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Lorico [155]
3 years ago
14

When baking a cake you have a choice of the following pans: a round cake that is 2 inches deep and has a 7-inch diameter, a 6 in

ches times 9-inch rectangular cake pan that is 2 inches deep.
Question: Excluding the bottom of the cake what is the surface area of each cake that will be covered by frosting?
I do not need an answer I just need help on this question. Thank you.
Mathematics
1 answer:
irga5000 [103]3 years ago
5 0
The answer is
round cake - 82.42 in²
rectangular cake - 114 in²

Round cake:
d = 7 in
r = d/2 = 7 in / 2 = 3.5 in
h = 2 in

The surface are of a cylinder is:
A = 2πr² + 2πrh

The surface are of the round cake (which is actually a cylindrical cake) excluding the bottom is:
A = 2πr² + 2πrh - πr²
A = πr² + 2πrh
A = 3.14 * 3.5² + 2 * 3.14 * 3.5 * 2
    = 38.46 + 43.96
    = 82.42 in²

Rectangular cake:
w = 6 in
l = 9 in
h = 2 in

The surface are of a rectangle is:
A = 2wl + 2wh + 2lh

The surface are of the rectangular cake excluding the bottom is:
A = 2wl + 2wh + 2lh - wl
A = wl + 2wh + 2lh
A = 6 * 9 + 2 * 6 * 2 + 2 * 9 * 2
    = 54 + 24 + 36
    = 114 in²

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Answer:

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Step-by-step explanation:

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3 years ago
Solve the equation. Round to the nearest hundredth. Show work.
Burka [1]

Answer:

x=-0.75

Step-by-step explanation:

The given equation is

4^{-5x-7}=6^{2x-1}

We take logarithm of both sides to base 10.

\log(4^{-5x-7})=\log(6^{2x-1})

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We expand the brackets to get;

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Group similar terms;

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-7\log(4)+\log(6)=(2\log(6)+5\log(4))x

\frac{-7\log(4)+\log(6)}{(2\log(6)+5\log(4))}=x

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To the nearest hundredth.

x=-0.75

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answer: 1/6
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