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3 years ago

Let u = ln x and v= In y. Write In (x^2y^5) in terms of u and y.

Mathematics

1 answer:

Oxana [17]3 years ago

7 0

Answer:

Step-by-step explanation:

u=㏑ x

x=e^u

v=㏑ y

y=e^v $x^{2} y^5 =(e^u)^{2} (e^v)^5=e^{2u}e^{5v}=e^{2u+5v}$

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3 years ago

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3 years ago

Jason consumes 1500 calories and needs to

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Answer:

100*150=15000 calories

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3 years ago

100% of what number is 61

Dominik [7]

" 100% " means "the whole thing",
and "is" can be replaced by "is equal to".

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or
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3 0

3 years ago

Read 2 more answers

Determine the equation of the circle graphed below. ( please help me )

Brums [2.3K]

Answer:

$Equation : \\(x+3)^2 + ( y -2)^2 = 36$

Step-by-step explanation:

For standard form the circle's equation we need the centre of the circle and the radius.

Step 1: Find the centre

If the centre is not given find the end points of the diameter

and then find the mid point.

Let the end points of the diameter be : ( - 3 , 8 ) and ( -3 , -4 )

The mid-point of the diameter is :

$Mid-point = (\frac{-3 + - 3}{2}, \frac{-4+8}{2}) = (-3, 2)$

Therefore, centre of the circle = ( -3 , 2 )

Step 2 : Find radius

$Radius = \frac{Diameter }{2}$

Diameter is the distance between the end points ( -3 , 8) and ( -3 , -4 )

That is ,

$Diameter = \sqrt{(-3-(-3))^2 + ( -4 -8)^2}\\$

$= \sqrt{(-3 + 3)^2 + (-12)^2}\\\\=\sqrt{0 + 144}\\\\=12$

Therefore ,

$Radius = \frac{12}{2} = 6$

Step 3 : Equation of the circle

Standard equation of the circle with centre ( h ,k )

and radius ,r is :

$(x - h)^2+(y -k)^2 = r^2$

Therefore, the equation of the circle with centre ( -3, 2)

and radius = 6 is :

$(x - (-3))^2 + (y - 2)^2 = 6^2\\\\(x + 3)^2 + (y - 2)^2 = 36$

5 0

3 years ago