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Vika [28.1K]
3 years ago
14

Two boats leave a port at the same​ time, one going north and the other traveling south. The northbound boat travels 18 mph fast

er than the southbound boat. If the southbound boat is traveling at 40 ​mph, how long will it be before they are 980 miles​ apart?
Mathematics
1 answer:
shutvik [7]3 years ago
3 0

Answer:

10 hours

Step-by-step explanation:

Two boats leave a port at the same​ time, one going north and the other traveling south.

<u>Southbound boat rate</u> = 40 mph

The northbound boat travels 18 mph faster than the southbound boat.

<u>Northbound boat rate</u> = 40 + 18 = 58 mph

Two boats rate = 40 + 58 = 98 mph (this means they are apart 98 km after one hour)

Total distance = 980 miles

Total rate = 98 mph

Time =980:98=10 hours

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1.20 x 10⁶<br><br>6.0 x 10⁵​
omeli [17]

Answer:

1.20 x 10^{6}=1,200,000\\\\6.0 x 10^{5}=600,000

Step-by-step explanation:

You move the decimal place over 6 times for the first equation and 5 for the second equation.

Hope this helps! Have a great day! :)

8 0
3 years ago
A restaurant manager recorded the number of people in different age groups who attended her food festival. She then created the
Andre45 [30]

Answer:

Option: B

There are three times as many participants in the 40-59 age group than in the 0-19 age group.

Step-by-step explanation:

From the histogram we have following observations:

Number of people in the age group  0-19= 20

Number of people in the age group 20-39=40

Number of people in the age group 40-59=60

Number of people in the age group 60-79=50

Hence, from the data we could say that  there are three times as many participants in the 40-59 age group( which is 60) than in the 0-19 age group(which is 20).

Hence, option B is correct.


4 0
3 years ago
Read 2 more answers
At time​ t, the position of a body moving along the​ s-axis is sequalsnegative t cubed plus 15 t squared minus 72 t m. a. Find t
larisa [96]

Answer:

Step-by-step explanation:

Given that at time t, the position of a body moving along the​ s-axis is sequalsnegative t cubed plus 15 t squared minus 72 t m

i.e. s(t) = -t^3+15t^2-72 t

Velocity is nothing but s'(t) = derivative of s

and acceleration is s"(t) = derivative of v(t)

v(t) = -3t^2+30t-72\\=-3(t^2-10t+24)\\= -3(t-6)(t-4)

a) v(t) =0 when t = 4 or 6

b) a(t) = -6t+30

a(t) =0 when t =5

c) Distance travelled by the body from 0 to 5 would be

s(5)-s(0)\\= -5^3+15(5^2)-72(5)\\= -125+375-360\\=-110

i.e. 110 miles (distance cannot be negative)

3 0
3 years ago
Quesuon 18 of 40
dimulka [17.4K]

Answer:

answer is a cylinder I hope it will help you please follow

7 0
3 years ago
The area of sandbox in park is represented by 2X^2-5x-3 find the dimensions of the sandbox in terms of
lora16 [44]

Answer:

The dimension of the sandbox is (2x+1) by (x - 3)

Step-by-step explanation:

It seems the complete question will be:

The area of sandbox in park is represented by 2X^2-5x-3 find the dimensions of the sandbox in terms of x.

Step-by-step explanation:

From the question, the given expression is 2X^2-5x-3. This can be rewritten as

2x^{2} -5x-3

If the area of the sandbox in park is represented by this expression, then the dimensions of the sandbox will be the product of the factors. To determine the factors, we will factorize the given quadratic expression.

Factorizing the expression 2x^{2} -5x-3, we get

2x^{2} -6x+x-3

2x(x-3)+1(x-3)

(2x+1)(x-3)

Hence, the dimension of the sandbox is (2x+1) by (x - 3)

6 0
3 years ago
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