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Ksivusya [100]
3 years ago
10

John and 3 friends are going out for pizza for lunch. They split one pizza and 4 large drinks. The pizza cost $13.00. After usin

g a $6.00 gift certificate, they spend a total of $13.20. Write an equation to model this situation, and find the cost of one drink
Mathematics
2 answers:
Oduvanchick [21]3 years ago
7 0

Answer:

4d +$13.00 - $6.00 = $13.20; $1.55

Step-by-step explanation:

S_A_V [24]3 years ago
3 0
A single drink would cost be $1.55. This is because:
13 + 4x - 6 = 13.20.
(13.20 - 13 + 6) = 6.20.
6.20 / 4 = 1.55

Hope this helped.
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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

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h = height,

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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

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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
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Answer:

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Step-by-step explanation:

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