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garik1379 [7]
3 years ago
7

A plumber had a pipe that was 2 yards long. He cut 18 inches off the end of the pipe. How many feet long was the pipe after the

plumber cut it?
1 yard = 3 feet
1 foot =12 inches
Mathematics
1 answer:
GaryK [48]3 years ago
8 0
2 yards = 6 feet
6 feet - 1.5 feet = 4.5 feet
The pipe was 4.5 feet long after being cut.
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The reason why is because they are $20 or more.  if it was $19.99 then no.  

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Brendan bought an iced tea that cost $4.06 plus 7% sales tax. If Brendan left an 18% tip on the $4.06, how much in total did he
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Answer:

Step-by-step explanation:

18% tip in $=18/100 * $4.06

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3 years ago
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6 groups of students sell 162 balloons at the school carnival. there are 3 students in each group. if each student sells the sam
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Assume that it costs Apple approximately E(x) 25,600 + 100x + 0.012 dollars to manufacture x 32GB iPods in a day. (a) The averag
uranmaximum [27]

Answer:

(a)C'(x)=\dfrac{x^2-2560000}{x^2}

(b)x=1600, Minimum Average Cost Per iPod=$132

(c)C''(x)=\dfrac{5120000}{x^3}

The result, C''(1600) is positive, which means that the average cost is Concave up at the critical point, and the critical point is a minimum.

Step-by-step explanation:

Given that it costs Apple approximately $ C(x) to manufacture x 32GB iPods in a day, where:

C(x)=25,600+100x+0.01x^2

(a)The average cost per iPod when they manufacture x iPods in a day is given by:

Cost \:Per \:iPod=\dfrac{C(x)}{x} =\dfrac{25,600+100x+0.01x^2}{x}

The average cost per iPod is therefore:

C'(x)=\dfrac{x^2-2560000}{x^2}

(b)To minimize average cost of x iPods per day, we set the average cost per iPod=0 and solve for x.

C'(x)=\dfrac{x^2-2560000}{x^2}=0\\x^2-2560000=0\\x^2=2560000\\x=\sqrt{2560000}=1600

The resulting minimum average cost (at x=1600) is given as:

Cost \:Per \:iPod=\dfrac{C(x)}{x} =\dfrac{25,600+100x+0.01x^2}{x}\\\dfrac{25,600+100(1600)+0.01(1600)^2}{1600}\\=\$132

<u>Second derivative test</u>

(c)The answer above is a critical point for the average cost function. To show it is a minimum, we calculate the second derivative of the average cost function.

C''(x)=\dfrac{5120000}{x^3}

At the critical point,  x=1600

C''(1600)=\dfrac{5120000}{1600^3}=0.00125

The result, C''(1600) is positive, which means that the average cost is Concave up at the critical point, and the critical point is a minimum.

3 0
3 years ago
Simplify this expression 4x^2y^3 x 2x^3y^4
sp2606 [1]

Answer:

8 x^5 y^7

Step-by-step explanation:

Simplify the following:

4×2 x^2 y^3 x^3 y^4

Hint: | Combine products of like terms.

4 x^2 y^3×2 x^3 y^4 = 4 x^(2 + 3) y^(3 + 4)×2:

4×2 x^(2 + 3) y^(3 + 4)

Hint: | Evaluate 3 + 4.

3 + 4 = 7:

4×2 x^(2 + 3) y^7

Hint: | Evaluate 2 + 3.

2 + 3 = 5:

4×2 x^5 y^7

Hint: | Multiply 4 and 2 together.

4×2 = 8:

Answer: 8 x^5 y^7

3 0
3 years ago
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