, P(x > 3) = P(z > 3) = 1 - P(z < 3) = 1 - 0.9987 = 0.0013 = 0.13%

Option 3 is incorrect

4)  For x < 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0

For x > 100:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{100-130}{20} =-1.5

From the normal table, P(100 < x < 130) = P(-1.5 < z < 0) = P(z < 0) - P(z < 1.5) = 0.5 - 0.0668 = 0.9332 = 93.32%

Option 4 is incorrect

5)  For x = 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0

Option 5 is incorrect

6)  For x = 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{160-130}{20} =1.5

Since 1.5 is between 1 and 2, option 6 is correct

5 0
3 years ago
Factor both quadratic expressions. (x 4 + 5x 2 - 36)(2x 2 + 9x - 5) = 0
Sonbull [250]

So focusing on x^4 + 5x^2 - 36, we will be completing the square. Firstly, what two terms have a product of -36x^4 and a sum of 5x^2? That would be 9x^2 and -4x^2. Replace 5x^2 with 9x^2 - 4x^2: x^4+9x^2-4x^2-36

Next, factor x^4 + 9x^2 and -4x^2 - 36 separately. Make sure that they have the same quantity inside of the parentheses: x^2(x^2+9)-4(x^2+9)

Now you can rewrite this as (x^2-4)(x^2+9) , however this is not completely factored. With (x^2 - 4), we are using the difference of squares, which is a^2-b^2=(a+b)(a-b) . Applying that here, we have (x+2)(x-2)(x^2+9) . x^4 + 5x^2 - 36 is completely factored.

Next, focusing now on 2x^2 + 9x - 5, we will also be completing the square. What two terms have a product of -10x^2 and a sum of 9x? That would be 10x and -x. Replace 9x with 10x - x: 2x^2+10x-x-5

Next, factor 2x^2 + 10x and -x - 5 separately. Make sure that they have the same quantity on the inside: 2x(x+5)-1(x+5)

Now you can rewrite the equation as (2x-1)(x+5) . 2x^2 + 9x - 5 is completely factored.

<h3><u>Putting it all together, your factored expression is (x+2)(x-2)(x^2+9)(2x-1)(x+5)=0</u></h3>
5 0
3 years ago
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