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2 years ago

Differentiate f and find the domain of f. f(x) = x/[1-ln(x-1)]

Mathematics

1 answer:

photoshop1234 [79]2 years ago

6 0

F(x) = [2x - 1 - (x-1) in (x - 1)] / [ 1 - in(x-1)]^2

domain of f = (1, 1+e) U (1 + e, infinty)

hope this helps

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The area of two similar triangles are 50dm^2 and 32dm^2. The sum of their perimeters is 117dm. What us the perimeters of each of

Kay [80]

Answer:

Perimeter of one triangle is 65 dm

Perimeter of other triangle is 52 dm

Step-by-step explanation:

Please remember the concept

If sides are in the ratio of a:b

Then the area in the ratio of $a^{2} :b^{2}$

It is given sum of their perimeter is 117.

Let the small triangle has perimeter as x.

So, perimeter of big triangle is 117-x.

So, we can set up equation as

$\frac{50}{32} =\frac{x^{2} }{(117-x)^{2} }$

Cross multiply

50(117-x)^2 =32x^2

Expand the left side

50 $(13689 -234x+x^{2} )$ = $32x^{2}$

Distribute the left side

684450-11700x+ $50x^{2}$ = $32x^{2}$

Subtract both sides $32x^{2}$ and rewrite it $18x^{2} -11700x+684450=0$

Solve this quadratic for x.

Divide both sides of the equation by 18 to simplify.

$x^{2}$ -650 x+38025=0

Now, if possible let's factor

Find two integers whose multiplication is 38025 but adds to -650.

-65 and -585 works.

So, we can rewrite it as

(x -65)(x-585) =0

Solve them using zero product property

x=65, x=585

So, x=65 works here.

So, perimeter of one triangle is 65 dm

Perimeter of other triangle is 117-65= 52 dm

5 0

2 years ago

-4x+3x=2<br> Solve linear equation

Sergeu [11.5K]

Answer:
X = -2
Suerte :)

4 0

2 years ago

the perimeter of the floor of a room is 18 metre and its height is 3 cm what is the area of 4 walls of the room

Lina20 [59]

Note: The height of the room must be 3 m instead of 3 cm because 3 cm is too small and it cannot be the height of a room.

Given:

Perimeter of the floor of a room = 18 metre

Height of the room = 3 metre

To find:

The area of 4 walls of the room.

Solution:

We know that, the area of 4 walls of the room is the curved surface area of the cuboid room.

The curved surface area of the cuboid is

$C.S.A.=2h(l+b)$

Where, h is height, l is length and b is breadth.

Perimeter of the rectangular base is 2(l+b). So,

$C.S.A.=\text{Perimeter of the base} \times h$

Putting the given values, we get

$C.S.A.=18\times 3$

$C.S.A.=54$

Therefore, the area of 4 walls of the room is 54 sq. metres.

7 0

3 years ago

Which polynomial is a product of (3x+2) and (x-7)

pogonyaev

(3x+2) (x-7)

FOIL

first 3x*x = 3x^2

outer 3x*-7 = -21x

inner = 2*x = 2x

last 2*-7 = -14

add them together = 3x^2 -21x+2x -14

combine like terms 3x^2 -19x -14

5 0

2 years ago

Who is correct science or religious

kati45 [8]

Answer:

it depends on what you believe, butttt i would say science

3 0

2 years ago

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