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# what is the unit rate for 490 in 3 1/2 hours & how do you get it please it's need

it just says find the unit rate for 490 in 3 1/2 or 3.5 i want answers & how you got it please...

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### Unit Rate Calculator

Unit Rate Calculator. Unit Rate Calculator. Unit Rate Calculator. quantity. units or items. ... 30 miles ÷ 2.5 hours We want a unit rate where 1 is in the denominator,

### Definition and examples of unit rate | define unit rate ...

Unit Rate is the ratio of two measurements in which the second term is 1. More About Unit of ... Step 2: So, Unit Rate = toys/hours =60/40; Step 3: ... Please provide ...

### MathSteps: Grade 6: Rates: What Is It? - eduplace.com

... they are called unit rates. If you have a multiple-unit rate such as 120 ... 120/3 = 40/1 The unit rate of 120 students ... works at the rate of 60 hours every 3 ...

### How to calculate battery run-time when design equipment using ...

How to calculate battery run-time: ... them and came up with the unauthorized unit of amp-hours ... at a slow rate you will get the rated number of amp-hours ...

### Solving unit rate problem (video) | Khan Academy

Solving unit rate problem. ... So you have 3 hours for every 189 ... So if you divide 189 by 3. Let's do it over on the side right here. 3 goes into 189. 3 goes and ...

### Unit rates and unit ratios | MathVillage

Unit rates and unit ratios. A unit rate is a ratio that has a denominator of 1. ... Write the following ratio as a unit rate (km per hr): 474 km in 6 hours [show answer]

### Battery Ratings | Batteries And Power Systems | Electronics ...

Since 1 amp is actually a flow rate of 1 ... exactly 1 hour, or 2 amps for 1/2 hour, or 1/3 amp ... be 20 hours (70 amp-hours / 3.5 amps). But let’s ...

### Clearing up the question of battery capacity in electronics ...

Clearing up the question of battery capacity in ... the Wh became a unit at all. A Watt is a rate of energy ... If you really need to know Watt-hours, ...

### How to Calculate 12 Volt Amp-Hours | It Still Runs

How to Calculate 12 Volt Amp-Hours ... in the same way you might rate a bucket's ... Divide the battery's 100-amp-hour capacity by 83.3, and you get 1.2 hours, ...

### Ratios & Proportional Relationships | Common Core State ...

Ratios & Proportional Relationships ... if it took 7 hours to mow 4 lawns, then at that rate, ... compute the unit rate as the complex fraction 1/2 / 1/4 miles per ...

## Suggested Questions And Answer :

### UNIT COST PER PLASTIC JAR

Polypropylene cost per jar=0.0495*18.9=R0.93555. Pigment usage per kg=10/5=2g; usage per g=0.002g; usage per 49.5g jar=0.002*49.5=0.99g. Pigment cost=0.99*95/1000=R0.09405. Total cost of one jar (one unit)=R1.0296. Labour per unit=60/120=0.5hr; labour rate=35/2=R17.5/hr; labour cost per jar=17.5*0.5=R8.75. Manufacturing cost per jar=1.0296+8.75=R9.7796 (without overheads). Cost of 15,000 units=15000*9.7796=R146,694. Time to make 15000 units @ 2 units per hour=7,500 hours. We don't have any figures for the number of workers (unless the hourly rate of 2 workers implies there are 2 workers). If there are two workers then they each take 7500/2=3,750 hours to make 15,000 units between them. We don't know the length of the working week, but if we assume 37.5 hours, then 15,000 units would take 100 weeks, or approximately 1 year. The overheads are R7500 a month so 12 months=R90,000 per year. Therefore the total annual cost is 146694+90000=R236,694. When this is divided by 15000 we get the cost of each jar including overheads: R15.78 approx.

