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3 years ago

Point Mis the midpoint of AB. AM = 3x + 3, and AB= 83 – 6.

Mathematics

2 answers:

Naddik [55]3 years ago

6 0

Answer:

$\Huge \boxed{21}$

Step-by-step explanation:

AM = 3x + 3

AB = 8x - 6

Point M is the midpoint of AB.

So, AM = AB/2

3x + 3 = (8x - 6)/2

Multiplying both sides by 2.

2(3x + 3) = 8x - 6

Expanding brackets.

6x + 6 = 8x - 6

Subtracting 6x from both sides.

6 = 2x - 6

Adding 6 to both sides.

12 = 2x

Dividing both sides by 2.

6 = x

Let x = 6 for the length of AM.

3(6) + 3

18 + 3

Ber [7]3 years ago

4 0

Answer:

AM= Half of AB

or, 3x+3=(8x-6)/2

or, 6x+6=8x-6

or, 2x=12

Therefore,x=6

so,AM=3*6+3=21

So the units is 21

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3 years ago

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2 years ago

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Paha777 [63]

Answer:

Step-by-step explanation:

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2 years ago

A right circular cylinder is inscribed in a sphere with diameter 4cm as shown. If the cylinder is open at both ends, find the la

SOVA2 [1]

Answer:

$8\pi\text{ square cm}$

Step-by-step explanation:

Since, we know that,

The surface area of a cylinder having both ends in both sides,

$S=2\pi rh$

Where,

r = radius,

h = height,

Given,

Diameter of the sphere = 4 cm,

So, by using Pythagoras theorem,

$4^2 = (2r)^2 + h^2$ ( see in the below diagram ),

$16 = 4r^2 + h^2$

$16 - 4r^2 = h^2$

$\implies h=\sqrt{16-4r^2}$

Thus, the surface area of the cylinder,

$S=2\pi r(\sqrt{16-4r^2})$

Differentiating with respect to r,

$\frac{dS}{dr}=2\pi(r\times \frac{1}{2\sqrt{16-4r^2}}\times -8r + \sqrt{16-4r^2})$

$=2\pi(\frac{-4r^2+16-4r^2}{\sqrt{16-4r^2}})$

$=2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})$

Again differentiating with respect to r,

$\frac{d^2S}{dt^2}=2\pi(\frac{\sqrt{16-4r^2}\times -16r + (-8r^2+16)\times \frac{1}{2\sqrt{16-4r^2}}\times -8r}{16-4r^2})$

For maximum or minimum,

$\frac{dS}{dt}=0$

$2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})=0$

$-8r^2 + 16 = 0$

$8r^2 = 16$

$r^2 = 2$

$\implies r = \sqrt{2}$

Since, for r = √2,

$\frac{d^2S}{dt^2}=negative$

Hence, the surface area is maximum if r = √2,

And, maximum surface area,

$S = 2\pi (\sqrt{2})(\sqrt{16-8})$

$=2\pi (\sqrt{2})(\sqrt{8})$

$=2\pi \sqrt{16}$

$=8\pi\text{ square cm}$

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