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3 years ago

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Two in-line skaters, Nell and Kristin, start from the same point and skate in the same direction. Nell skates at 16 miles per ho

ur and Kristin skates at 18 miles per hour. If they can keep up that pace for 3.5 hours, how far apart will they be at the end of that time?

1 answer:

Paha777 [63]3 years ago

8 0

First, we count how far reached the Nell:
3,5 * 16 = 56 miles

Secondly, we count how far reached the Kristin:
3,5 * 18 = 63 miles

Now we expect the difference in the distance between them:
63 - 56 = 7 miles

The results that Kristin about seven miles driven more than Nell.

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tigry1 [53]

Answer:

$=-12$

$d(u,v)=2\sqrt{31}$

Step-by-step explanation:

We are given that inner product defined on $R_n$

$=3u_1v_1+u_2v_2$

u=(0,-4),v=(5,3)

We have to find the value of <u,v> and d(u,v)

We have $u_1=0,u_2=-4,v_1=5,v_2=3$

Substitute the value then we get

$=3(0\cdot5)+(-4)(3)=-12$

Now, $d(u,v)=\left \|v-u\right \|$

Using this formula we get

$d(u,v)=\left \| (5,7) \right \|=\sqrt{3(5)^2+(7)^2}=\sqrt{75+49}=\sqrt{124}$

$d(u,v)=2\sqrt{31}$

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3 years ago

A culture started with 5000 bacteria after 5 hours it grew to 6500 bacteria predict how many bacteria will be present after 16 h

postnew [5]

Answer:

After 16 hours bacterial count will be 11600.

Step-by-step explanation:

Bacterial growth is always an exponential growth in a given culture.So the formula for the growth will be

Final counts of bacteria = Initial counts × $(rate of growth)^{time}$

Given from first part Initial counts = 5000

Final counts= 6500 and time of growth = 5 hours

Then we will get the value of rate of growth of bacteria from the bacteria

⇒ 6500 = 5000× $(rate)^{5}$

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$Rate = \sqrt[5]{1.3} = 1.05$

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Final Count = 5000× $(1.05)^{16}$

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4 years ago

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alukav5142 [94]

Answer:

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Step-by-step explanation:

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3 years ago

In the 1970s, due to world events, there was a gasoline shortage in the United States. There were often long lines of cars wai

andrezito [222]

Answer:

Part A: The average car length is 9.1 feet to the nearest tenth foot

Part B: The line would be 9100 feet to contained 1000 cars

Step-by-step explanation:

* Lets explain how to solve the problem

# Part A:

- There were 62 cars in a line that stretched 567 feet

- The cars are lined up bumper-to-bumper

- That means there is no empty spaces between the cars

* To find the average length of the car we will divide the length of

the line by the numbers of the cars

∵ The average car length = length of the line/number of cars

∵ The length of the line is 567 feet

∵ The numbers of the cars is 62 cars

∴ The average car length = 567/62 = 9.1 feet

* The average car length is 9.1 feet to the nearest tenth foot

# Part B:

- There are 1000 cars

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∵ The average car length = length of the line/number of cars

∵ The average of car length is 9.1 feet

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4 years ago

One side of a triangle is increasing at a rate of 9 cm/s and a second side is decreasing at a rate of 2 cm/s. if the area of the

LenKa [72]

Use the attached figure to guide yourself through the explanation. In the figure:
$a=21cm, b=36cm, \widehat{C}=\frac{\pi}{6}$ .
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Also remmeber that the area of a triangle is:
$A=\frac{Base\times Height}{2}$
The hegiht is shown in the figure in red, the base is just the b side.
Using the angle C the area of the triangle is:
$A=\frac{1}{2}ab\cos(\widehat{C})\implies2A=ab\cos(\widehat{C})$
Now taking the derivative of the area 2A with respect to t we have:
$2\frac{dA}{dt}=b\cos(\widehat{C})\frac{da}{dt}+a\cos(\widehat{C})\frac{db}{dt}-ab\sin(\widehat{C})\frac{d\widehat{C}}{dt}$

Replacing with the right values:
$\frac{dA}{dt}=0\\
\\
\cos({\widehat{C}})=\frac{\sqrt{3}}{2}\\ \\
\sin({\widehat{C}})=\frac{1}{2}\\ \\

$
yields:
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4 years ago