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soldi70 [24.7K]
2 years ago
6

Let u = PQ be the directed line segment from P(0,0) to Q(9,12), and let c be a scalar such that c < 0. Which statement best d

escribes cu?
A) the terminal point of vector cu lies in quadrant II
B) the terminal point of vector cu lies in quadrant III
C) the terminal point of vector cu lies in quadrant I
D) the terminal point of vector cu lies in quadrant IV
Mathematics
1 answer:
Angelina_Jolie [31]2 years ago
7 0
The vector u is <9, 12> which means "start at any point, go to the right 9 units, and then up 12 units". If the starting point is point P = (0,0), then the terminal point is in quadrant I. For vector cu, the terminal point is now in quadrant III because we multiply the coordinates of (9,12) by some negative number c (eg: c = -1 so the terminal point is (-9,-12))

Answer: Choice B) terminal point of vector cu is in quadrant 3
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Water is poured into a conical paper cup at the rate of 3/2 in3/sec (similar to Example 4 in Section 3.7). If the cup is 6 inche
aliya0001 [1]

Answer:

The water level rising when the water is 4 inches deep is \frac{3}{8\times \pi} inch/s.

Step-by-step explanation:

Rate of water pouring out in the cone = R=\frac{3}{2} inch^3/s

Height of the cup = h = 6 inches

Radius of the cup = r = 3 inches

\frac{r}{h}=\frac{3 inch}{6 inch}=\frac{1}{2}

r = h/2

Volume of the cone = V=\frac{1}{3}\pi r^2h

V=\frac{1}{3}\pi r^2h

\frac{dV}{dt}=\frac{d(\frac{1}{3}\pi r^2h)}{dt}

\frac{dV}{dt}=\frac{d(\frac{1}{3}\pi (\frac{h}{2})^2h)}{dt}

\frac{dV}{dt}=\frac{1}{3\times 4}\pi \times \frac{d(h^3)}{dt}

\frac{dV}{dt}=\frac{1\pi }{12}\times 3h^2\times \frac{dh}{dt}

\frac{3}{2} inch^3/s=\frac{1\pi }{12}\times 3h^2\times \frac{dh}{dt}

h = 4 inches

\frac{3}{2} inch^3/s=\frac{1\pi }{12}\times 3\times (4inches )^2\times \frac{dh}{dt}

\frac{3}{2} inch^3/s=\pi\times 4\times \frac{dh}{dt} inches^2

\frac{dh}{dt}=\frac{3}{8\times \pi} inch/s

The water level rising when the water is 4 inches deep is \frac{3}{8\times \pi} inch/s.

6 0
2 years ago
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Answer:

3 1/3 cups of flour

Step-by-step explanation:

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20 servings is 5 times 4 servings, so you need 5 times the amount of ingredients.

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Answer: 3 1/3 cups of flour

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25xy=225 convert to polar form
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6 0
2 years ago
Recall the scenario about Eric's weekly wages in the lesson practice section. Eric's boss have been very impressed with his work
Alona [7]

Answer:  

1)\quad f(x)=\bigg\{\begin{array}{ll}12x&0\leq x

2) D: x = [0, 24]

3) R: y = [0, 384]

4) see graph

<u>Step-by-step explanation:</u>

Eric's regular wage is $12 per hour for all hours less than 9 hours.

The minimum number of hours Eric can work each day is 0.

f(x) = 12x    for   0 ≤ x < 9

Eric's overtime wage is $18 per hour for 9 hours and greater.

The maximum number of hours Eric can work each day is 24 (because there are only 24 hours in a day).

f(x) = 18(x - 8) + 12(8)

    = 18x - 144 + 96

    = 18x - 48           for 9 ≤ x ≤ 24

The daily wage where x represents the number of hours worked can be displayed in function format as follows:

f(x)=\bigg\{\begin{array}{ll}12x&0\leq x

2) Domain represents the x-values (number of hours Eric can work).

The minimum hours he can work in one day is 0 and the maximum he can work in one day is 24.

D:  0 ≤ x ≤ 24        →        D: x = [0, 24]

3) Range represents the y-values (wage Eric will earn).

Eric's wage depends on the number of hours he works. Use the Domain (given above) to find the wage.

The minimum hours he can work in one day is 0.

f(x) = 12x

f(0) = 12(0)

     =  0

The maximum hours he can work in one day is 24 <em>(although unlikely, it is theoretically possible).</em>

f(x) = 18x - 48

f(24) = 18(24) - 48

       = 432 - 48

       = 384

D:  0 ≤ y ≤ 384        →        D: x = [0, 384]

4) see graph.

Notice that there is an open dot at x = 9 for f(x) = 12x

and a closed dot at x = 9 for f(x) = 18x - 48

4 0
2 years ago
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