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faltersainse [42]

3 years ago

10

Jethro wants a swimming pool in his backyard, so he digs a rectangular hole with dimensions 40 feet long, 20 feet wide, and 5 fe

et deep. How many cubic feet of water will the pool hold?

1 answer:

KiRa [710]3 years ago

8 0

4000 feet.
Solve by multiplying 40 20 and 5.

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siniylev [52]

Answer:

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59 *2 = 118

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Step-by-step explanation:

3 0

3 years ago

A leading bank is coming up with an investment that pays 8 percent interest compounded semiannually. What is the investment's ef

uranmaximum [27]

The effective rate is calculated in the following way:
$r = {(1 + \frac{i}{n} )}^{n} - 1$
where r is the effective annual rate, i the interest rate, and n the number of compounding periods per year (for example, 12 for monthly compounding).
our compounding period is 2 since the bank pays us semiannually(two times per year) and our interest rate is 8%
so lets plug in numbers:
$r = {(1 + \frac{8\%}{2}) }^{2} - 1 \\ r = {(1 + \frac{1}{25}) }^{2} - 1 \\ r = \frac{676}{625} - 1 \\ r = 0.0816 \: or \: 8.16\%$

5 0

3 years ago

There is a balanced scale. On one side of the scale there are 5 spheres and 1 cube. On the other side of the scale there is 2 sp

Debora [2.8K]

The scale has two sides:
The Left Hand Side (LHS) and the Right Hand Side (RHS).
Let the weight of the sphere be represented as s
Let the weight of the cube be represented as c

Let the LHS have the 5 spheres and 1 cube, represented as = (5s + 1c)
Let the RHS have the 2 spheres and 3 cubes, represented as = (2s + 3c)

If the cube c, weighs 150 gram, c = 150g

Since the scale is balanced, therefore the weights on the LHS would be equal to the weights on the RHS.

Therefore: (5s + 1c) = (2s + 3c), c = 150.
   (5s + 1*150) = (2s + 3*150)
   (5s + 150) = (2s + 450) collect like terms.
      5s - 2s = 450 - 150
      3s = 300 Divide both sides by 3
   s =   300/3
      s = 100
Therefore the sphere weighs 100 gram each.

4 0

3 years ago

A and B are supplementary angles. If A= (4x – 30)° and m

AfilCa [17]

Answer:

82 degrees

Step-by-step explanation:

Hope this helps! ^^

3 0

3 years ago

The fish tank has side lengths 20in, 10in and height 15in. The water level is two inches below the top of the tank. A glass sphe

Kitty [74]

Answer:

Part 1) The new distance from the water to the top of the tank is $1.979\ in$

Part 2) The maximum number of balls that can be put into the tank with the tank not overflowing is 95

Step-by-step explanation:

step 1

Find the total volume of the tank

$V=20*10*15=3,000\ in^{3}$

step 2

Find the volume of the tank if the water level is two inches below the top of the tank

$V=20*10*(15-2)=2,600\ in^{3}$

step 3

Find the volume of the glass sphere

The volume of the glass sphere is equal to

$V=\frac{4}{3}\pi r^{3}$

we have

$r=1\ in$

assume

$\pi=3.14$

substitute

$V=\frac{4}{3}(3.14)(1)^{3}$

$V=4.2\ in^{3}$

step 4

What is the new distance from the water to the top of the tank?

we know that

$2 in -----> represent (3,000-2,600)=400\ in^{3}$

so

using proportion

Find how many inches correspond a volume of $4.2\ in^{3}$

$\frac{2}{400}\frac{in}{in^{3}}=\frac{x}{4.2}\frac{in}{in^{3}}\\ \\x=4.2*2/400\\ \\x=0.021\ in$

The new distance from the water to the top of the tank is

$2-0.021=1.979\ in$

step 5

Find how many of these balls can be put into the tank with the tank not overflowing

we know that

The volume of one ball is equal to $4.2\ in^{3}$

using proportion

$\frac{1}{4.2}=\frac{x}{400}\\ \\x=400/4.2\\ \\x=95.23\ balls$

therefore

The maximum number of balls that can be put into the tank with the tank not overflowing is 95

4 0

3 years ago