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3 years ago

Help please

Mathematics

1 answer:

user100 [1]3 years ago

4 0

Answer:

31 (i.e 3 tens and 1 unit)

Step-by-step explanation:

217 can be split up as,

2 hundreds, 1 tens, and 7 units.

186 can be split up as,

1 hundred, 8 tens and 6 units

If we solve straight we'll have

(2-1)hundreds + (1-8)tens + (7-6)units.

(1-8)tens will give a negative value, complicating the operation.

To make our operation easier we can use the fact that 1 hundred is equal to 10 ten to further split the 217, and it becomes

1 hundred, 11 tens and 7 units.

The operation becomes

(1-1)hundreds + (11 - 8)tens + (7-6)units

= 0 hundred + 3tens + 1 unit

= 3 tens + 1unit

= (3 x 10) + 1 = 31

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Compute the loss of head and pressure drop in 200 ft of horizontal 6-in diameter asphalted cast iron pipe carrying water with a

laiz [17]

Answer:

The answer is below

Step-by-step explanation:

12 inches = 1 ft.

6 inches = 6 inches * (1 ft./12 inches) = 0.5 ft.

Therefore the diameter of the cast iron (D) = 6 inches = 0.5 ft.

The area of cast iron (A) = πD²/4 = π(0.5)²/4 = 0.196 ft²

The velocity (V) = 6 ft./s, the acceleration due to gravity (g) = 32.2 ft./s²

ε/D = 0.0004 ft./ 0.5 ft. = 0.0008

Using the moody chart, find the line ε/D = 0.0008 and determine the point of intersection with the vertical line R = 2.7 * 10⁵. Hence we get f = 0.02.

The head loss (h) is:

$h_f=f*\frac{L}{d}* \frac{V^2}{2g}=0.02*\frac{200\ ft}{0.5\ ft} *\frac{(6\ ft/s)^2}{2*32.2\ ft/s^2}=4.5\ ft$

The pressure drop (Δp) is:

Δp = ρg $h_f$ = $(62.4\ lbf/ft^3)(4.5\ ft)= 280\ lbf/ft^2$

5 0

3 years ago

Which of the following, to the nearest tenth, is a solution of f(x) = g (x) if f (x) = in (x + 5 ) - 1 and g(x) =x^3 - 2x+1

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Answer:

Step-by-step explanation:

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3 years ago

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andre [41]

Check the picture below. So, more or less looks like so.

notice, the center is clearly at the origin, and notice how long the "a" component is, also, bear in mind that, is opening towards the y-axis, that means the fraction with the "y" variable is the positive one.

Also notice, the "c" distance from the center to either foci, is just 5 units.

$\bf \textit{hyperbolas, vertical traverse axis }\\\\
\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}, {{ k}}\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{{{ a }}^2+{{ b }}^2}
\end{cases}\\\\
-------------------------------\\\\$

$\bf \begin{cases}
h=0\\
k=0\\
a=4\\
c=5
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