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KiRa [710]
3 years ago
9

A given line has the equation 10x + 2y = −2.

Mathematics
2 answers:
Cerrena [4.2K]3 years ago
6 0
I think is
y = ( ______)x + 12

y = -5x + 12
sergij07 [2.7K]3 years ago
5 0
Firstly, let's convert 10x + 2y = -2 to slope-intercept form (y = mx + b):

10x + 2y = -2
2y = -10x - 2
y = -5x - 2/5

The slope is -5. Parallel lines have the same slope; the question provides the y-int (12), so our equation is:

y = -5x + 12

Hope this helps.
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Find the value of x to the nearest tenth.
vladimir1956 [14]

Answer:

The answer to your question is  x = 14.7  

Step-by-step explanation:

Data

∠A = 20°

∠B = 46

a = 7

b = x

Process

To solve this problem use, the law of sines. This law states that the ratio of a side of a triangle to the sine of the opposite angle is the same for all three sides.

The law of sines for this problem is

                   x / sin 46 = 7 / sin 20

-Solve for x

                          x = 7 sin 46 / sin 20

-Simplification

                          x = 7 (0.719) / 0.342

                          x = 5.035/0.342

-Result

                         x = 14.7  

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4 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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Answer:

Step-by-step explanation:

let x be the length of third side.

10-5<x<10+5

or 5<x<15

so third side is between 5 and 15 .

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3 years ago
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