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Aleksandr [31]
3 years ago
14

A car traveling at a speed of 63 kilometers per hour. What is the car’s speed in meters per second? How many meters will the car

travel in 10 seconds? Don’t round answer
Mathematics
1 answer:
dezoksy [38]3 years ago
5 0

Answer:

Speed = 1050 meters per seconds

Distance =10500 meters

Step-by-step explanation:

Car traveling at a speed of 63 kilometers per hour

Let us convert this into meter per second

Speed = \frac{63 }{1}  kilometers per hour

Speed = \frac{63 \ kilometers }{1 \ hour}  --------(A)

1 kilometer = 1000 meters

1 hours = 3600 seconds

Replacing them  in (A)

Speed = \frac{63 \times 1000 \ meters }{1 \times 60 \ seconds}

Speed = \frac{63 \times 1000}{1 \times 60} meters per seconds

Speed = \frac{63000}{60} meters per seconds

Speed = 1050 meters per seconds

Distance = speed x time

Speed = 1050 mt per sec

time = 10 sec

Distance = 1050 x 10 = 10500 meters

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Answer:

Step-by-step explanation:

-3x=16+5

-3x=21

x=21/-3

x= -7

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3 years ago
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irga5000 [103]
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2 years ago
Multiply: -12y(y - 6) Enter the correct answer. ​
bogdanovich [222]

Answer:

-13y - (-72y)

59y is answer

Step-by-step explanation:

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3 years ago
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What is the effect in the time required to solve a prob- lem when you double the size of the input from n to 2n, assuming that t
Inga [223]

Answer:

f(2n)-f(n)=log2

b.lg(lg2+lgn)-lglgn

c. f(2n)/f(n)=2

d.2nlg2+nlgn

e.f(2n)/(n)=4

f.f(2n)/f(n)=8

g. f(2n)/f(n)=2

Step-by-step explanation:

What is the effect in the time required to solve a prob- lem when you double the size of the input from n to 2n, assuming that the number of milliseconds the algorithm uses to solve the problem with input size n is each of these function? [Express your answer in the simplest form pos- sible, either as a ratio or a difference. Your answer may be a function of n or a constant.]

from a

f(n)=logn

f(2n)=lg(2n)

f(2n)-f(n)=log2n-logn

lo(2*n)=lg2+lgn-lgn

f(2n)-f(n)=lg2+lgn-lgn

f(2n)-f(n)=log2

2.f(n)=lglgn

F(2n)=lglg2n

f(2n)-f(n)=lglg2n-lglgn

lg2n=lg2+lgn

lg(lg2+lgn)-lglgn

3.f(n)=100n

f(2n)=100(2n)

f(2n)/f(n)=200n/100n

f(2n)/f(n)=2

the time will double

4.f(n)=nlgn

f(2n)=2nlg2n

f(2n)-f(n)=2nlg2n-nlgn

f(2n)-f(n)=2n(lg2+lgn)-nlgn

2nLg2+2nlgn-nlgn

2nlg2+nlgn

5.we shall look for the ratio

f(n)=n^2

f(2n)=2n^2

f(2n)/(n)=2n^2/n^2

f(2n)/(n)=4n^2/n^2

f(2n)/(n)=4

the time will be times 4 the initial tiote tat ratio are used because it will be easier to calculate and compare

6.n^3

f(n)=n^3

f(2n)=(2n)^3

f(2n)/f(n)=(2n)^3/n^3

f(2n)/f(n)=8

the ratio will be times 8 the initial

7.2n

f(n)=2n

f(2n)=2(2n)

f(2n)/f(n)=2(2n)/2n

f(2n)/f(n)=2

5 0
2 years ago
Given the equation square root of quantity 2x plus 8 end quantity equals 6, solve for x and identify if it is an extraneous solu
Alex_Xolod [135]

Answer:

Solution: x = 6

Step-by-step explanation:

Given equation is:

\sqrt{2x+8}=6

In order to solve the equation both sides will be squared

(\sqrt{2x+8})^{2} = (6)^2 \\2x+8 = 36\\2x = 36-8\\2x = 28\\x = 14

Verifying the solution

\sqrt{2(14)+8} = 6\\ \sqrt{28+8} = 6\\\sqrt{36} = 6\\6 = 6

3 0
2 years ago
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