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4 years ago

Find the value of x. Please help

Mathematics

1 answer:

strojnjashka [21]4 years ago

3 0

In a circle, the measure of an inscribed angle is half the measure of the intercepted arc.
16x-10 = 67×2
16x-10 = 134
16x = 134+10
16x = 144
x = 144/16
x = 9

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Step-by-step explanation:

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$y = (2ax^2 + c)^2 (bx^2 - cx)^{-1}$

Product rule:

$y' = \bigg((2ax^2+c)^2\bigg)' (bx^2-cx)^{-1} + (2ax^2+c)^2 \bigg((bx^2-cx)^{-1}\bigg)'$

Chain and power rules:

$y' = 2(2ax^2+c)\bigg(2ax^2+c\bigg)' (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} \bigg(bx^2-cx\bigg)'$

Power rule:

$y' = 2(2ax^2+c)(4ax) (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} (2bx - c)$

Now simplify.

$y' = \dfrac{8ax (2ax^2+c)}{bx^2 - cx} - \dfrac{(2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}$

$y' = \dfrac{8ax (2ax^2+c) (bx^2 - cx) - (2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}$

$y = \dfrac{3bx + ac}{\sqrt{ax}}$

Quotient rule:

$y' = \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{\left(\sqrt{ax}\right)^2}$

$y'= \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{ax}$

Power rule:

$y' = \dfrac{3b \sqrt{ax} - (3bx+ac) \left(-\frac12 \sqrt a \, x^{-1/2}\right)}{ax}$

Now simplify.

$y' = \dfrac{3b \sqrt a \, x^{1/2} + \frac{\sqrt a}2 (3bx+ac) x^{-1/2}}{ax}$

$y' = \dfrac{6bx + 3bx+ac}{2\sqrt a\, x^{3/2}}$

$y' = \dfrac{9bx+ac}{2\sqrt a\, x^{3/2}}$

$y = \sin^2(ax+b)$

Chain rule:

$y' = 2 \sin(ax+b) \bigg(\sin(ax+b)\bigg)'$

$y' = 2 \sin(ax+b) \cos(ax+b) \bigg(ax+b\bigg)'$

$y' = 2a \sin(ax+b) \cos(ax+b)$

We can further simplify this to

$y' = a \sin(2(ax+b))$

using the double angle identity for sine.

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2 years ago

Factorise a)2x+8 b)3x-12 c)6x+4 d) 18x-9

uranmaximum [27]

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3 years ago

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kakasveta [241]

Answer:

Exact height = 8*sqrt(3) mm

Approximate height = 13.856 mm

=============================================================

Explanation:

If you do a vertical cross section of the cylinder, then the 3D shape will flatten into a rectangle as shown in the diagram below.

After flattening the picture, I've added the points A through F

point A is the center of the sphere and cylinder
points B through E are the corner points where the cylinder touches the sphere
point F is at the same horizontal level as point A, and it's on the edge of the cylinder.

Those point labels will help solve the problem. We're told that the radius of the sphere is 8 mm. So that means segment AD = 8 mm.

Also, we know that FA = 4 mm because this is the radius of the cylinder.

Focus on triangle AFD. We need to find the height x (aka segment FD) of this triangle so we can then double it later to find the height of the cylinder. This in turn will determine the height of the bead.

------------------------------

As the hint suggests, we'll use the pythagorean theorem

a^2 + b^2 = c^2

b = sqrt(c^2 - a^2)

x = sqrt(8^2 - 4^2)

x = sqrt(48)

x = sqrt(16*3)

x = sqrt(16)*sqrt(3)

x = 4*sqrt(3)

This is the distance from D to F

The distance from D to E is twice that value, so DE = 2*(FD) = 2*4*sqrt(3) = 8*sqrt(3) is the exact height of the bead (since it's the exact height of the cylinder).

Side note: 8*sqrt(3) = 13.856 approximately.

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3 years ago