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Ivahew [28]
2 years ago
14

Jacobs sand bucket has a top diameter of 10 inches and a bottom diameter of 7 inches. the bucket is 12 inches tall. What’s the v

olume
Mathematics
1 answer:
Elanso [62]2 years ago
4 0
We know that
The volume of a circular truncated cone=(1/3)*pi*[r1²+r1*r2+r2²]*h
where
r1=7/2-----> 3.5 in
r2=10/2----> 5 in
h=12 in
volume of a circular truncated cone=(1/3)*pi*[3.5²+3.5*5+5²]*12
volume=688 in³

the answer is
688 in³

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Diddy Corp. Stock has a beta of 1.2, the current risk-free rate is 6 percent, and the expected return on the market is 14.50 per
ryzh [129]

Answer: 16.2%

Step-by-step explanation:

You can find the cost of equity using the Capital Asset Pricing Model (CAPM).

Cost of equity = Risk free rate + Beta * (Expected return on market - Risk free rate)

= 6% + 1.2 * (14.50 - 6%)

= 6% + 10.2%

= 16.2%

8 0
2 years ago
Dee's credit card has an APR of 17% calculated on the previous monthly balance, and Dee makes a payment of $50 every month. Her
Bess [88]

Where is the table..? You didn't provide any pictures..

8 0
2 years ago
Read 2 more answers
Phil has 3 times as many rocks as Peter. Together, they have 48 rocks. How many more rocks does Phil have than Peter?
Ivan
Let x = the number of marbles for Phil
Then 3x = the number of marbles for Peter
Together they have 3x+x=48 or 4x=48 and x=12
So Phil has x = 12 marbles and Peter has 3x, let x = 12, 3(12) = 36 marbles.
Peter has 24 more marbles than Phil.
3 0
2 years ago
Read 2 more answers
Find the value of cosAcos2Acos3A...........cos998Acos999A where A=2π/1999
Lady bird [3.3K]
Hello,

Here is the demonstration in the book Person Guide to Mathematic by Khattar Dinesh.

Let's assume
P=cos(a)*cos(2a)*cos(3a)*....*cos(998a)*cos(999a)
Q=sin(a)*sin(2a)*sin(3a)*....*sin(998a)*sin(999a)

As sin x *cos x=sin (2x) /2

P*Q=1/2*sin(2a)*1/2sin(4a)*1/2*sin(6a)*....
         *1/2* sin(2*998a)*1/2*sin(2*999a) (there are 999 factors)
= 1/(2^999) * sin(2a)*sin(4a)*...
     *sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
 as sin(x)=-sin(2pi-x) and 2pi=1999a

sin(1000a)=-sin(2pi-1000a)=-sin(1999a-1000a)=-sin(999a)
sin(1002a)=-sin(2pi-1002a)=-sin(1999a-1002a)=-sin(997a)
...
sin(1996a)=-sin(2pi-1996a)=-sin(1999a-1996a)=-sin(3a)
sin(1998a)=-sin(2pi-1998a)=-sin(1999a-1998a)=-sin(a)

So  sin(2a)*sin(4a)*...
     *sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
= sin(a)*sin(2a)*sin(3a)*....*sin(998)*sin(999) since there are 500 sign "-".

Thus
P*Q=1/2^999*Q or Q!=0 then
P=1/(2^999)

       








7 0
2 years ago
Read 2 more answers
Find the arc length of AB. Round your answer to the nearest hundredth.
Fantom [35]

The arc length of AB is 8 m (app.)

Explanation:

Given that the radius of the circle is 8 m.

The central angle is 60°

We need to determine the arc length of AB

The arc length of AB can be determined using the formula,

arc \ length=\frac{central \ angle}{360^{\circ}} \times circumference

Substituting central angle = 60° and circumference = 2πr in the above formula, we get,

arc \ length=\frac{60^{\circ}}{360^{\circ}} \times 2 \pi(8)

Simplifying the terms, we get,

arc \ length=\frac{8 \pi }{3}

Dividing, we get,

arc \ length=8.37758041

arc \ length=8(app.)

Hence, the arc length is approximately equal to 8.

Therefore, the arc length of AB is 8 m

5 0
2 years ago
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