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Brilliant_brown [7]
3 years ago
8

A bakery offers a small circular cake with a diameter of 8 inches. It also offers a large circular cake with a diameter of 24 in

ches. Complete the explanation for whether the top of the large cake has three times the area of that of the small cake.
Mathematics
2 answers:
drek231 [11]3 years ago
8 0
Large cake:
24÷2=12
12 is the radius of the large cake.
12×12×3.14=452.16

Small cake:
8÷2=4
4 is the radius of the small cake.
4×4×3.14=50.24

452.16÷3=150.72
So, The area of the large cake is not three times the area of the small cake.
Elanso [62]3 years ago
7 0
We must find the area of each cake's top.

Formula for area of a circle:

A =  \pi  r^{2} where r is the radius.

<span>small cake:
</span>Plug in 4 for r because the radius is 4. (They give us the diameter, which is 8, and the radius is half that)

A =  \pi  4^{2}

A = 16π

<span>big cake:
</span>Repeat the process: 
Plug in 12 for r because the radius is 12. (They give us the diameter, which is 24, and the radius is half that)

A = \pi 12^{2}

A = 144π

So our two radii are 144π and 16π. 
The large cake's top is not 3 times the area of the small cake's.

This makes sense because you are squaring the radius, which makes the fact that the larger cake's diameter is triple the smaller cake diameter irrelevant. 

Hope this helped! ^-^
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triumphant college is planning to increase the registration fee by 11.5%.if the registration fee is currently N$ 1000, how much
frozen [14]

Current registration fee at triumphant college = N$ 1000

Percentage by which the registration fee will be increased next year = 11.5%

Registration fee at triumphant college next year :

= Current registration fee + 11.5% of current registration fee

=\tt1000 +  \frac{115}{10 \times 100}  \:  \: of \:  \: 1000

= \tt1000 +  \frac{115}{1000}  \times 1000

=\tt 1000 +  \frac{115000}{1000}

=\tt 1000 + 115

\color{plum} = \tt \:\bold{ N\$ \: 1115}

Therefore, the registration fee at triumphant college next year = <u>N$ 1115</u>

5 0
3 years ago
Water is being pumped into a conical tank that is 8 feet tall and has a diameter of 10 feet. If the water is being pumped in at
Deffense [45]

The rate of change of the depth of water in the tank when the tank is half

filled can be found using chain rule of differentiation.

When the tank is half filled, the depth of the water is changing at  <u>1.213 × </u>

<u>10⁻² ft.³/hour</u>.

Reasons:

The given parameter are;

Height of the conical tank, h = 8 feet

Diameter of the conical tank, d = 10 feet

Rate at which water is being pumped into the tank, = 3/5 ft.³/hr.

Required:

The rate at which the depth of the water in the tank is changing when the

tank is half full.

Solution:

The radius of the tank, r = d ÷ 2

∴ r = 10 ft. ÷ 2 = 5 ft.

Using similar triangles, we have;

\dfrac{r}{h} = \dfrac{5}{8}

The volume of the tank is therefore;

V = \mathbf{\dfrac{1}{3} \cdot \pi \cdot r^2 \cdot h}

r = \dfrac{5}{8} \times h

Therefore;

V = \dfrac{1}{3} \cdot \pi \cdot \left(  \dfrac{5}{8} \times h\right)^2 \cdot h = \dfrac{25 \cdot h^3 \cdot \pi}{192}

By chain rule of differentiation, we have;

\dfrac{dV}{dt} = \mathbf{\dfrac{dV}{dh} \cdot \dfrac{dh}{dt}}

\dfrac{dV}{dh}=\dfrac{d}{h} \left(  \dfrac{25 \cdot h^3 \cdot \pi}{192} \right) = \mathbf{\dfrac{25 \cdot h^2 \cdot \pi}{64}}

\dfrac{dV}{dt} = \dfrac{3}{5}  \ ft.^3/hour

Which gives;

\dfrac{3}{5} =  \mathbf{\dfrac{25 \cdot h^2 \cdot \pi}{64} \times \dfrac{dh}{dt}}

When the tank is half filled, we have;

V_{1/2} = \dfrac{1}{2} \times  \dfrac{1}{3} \times \pi \times 5^2 \times 8 =\mathbf{ \dfrac{25 \cdot h^3 \cdot \pi}{ 192}}

Solving gives;

h³ = 256

h = ∛256

\dfrac{3}{5} \times \dfrac{64}{25 \cdot h^2 \cdot \pi} = \dfrac{dh}{dt}

Which gives;

\dfrac{dh}{dt} = \dfrac{3}{5} \times \dfrac{64}{25 \cdot (\sqrt[3]{256}) ^2 \cdot \pi} \approx \mathbf{1.213\times 10^{-2}}

When the tank is half filled, the depth of the water is changing at  <u>1.213 × 10⁻² ft.³/hour</u>.

Learn more here:

brainly.com/question/9168560

6 0
3 years ago
3/4(8x + 20) - 4x &gt; -11
den301095 [7]

In this case, we'll have to carry out several steps to find the solution.

Step 01:

Data

3/4(8x + 20) - 4x > -11​

x ==> ?

Step 02:

Inequality :

\begin{gathered} \frac{3}{4}\cdot(8x\text{ + 20) - 4x > -11} \\ 6x\text{ + }15\text{ -4x > -11} \end{gathered}

2x + 15 > -11

2x + 15 - 15 > -11 - 15

2x > - 26

2x / 2 > -26 / 2

x > - 13

The answer is:

x > -13

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1 year ago
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Stolb23 [73]

Answer:

Step-by-step explanation:

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6 0
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