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3 years ago

Answer any problem and I will be happy (30) pts

Mathematics

1 answer:

KatRina [158]3 years ago

6 0

Problem 4

c = speed of current
p = paddling speed in still water

Against current:
speed = p-c (start with paddling speed and subtract off current's speed)
speed = 2 (given)
p-c = 2
p = 2+c

With the current
speed = p+c (now the current is speeding things up, so we add on c)
speed = 3 (given)
p+c = 3
2+c+c = 3 ... plug in p = 2+c; solve for c
2+2c = 3
2+2c-2 = 3-2
2c = 1
c = 1/2
c = 0.5

Since c = 0.5, we can use this to find p
p = 2+c
p = 2+0.5
p = 2.5

So,
The speed of the river current is 0.5 mph
Rita's paddling speed in still water is 2.5 mph

==================================================

Problem 5

Terms to know:
headwind = wind that slows the plane down (wind is coming from the head of the plane flowing in the opposite direction of the plane's intended direction)
tailwind = wind that speeds the plane up (wind is coming from the tail of the plane flowing in the same direction of the plane's intended direction)

d = distance = 255 miles
p = speed of plane in still air
w = speed of wind

Against the wind (headwind), the plane travels 1.7 hours at some speed p-w. We start with the plane's speed in still air (p) and subtract off the wind speed because the wind is slowing the plane down. So the first equation is
(p-w)*1.7 = 255
since the plane travels 255 miles
I'm using the formula d = r*t
d = distance
r = rate or speed
t = time

Divide both sides of (p-w)*1.7 = 255 by 1.7 and we get p-w = 150
Then add w to both sides and we have p = 150+w

Similarly, the second equation is
(p+w)*1.5 = 255
since the tailwind speeds the plane up from p to p+w, the time is 1.5 hrs and the distance is the same (255 mi)

Plug the equation p = 150+w into the second equation
(p+w)*1.5 = 255
(150+w+w)*1.5 = 255
(150+2w)*1.5 = 255
150*1.5+2w*1.5 = 255
225+3w = 255
225+3w-225 = 255-225
3w = 30
3w/3 = 30/3
w = 10

Use w = 10 to find p
p = 150+w
p = 150+10
p = 160

So,
wind speed = 10 mph
speed of plane in still air = 160 mph

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6+5 x 2 +5+ = <br><br> 67 +84 - 12 x 4 = 16

lozanna [386]

Answer:

6+(5 x 2)+5=21

the way you wrote this was kinda weird so I don't know what you expected as an answer

Step-by-step explanation:

5 x 2=10

10+6=16

16+5=21

3 0

3 years ago

PSEG Long Island charges its residential customers a little monthly service charge plus an energy charge based on the amount of

german

Using linear function concepts, it is found that:

a) It costs $0.1 for each kilowatt hour of electricity used in excess of 250 kWh.
b) f(90) = 46.6, which is the cost of 340 kWh of consumption in a month.

------------------------------

A <em>linear function </em>has the format given by:

$y = mx + b$

In which:

m is the slope, which is the rate of change, that is, how much y changes when x changes by 1.
b is the y-intercept, which is the value of y when x = 0.

The equation for the cost of h kilowatt hours (kWh) of electricity used in excess of 250 kWh is of:

$C = f(h) = 37.6 + 0.1h$

Item a:

The slope is of $m = 0.1$ , which means that it costs $0.1 for each kilowatt hour of electricity used in excess of 250 kWh.

Item b:

$f(90) = 37.6 + 0.1(90) = 37.6 + 9 = 46.6$

250 + 90 = 340.

f(90) = 46.6, which is the cost of 340 kWh of consumption in a month.

A similar problem is given at brainly.com/question/24808124

8 0

2 years ago

Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer

agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming: the function is $f(x)=x^{2}$ in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

$g'(0)=g'(1)$

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

$g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0$

$g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2$

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

$f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]$