Pleasee i need help very bad

Mathematics

1 answer:

Anon25 [30]2 years ago

3 0

Angles UST and TSU are equivalent.

The answer is TSU

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Work out the surface area of a sphere with the diameter of 13cm

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Answer:

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Step-by-step explanation:

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2 years ago

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The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is ...?

Lina20 [59]

$\frac{d}{dx}(y^2 + (xy+1)^3 = 0)
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\\\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0
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\\\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0
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\\\frac{dy}{dx}2y + \left(\frac{dx}{dx}y+x\frac{dy}{dx}+\frac{dx}{dx}0\right)\ * 3(xy+1)^2 = 0
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\\\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0$
$\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
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\\2y\ y' + \left(3y+3x\ y'\right) (xy+1)^2 = 0
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\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
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\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
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\\2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0
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\\3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0
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\\y'(3x +6x^2y+3x^2y+2y ) = -3x^2y^3 -6xy^2-3y

$
$y' = \frac{-3x^2y^3 -6xy^2-3y}{(3x +6x^2y+3x^2y+2y )};\ x=2,y=-1
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\\y' = \frac{-3(2)^2(-1)^3 -6(2)(-1)^2-3(-1)}{(3(2) +6(2)^2(-1)+3(2)^2(-1)+2(-1) )}
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2 years ago

Suppose that w and t vary inversely and that t = 5/12

Svetlanka [38]

Answer:

Option B. t=5/3w ;1/3

Step-by-step explanation:

We are told that $w$ varies Inversely with $t$ thus generally speaking $w$ ∝ $\frac{1}{t}$ since they are inverted, <u>otherwise</u> if they were proportionally varied then $w$ ∝ $t$ . This means that there is a constant value ( $a$ ) for which $w$ is inversely proportional to $t$ and can be mathematically expressed as:

$w=\frac{a}{t}$ Eqn. (1)

Now since we are given the values of $w=4$ and $t=\frac{5}{12}$ , we can plug them in Eqn. (1) and find our constant of proportionality $a$ as follow: