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zhuklara [117]
3 years ago
11

Solve for t literal equation

\frac{v-u}{t}" alt="y = \frac{v-u}{t}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
balu736 [363]3 years ago
4 0

Answer:

t = (v-u)/y

Step-by-step explanation:

yt = v - u

t = (v-u)/y

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Lisa makes $78.37 per day. If she works 8 12<br> hours, how much does she make in 1 hour?
Leni [432]
I’m assuming the time is eight hours and 12 minutes. let me know if that’s not what it’s supposed to be

78.37 / 8.12 = 9.65

lisa makes $9.65 in 1 hour
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Lidocaine (2 grams) was prescribed in 500 mL of fluid at 2 mg/ minute, to be administered via an infusion pump. What is the flow
Luden [163]

9514 1404 393

Answer:

  1/2 mL/min

Step-by-step explanation:

  \dfrac{2\text{ mg}}{\text{min}}\div\left(\dfrac{2\text{ g}}{500\text{ mL}}\times\dfrac{1000\text{ mg}}{\text{g}}\right)=\dfrac{2\times500\text{ mL}}{2\times1000\text{ min}}=\boxed{\dfrac{1}{2}\text{ mL/min}}

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<em>Additional comment</em>

The units analysis tells you how to work this problem. In order to get from mg/min to mL/min, you need to multiply by a factor with units mL/mg. Those units are the inverse of the concentration in mg/mL, which is found from g/mL.

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The graph of the continuous function g, the derivative of the function f, is shown above. The function g is piecewise linear for
4vir4ik [10]
A) g=f' is continuous, so f is also continuous. This means if we were to integrate g, the same constant of integration would apply across its entire domain. Over 0, we have g(x)=2x. This means that


f_{0


For f to be continuous, we need the limit as x\to1^- to match f(1)=3. This means we must have


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b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,


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c) f is increasing when f'=g>0, and concave upward when f''=g'>0, i.e. when g is also increasing.

We have g>0 over the intervals 0 and x>4. We can additionally see that g'>0 only on 0 and x>4.


d) Inflection points occur when f''=g'=0, and at such a point, to either side the sign of the second derivative f''=g' changes. We see this happening at x=4, for which g'=0, and to the left of x=4 we have g decreasing, then increasing along the other side.


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