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3 years ago

8

Six times the square of a number x plus 11 times the number equals 2. What are possible values of x?

1 answer:

Bas_tet [7]3 years ago

7 0

Answer:

-2, 1/6

Step-by-step explanation:

6x²+11x=2

6x²+11x-2=0

Delta=11²-4×6×(-2)=121+48=169

x1,2=(-11 +- sqrt(169))/(2×6)=

(-11 +- 13)/12

x1=(-11+13)/12=2/12=1/6

x2=(-11-13)/12=-24/12=-2

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What is the decimal equivalent of 17/20?<br> A. 0.085<br> B. 0.117<br> C. 0.85<br> D. 8.5

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Read 2 more answers

14/2 three equivalent ratios

Anna71 [15]

Answer:

Three equivalent ratios of $\frac{14}{2}$ are $\frac{7}{1}$ , $\frac{21}{3}$ and $\frac{28}{8}$ .

Step-by-step explanation:

We are given a fraction $\frac{14}{2}$ .

We need to find the three equivalent ratios/fractions of $\frac{14}{2}$ .

The common factor of 14 and 2 is 2.

So, dividing top and bottom by 2, we get

14÷2 = 7 and 2÷2 that is $\frac{7}{1}$ .

Multiplying $\frac{7}{1}$ fraction by a common number 3 in top and bottom, we get

7 × 3 = 21

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So, we get another equivalent fraction $\frac{21}{6}$ .

Multiplying $\frac{7}{1}$ fraction by a common number 3 in top and bottom, we get

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3 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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3 years ago