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2 years ago

4. (a)(i)Show that log4x=2log16x. (ii)Show that log x=3logb³ x. (iii) Show that log₂x=(1+log₂3)logix.

Mathematics

1 answer:

alexdok [17]2 years ago

3 0

that log4x=2log16x. (ii)Show that log x=3logb³ x. (iii) Show that log₂x=(1+log₂3)logix

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Find dy/dx for 4 - xy = y^3

storchak [24]

Answer:

$\frac{dy}{dx}=-\frac{y}{3y^2+x}$

Step-by-step explanation:

4-xy=y^3

dy/dx=?

$\frac{d(4-xy)}{dx}=\frac{d(y^3)}{dx}\\ \frac{d(4)}{dx}-\frac{d(xy)}{dx}=3y^{3-1}\frac{dy}{dx}\\ 0-(\frac{dx}{dx}y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -(1y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -(y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -y-x\frac{dy}{dx}=3y^2\frac{dy}{dx}$

Solving for dy/dx: Addind x dy/dx both sides of the equation:

$-y-x\frac{dy}{dx}+x\frac{dy}{dx}=3y^2\frac{dy}{dx}+x\frac{dy}{dx} \\ -y=3y^2\frac{dy}{dx}+x\frac{dy}{dx}$

Common factor dy/dx on the right side of the equation:

$-y=(3y^2+x)\frac{dy}{dx}$

Dividing both sides of the equation by 3y^2+x:

$\frac{-y}{3y^2+x}=\frac{(3y^2+x)}{3y^2+x}\frac{dy}{dx}\\ -\frac{y}{3y^2+x}=\frac{dy}{dx}\\ \frac{dy}{dx}=-\frac{y}{3y^2+x}$

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3 years ago

Prove <4 and <1 are supplementary.

nignag [31]

Answer:

the = 180

Step-by-step explanation:

8 0

3 years ago

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Which of these choices show a pair of equivalent expressions? A.7^5/7 and (√ 7)^5 B.6^7/2 and (√ 6)^7 C.(4√ 81)^7 and 81^7/4 D.5

OverLord2011 [107]

We will use the following law of indices (or 'index law') to check each pair of expression

$x^{ \frac{m}{n}} = ( \sqrt[n]{x} )^{m}$

With fractional power, the denominator is the root and the numerator is the power of the term. When the denominator is 2, we usually only write the normal square symbol (√). Denominator other than 2, we usually write the value of the root, for example, the cubic root ∛

Option A - Incorrect
$7^{ \frac{5}{7}}$ should equal to $( \sqrt[7]{7} )^{5}$

Option B - Correct
$6^{ \frac{7}{2} }$ does equal to $( \sqrt{6}) ^{7}$

Option C - Incorrect
$4( \sqrt{81} )^{7}$ should equal to $(4)( 81^{ \frac{7}{2} } )$

Option D - Incorrect
$5^{ \frac{2}{3} }$ should equal to $( \sqrt[3]{5} )^{2}$

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3 years ago

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What are the coordinates of this point?<br><br><br> PLLEASS ANSWER IM GVING BRAINLIEST

mafiozo [28]

Answer:

Step-by-step explanation:

they are (-5,-4) hope this helped

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3 years ago

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The length of the rectangle below is (2x-3) and the width is (x+7). Find the area of the rectangle in terms of x.

Alex777 [14]

Answer:

area = 2x² - 17x - 21

Step-by-step explanation:

Area of the rectangle is:

area = (width*length)

area = (x+7)*(2x-3)

= x*2x + x*-3 + 7*2x + 7*-3

= 2x² - 3x - 14x - 21

= 2x² - 17x - 21

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3 years ago

4. (a)(i)Show that log4x=2log16x. (ii)Show that log x=3logb³ x. (iii) Show that log₂x=(1+log₂3)logix.​

4. (a)(i)Show that log4x=2log16x. (ii)Show that log x=3logb³ x. (iii) Show that log₂x=(1+log₂3)logix.