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Shtirlitz [24]
3 years ago
9

A local diner uses 4 2/3 pounds of meat to make five hamburgers. How much meat is used to make one hamburger?

Mathematics
1 answer:
hodyreva [135]3 years ago
8 0

Answer:14/15 pounds of meat

Step-by-step explanation:

<h3>SO, 4 2/3=14/3</h3><h3>5 HAMBURGERS ARE MADE BY USING 14/3 POUNDS OF MEAT</h3><h3>1 HAMBURGER IS MADE BY USING 14/3×1/5 OR,14/15 POUNDS OF MEAT.</h3><h3>BASICALLY THE ANSWER IS:14/15 pounds of meat is used to make one hamburger.</h3>

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2(x+x+5)=38

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Step-by-step explanation:

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What is pi and why do we need it
blsea [12.9K]

Answer:

See below

Step-by-step explanation:

π, pronounced as <em>pi</em> is the ratio between 2πr which is circumference to 2r which is diameter.

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π is often used to be defined as 3, 3.14 or 22/7 [Keep in mind, these numbers are only approximation.] in some fields. In trigonometry, π is mostly used to refer as <u>π radians</u> or <u>π rad</u> which is equivalent to 180°.

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\displaystyle \large{V =\pi \int\limits^a_b {[f(x)]^2} \ dx }\\\\\displaystyle \large{V = \pi \int\limits^a_b {[g(y)]^2} \ dy}

In trigonometry, you need it for measurement/angle.  

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7 0
2 years ago
Find the point P on the graph of the function y=√x closest to the point (9,0)
Sphinxa [80]

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\displaystyle \frac{17}{2}.

Step-by-step explanation:

Let the x-coordinate of P be t. For P\! to be on the graph of the function y = \sqrt{x}, the y-coordinate of \! P would need to be \sqrt{t}. Therefore, the coordinate of P \! would be \left(t,\, \sqrt{t}\right).

The Euclidean Distance between \left(t,\, \sqrt{t}\right) and (9,\, 0) is:

\begin{aligned} & d\left(\left(t,\, \sqrt{t}\right),\, (9,\, 0)\right) \\ &= \sqrt{(t - 9)^2 +\left(\sqrt{t}\right)^{2}} \\ &= \sqrt{t^2 - 18\, t + 81 + t} \\ &= \sqrt{t^2 - 17 \, t + 81}\end{aligned}.

The goal is to find the a t that minimizes this distance. However, \sqrt{t^2 - 17 \, t + 81} is non-negative for all real t\!. Hence, the \! t that minimizes the square of this expression, \left(t^2 - 17 \, t + 81\right), would also minimize \sqrt{t^2 - 17 \, t + 81}\!.

Differentiate \left(t^2 - 17 \, t + 81\right) with respect to t:

\displaystyle \frac{d}{dt}\left[t^2 - 17 \, t + 81\right] = 2\, t - 17.

\displaystyle \frac{d^{2}}{dt^{2}}\left[t^2 - 17 \, t + 81\right] = 2.

Set the first derivative, (2\, t - 17), to 0 and solve for t:

2\, t - 17 = 0.

\displaystyle t = \frac{17}{2}.

Notice that the second derivative is greater than 0 for this t. Hence, \displaystyle t = \frac{17}{2} would indeed minimize \left(t^2 - 17 \, t + 81\right). This t\! value would also minimize \sqrt{t^2 - 17 \, t + 81}\!, the distance between P \left(t,\, \sqrt{t}\right) and (9,\, 0).

Therefore, the point P would be closest to (9,\, 0) when the x-coordinate of P\! is \displaystyle \frac{17}{2}.

8 0
3 years ago
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