### it is for homework,i really need help, plz! Find the shortest distance from the orgin to L

4x + 7y - 5 = 0 7y = -4x + 5 y = (-4/7)x + 5/7 Distance to origin = sqrt( x^2 + y^2 ) D = sqrt( x^2 + ( (-4/7)x + 5/7 )^2 ) D = sqrt( x^2 + (16/49)x^2 + 25/49 - (40/49)x ) D = sqrt( (65/49)x^2 - (40/49)x + 25/49 ) I don't know what level of math you're in, so I don't know if you should know about the following or not. There's a thing called a "derivative" that tells you the rate of change (the slope) of an equation.  If you want to find a minimum or maximum for a function, you can figure out the derivative, set that = to 0, then figure out the x value that makes the derivative = 0. With this problem we can look at the graph (here: https://www.google.com/?gws_rd=ssl#q=plot(+sqrt(+x%5E2+%2B+(+(-4%2F7)x+%2B+5%2F7+)%5E2+)+) ) and see that we should get a minimum around x = 0.3. The derivative D ' of D: D ' = (1/2)( (130/49)x - 40/49 )/sqrt( (65/49)x^2 - (40/49)x + 25/49 ) We want D ' = 0 The (1/2) can't make it 0 and the sqrt stuff on the bottom can't make it 0 (can't make 0 from the bottom of a fraction), so the only part of D ' that can make the whole thing 0 is: (130/49)x - 40/49 = 0 130x - 40 = 0 130x = 40 x = 4/13  (about 0.3077) . But there might be a problem if x = 4/13 makes the bottom = 0, so we need to check that. Does sqrt( (65/49)x^2 - (40/49)x + 25/49 ) = 0 if x = 4/13 ? sqrt( (65/49)x^2 - (40/49)x + 25/49 ) = 0 (65/49)x^2 - (40/49)x + 25/49 = 0 (65/49)(4/13)^2 - (40/49)(4/13) + 25/49 = 0 (65*16)/(49*169) - 160/(49*13) + 25/49 = 0 1040/8281 - 160/637 + 25/49 = 0 1040/8281 - 2080/8281 + 4225/8281 = 0 1040 - 2080 + 4225 = 0 3185 = 0 Not true, so x = 4/13 does not make the bottom of D ' = 0. . So x = 4/13 makes D ' = 0 (a "local minimum" and a "global minimum").  Now we need to know what value for y goes with x = 4/13. y = (-4/7)x + 5/7 y = (-4/7)(4/13) + (5/7)(13/13) y = -16/91 + 65/91 y = 49/91 y = 7/13 Now we know x = 4/13 and y = 7/13.  This should give us the shortest distance from the line L to the origin (0,0). Distance to origin = sqrt( x^2 + y^2 ) D = sqrt( (4/13)^2 + (7/13)^2 ) D = sqrt( 16/169 + 49/169 ) D = sqrt( 65/169 ) D = sqrt( 5/13 ) D = about 0.6202

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### shaney receives an hourly wage of \$29.10 and hour as and emergency room nurse.

Let's put this into some equation. We need to know the nightly rate which would be 29.10 times 1.5 (1.5 because it is 1 and one half the pay for night time hours) We can say that 29.10*1.5= \$43.65 \$43.65 for every night hour worked. So if we set up the equation this way 34r+17n=x where r is the regular hourly rate of 29.10 and n is the nighttime rate of 43.65 plug in the numbers and you get 34*29.10 + 17*43.65= \$989.40 + \$742.05= \$1731.45

### what is the unit rate for 490 in 3 1/2 hours & how do you get it please it's need

490/3.5=140 ..................

### what is 31-48

31 - 48 is the same as 31 + (-48) lets think of this this way, if we draw a huge numberline from -50 to +50. We first go in the + direction to the right 31 units then we back track to the left 48 units, we will have gone the 31 units we already went which would bring us back to 0 then another 17 to the left in the - direction, therefore we get -17.

### Interpolate the data set (1, 150), (3, 175), (4, 185), (6, 200), (8, 300) to estimate the amount of money Gracie may earn if she displays her items for 7 hours

