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3 years ago

Find a pair of square numbers which give a difference of 39

Mathematics

1 answer:

nadya68 [22]3 years ago

5 0

64-25=39 All the square numbers are 1,4,9,16,25,36,49,64,81,100

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Calculate the percentage increase or decrease in each of

horrorfan [7]

Step-by-step explanation:

hey nidha the answer of ur question

5 0

3 years ago

1. Solve for y 4x + 2y = 6

xenn [34]

Answer:

y = -2x + 3

Step-by-step explanation:

4x + 2y = 6

⇔ 2y = $\frac{6 - 4x}{2} = \frac{6}{2} - \frac{4x}{2} = 3 - 2x = -2x + 3$

5 0

3 years ago

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:

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

4 0

3 years ago

X^2 + 7x – 18 = 0 a {–2, 9} b {–9, 2} c {2, 9} d {–2, –9}

Sedaia [141]

Answer:

b)-9,2

Step-by-step explanation:

x^2+7x-18=0

x^2+9x-2x-18=0

x(x+9)-2(x+9)=0

(x+9)(x-2)=0

therefore

x=-9

x=2

6 0

3 years ago

What is the precent of 2 is 8?

Vlad [161]

2 * x/100 = 8

x/100 = 4

x = 400

400%

3 0

3 years ago