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3 years ago

Please help! really appreciate it

Mathematics

1 answer:

Rama09 [41]3 years ago

3 0

rugby 7 has 3 out of 5 sold out = 3/5 = 0.6

Junior Athletics has 3 out of 5 sold out = 3/5 = 0.6

Volley ball has 2 out of 5 sold out = 2/5 = 0.4

Volleyball is the answer for the first part

Part 2 = 3/5 x 3/5 =9/25

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vichka [17]

Answer:

The answer is (3,2) or x=3 and y=2

Step-by-step explanation:

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3 years ago

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What decimal is equivalent to 41 ?<br> 100

BlackZzzverrR [31]

Answer:

In the fraction 41/100, 41 is the numerator and 100 is the denominator, the fraction bar means "divided by". So to convert 41/100 as decimal number we simply divide the numerator by denominator.

When we calculated 41 divided by 100, we found that 41/100 in decimal form is:

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Step-by-step explanation:

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How do I differentiate this?

disa [49]

Answer:

$\frac{8}{\left(-x+2\right)^2}$ & $x = 1, x = 3$

Step-by-step explanation:

I'm sure you are familiar with the product rule,

If y = u*v => dy/dx = u * dv/dx + v * dy/dx <----- product rule

In this case:

$u=3x+2,\:v=\left(2-x\right)^{-1},\\=>\frac{d}{dx}\left(3x+2\right)\left(2-x\right)^{-1}+\frac{d}{dx}\left(\left(2-x\right)^{-1}\right)\left(3x+2\right)$

Now remember the sum rule:

$\frac{d}{dx}\left(3x+2\right) = \frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(2\right),\\\frac{d}{dx}\left(3x\right) = 3,\\\frac{d}{dx}\left(2\right) = 0\\\frac{d}{dx}\left(3x+2\right) = 3$

For this second bit we apply the chain rule:

$\frac{d}{dx}\left(\left(2-x\right)^{-1}\right) = -\frac{1}{\left(2-x\right)^2}\frac{d}{dx}\left(2-x\right),\\\frac{d}{dx}\left(2-x\right) = -1,\\\\=> -\frac{1}{\left(2-x\right)^2}\left(-1\right)\\=> \frac{1}{\left(2-x\right)^2}$

If we substitute these values back into the expression... $\frac{d}{dx}\left(3x+2\right)\left(2-x\right)^{-1}+\frac{d}{dx}\left(\left(2-x\right)^{-1}\right)\left(3x+2\right)$

...we get the following:

$3\left(2-x\right)^{-1}+\frac{1}{\left(2-x\right)^2}\left(3x+2\right)$

The rest is just pure simplification:

$3\left(2-x\right)^{-1}+\frac{1}{\left(2-x\right)^2}\left(3x+2\right)\\= \frac{3}{-x+2}+\frac{3x+2}{\left(-x+2\right)^2}\\= \frac{3\left(-x+2\right)}{\left(-x+2\right)^2}+\frac{3x+2}{\left(-x+2\right)^2}\\\\= \frac{3\left(-x+2\right)+3x+2}{\left(-x+2\right)^2}\\\\= \frac{8}{\left(-x+2\right)^2}$

Now let's equate this to equal 8 for the second bit and solve for x:

$\frac{8}{\left(-x+2\right)^2}=8,\\\frac{8}{\left(-x+2\right)^2}\left(-x+2\right)^2=8\left(-x+2\right)^2,\\8=8\left(-x+2\right)^2,\\\left(-x+2\right)^2=1,\\x = 1, x = 3$