Since 7 is halfway between 6 and 8, Gracie should earn an amount about halfway between 200 and 300, that is, 250 (triangular interpolation). This is the simplest interpolation, but see later. Interpolating for 2 and 5 we get 162.5 and 192.5, that is, respectively, a difference of 12.5 (162.5-150 or 175-162.5) and 7.5 (192.5-185 or 200-192.5), while 250 is a difference of 50 from 200 and 300. So the interpolated figures fluctuate. For a more sophisticated approach, we need to take the whole dataset and look for a formula that best fits. One way to do this is to fit a polynomial F(x)=ax^4+bx^3+cx^2+dx+e into the five given points. This polynomial has 5 unknown coefficients, so with 5 simultaneous equations we should be able to find them. The process can be simplified slightly by taking the lowest "x" coord and using that as the zero starting point. In this case the lowest coord is 1 (hour) so we subtract 1 from the first coord of each pair to get: (0,150), (2,175), etc. F(0)=e=150. So we have the constant 150. The next step is to subtract 150 from each of the other "y" coords so we arrive at the following set of equations: (1) F(2)=16a+8b+4c+2d=25 (2) F(3)=81a+27b+9c+3d=35 (3) F(5)=625a+125b+25c+5d=50 (4) F(7)=2401a+343b+49c+7d=150 and we already have F(0)=150=e. We can now eliminate d from (1) and (2): 2F(3)-3F(2): (162-48)a+(54-24)b+(18-12)c=70-75=-5; (5) 114a+30b+6c=-5. and we can eliminate d from (3) and (4): 5F(7)-7F(5): (12005-4375)a+(1715-875)b+(245-175)c=750-350; 7630a+840b+70c=400 which simplifies to (6) 763a+84b+7c=40 or 109a+12b+c=40/7 We can eliminate c between (5) and (6): 6(6)-(5): (654-114)a+(72-30)b=240/7+5=275/7; (7) 540a+42b=275/7 or 90a+7b=275/42. So b=(275/42-90a)/7. From (6) we have: 109a+12b+c=109a+12(275/42-90a)/7+c=40/7, so c=40/7-109a-12(275/42-90a)/7; c=40/7-550/7+(1080/7-109)a=-510/7+317a/7=(317a-510)/7. We now have b and c in terms of a. We can continue to find d in terms of a. From (1) d=(25-16a-8b-4c)/2=25-16a-8(275/42-90a)/7-4(317a-510)/7; d=25-1100/147+2040/7+(-16+720/7-1268/7)a= (3675-1100+42840)/147+(-112+720-1268)a/7; d=45415/147-660a/7. We have b, c and d in terms of a, so we can find a by substituting into an equation containing all four coefficients (but not (1), because we used it to find d). Let's pick (2) and hope we get a sensible result! 81a+27(275/42-90a)/7+9(317a-510)/7+3(45415/147-660a/7)=35. From this a=5143/2772=1.855. Therefore b=-22.919, c=78.510, d=-67.687, e=150. And F(x)=1.855x^4-22.919x^3+78.510x^2-67.687x+150. This results need to be checked before we use F to find an interpolated value. Unfortunately, this polynomial approach produces inconsistent results, and needs to be discarded. Lagrange's method seems the obvious choice, even if it is tedious to do. We have 5 x values which we'll symbolise as x0, x1, x2, x3, x4 and 5 function values f0, f1, f2, f3, f4. If the function we're looking for is f(x) then: f(x)=(x-x1)(x-x2)(x-x3)(x-x4)f0/((x0-x1)(x0-x2)(x0-x3)(x0-x4))+         (x-x0)(x-x2)(x-x3)(x-x4)f1/((x1-x0)(x1-x2)(x1-x3)(x1-x4))+         (x-x0)(x-x1)(x-x3)(x-x4)f2/((x2-x0)(x2-x1)(x2-x3)(x2-x4))+... x0=1, x1=3, x2=4, x3=6, x4=8; f0=150, f1=175, f2=185, f3=200, f4=300. We want x=7, so f(7) is given by: 4.3.1.-1.150/(-2.-3.-5.-7)+6.3.1.-1.175/(2.-1.-3.-5)+ 6.4.1.-1.185/(3.1.-2.-4)+6.4.3.-1.200/(5.3.2.-2)+ 6.4.3.1.300/(7.5.4.2) This comes to: -60/7+105-185+240+540/7=1600/7=228.57 (229 to the nearest whole number) compared with 250 from the simple interpolation.

