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3 years ago

If F ( x ) = 2 x - 1, find the zero of F -1( x ). A) -1 B) 1/2 C) 1

Mathematics

1 answer:

tigry1 [53]3 years ago

7 0

Answer:

Step-by-step explanation:

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Find the lateral area and the surface area of the right cone. Round your nearest answers to the nearest hundredth

S_A_V [24]

Step-by-step explanation:

Given that,

The height of the cone, h = 24 cm

The radius of the cone, r = 10 cm

The slant height of the cone is :

$l=\sqrt{h^2+r^2} \\\\l=\sqrt{24^2+10^2} \\\\l=26\ cm$

The lateral area of the cone is given by :

$A=\pi rl\\\\=3.14\times 10\times 26\\\\=816.4\ cm^2\approx 816\ cm^2$

The surface area of the cone is given by :

$A=2\pi rh\\\\=2\times 3.14\times 10\times 24\\\\=1507.2 \approx 1507\ cm^2$

5 0

3 years ago

Lim x->pi/3 ((2cosx-1)/(tan^2x-3))

beks73 [17]

You can check that the limit comes in an undefined form:

$\displaystyle \lim_{x\to\frac{\pi}{3}} \frac{2\cos(x)-1}{\tan^2(x)-3}=\dfrac{0}{0}$

In these cases, we can use de l'Hospital rule, and evaluate the limit of the ratio of the derivatives. We have:

$\dfrac{\text{d}}{\text{d}x}2\cos(x)-1 = -2\sin(x)$

and

$\dfrac{\text{d}}{\text{d}x}\tan^2(x)-3 = 2\dfrac{\tan(x)}{\cos^2(x)}=\dfrac{2\sin(x)}{\cos^3(x)}$

So, we have

$\displaystyle \lim_{x\to\frac{\pi}{3}} \frac{2\cos(x)-1}{\tan^2(x)-3}=\lim_{x\to\frac{\pi}{3}} \dfrac{-2\sin(x)}{\frac{2\sin(x)}{\cos^3(x)}}=\lim_{x\to\frac{\pi}{3}}-\cos^3(x)=-\cos^3\left(\dfrac{\pi}{3}\right)$

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3 years ago

75/100 in its simplest form

Oxana [17]

3/4 is the answer for your question

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3 years ago

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The diagram shows the first four patterns of a sequence.

Ghella [55]

Answer:

$n {}^{2} + 3$

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3 years ago

Addmaths probability. I need helpp for question a and b thank youu

motikmotik

Answer:

sorry this is soo blur

please follow me and mark has brilliant please

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3 years ago