### I need help, I really need help pleeease.

I assume that (3) is a graph, because it doesn't display on my tablet. Question 1 a) 1 because the daily operating cost is stated specifically. b) 3 because the profit of \$500 will be a point on the graph corresponding to a value of t; 2 because if we put P=500 and solve for t we get t=1525/7.5=203.3 so the graph would show t=203-204. c) 2 because we set P=0 and solve for t=1025/7.5=136.7, so 137 tickets sold would give a profit of \$2.5, but 136 tickets would make a loss of \$5; alternatively, if (3) is a graph then P=0 is the t-axis so it's where the line cuts the axis between 136 and 137. d) The rate of change is 7.5 from (2).  e) 2, because the format shows it to be a linear relationship; a straight line graph for (3) also shows linearity. Question 2 a) profit=sales- operating costs so 1025 in (2) is the negative value representing these costs. If (3) is a graph, it's the intercept on the P axis at P=-1025; (4) is the P value when t=0. b) For (1) you would need to find out how many tickets at \$7.50 you would need to cover the operating costs of \$1025 plus the profit of \$500. That is, how many tickets make \$1525? Divide 1525 by 7.5; 2 and 3 have already been dealt with; to use (4) you would note that P=\$500 somewhere between t=200 and 250 in the table. c) For (1), work out how many tickets cover the operating costs. 1025/7.5=136.7, so pick 137 which gives the smallest profit to break even; 2 and 3 already given; in (4) it's where P goes from negative to positive, between 100 and 150. d) For (1) the only changing factor is the number of tickets sold. The rate is simply the price of the ticket, \$7.50; if (3) is a graph, the rate of change is the slope of the graph, to find it make a right-angled triangle using part of the line as the hypotenuse, then the ratio of the vertical side (P range) and the horizontal side (t range) is the slope=rate of change; for (4) take two P values and subtract the smallest from the biggest, then take the corresponding t values and subtract them, and finally divide the two differences to give the rate of change: example: (475-(-275))/(200-100)=750/100=7.5. e) For (1) it's clear that the profit increases (or loss decreases) with the sale of each ticket by the same amount as the price of a ticket, so there is a linear relationship; 2 and 3 already dealt with; in the table in (4) the profit changes by the fixed value of 50 tickets=\$375, showing that a linear relationship applies between P and t: for every 50 tickets we just add \$375 to the profit.

### I just need help getting started on this problem. I don't even know how to begin!

The graph is centered over the line x=-3. This line acts like a mirror so that the two halves of the inverted U-shaped graph are reflected in it. The graph has to be shifted 3 units to the right so that the line x=-3 becomes the line x=0, which is the y (f(x)) axis. The equation of the graph changes to f(x)=-0.5x^2+8. When x=0 f(0)=8, and this value on the vertical axis is the highest point of the curve (vertex). To picture f(x+k) you need a picture, a moving picture. Think of the horizontal axis, the x axis, and the line f(x)=8, which is a line parallel to the x axis a distance of 8 units above it. The two lines resemble a track, like a rail track. The curve  is, as we've established, an inverted U shape where the arms of the U are moving further apart the further away the curve is from its vertex. The curve has a constant shape and remains in contact with the line f(x)=8. We start with the curve right in the middle so that the vertical axis bisects the U curve. This is when k=0.  Now, we're going to slide the curve so that its vertex runs along the track f(x)=8 left and right. The curve cuts the axis at two points separated by 8 units, the gap between the zeroes of the function. When the curve is centrally positioned, the points on the x axis are -4 and 4 and k=0. Slide the curve to the left (negative side) and this is equivalent to positive values of k; slide it to the right and we're into negative values of k. If we move the y (f(x)) axis with the curve it becomes the movable axis of symmetry, the mirror I mentioned earlier. The points where the curve cuts the x axis remain 8 units apart. When the curve moves leftward one unit, the value of k increase by 1, and rightward one unit when k decreases by 1. The vertical line x=2 is unmovable, but as the curve slides from left to right this line touches or cuts through the curve. Consider only the part of the curve lying on or above the x axis. When does it touch the vertical line? Move the curve so that the right part of the curve just touches the line. This is the zero with the higher value. So it must be when x=2 is a root, i.e., -0.5(2+k)^2+8=0. 0.5(2+k)^2=8, so (2+k)^2=16, 2+k=+4, and k=4-2 or -4-2, which is 2 or -6. If we slide the curve past the line x=2, till the left part of the curve touches the line on the right side, the point where it touches is the other root, 8 lower than the the right-hand root. So if 2 was the right root then 2-8=-6 is the left root, and if 2 was the left root then 10 is the right root. So we have the functions f(x)=-0.5(x+2)^2+8 and f(x)=-0.5(x-6)^2+8, where the values of k have been substituted. When the curve is on the left of x=2, the roots of f(x+2) are -6 and 2 and when on the right the roots of f(x-6) are 2 and 10. In between these values of k, 2 and -6, the curve touches or is above the x axis, so this the range for k: -6 Read More: ...

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