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Ulleksa [173]
3 years ago
12

A ladybug and ant move at constant speeds. The diagrams with tick marks show their positions at different times. Each tick mark

represents 1 centimeter.
In your textbook pg. 111, scale the vertical and horizontal axes by labeling each grid line with a number. You will need to use the time and distance information in the tick-mark diagrams.

Which line shows the ladybugs movement? The ant?
Unit rate for the Ladybug? The ant?
How long does it take for the ladybug to travel 12 cm? The ant?

Mathematics
1 answer:
MissTica3 years ago
4 0

Answer:

From the graph we can say that ant is moving faster than the ladybug

Slope of the ant's movement will be steeper than the ladybug

  • Hence<u>, </u><u>line "</u>v" denotes the speed of ant and <u>line "</u>u" denotes the speed of ladybug.

Unit rate/slope of the graph is (distance traveled/elapsed time)

  • <u>Unit rate of ant is 3/2 cm/sec</u> and <u>that of ladybug is 1 cm/sec.</u>
  • Ant takes (12×2/3) =<u> </u><u>8</u><u> </u><u>sec</u> and ladybug takes (12×1) =<u> </u><u>12</u><u> </u><u>sec</u> to travel 12 cm.

Remember ant will take less time to travel a certain distance than ladybug.

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5 0
3 years ago
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What is the GCF of two prime numbers?
ehidna [41]
Prime numbers: numbers which only have the factors of themselves and 1.

They include numbers such as 17, 19, 23, etc. These are numbers which only have two factors: such as 17, with the factors only being 17 and 1.

So therefore, this means, that for any two prime numbers, unless they are the same number, the only factor they will share is 1. So therefore the GCF of two prime numbers is 1.

Hope this helped!
4 0
3 years ago
Consider the differential equation y'' − y' − 30y = 0. Verify that the functions e−5x and e6x form a fundamental set of solution
Alex Ar [27]

Answer:

Step-by-step explanation:

We have to take the derivatives for both functions and replace in the differential equation. Hence

for y=e^{-5x}:

y(x)=e^{-5x}\\y'(x)=-5e^{-5x}\\y''(x)=25e^{-5x}\\

for y=e^{6x}:

y(x)=e^{6x}\\y'(x)=6e^{6x}\\y''(x)=36e^{6x}\\

Now we replace in the differential equation  y'' − y' − 30y = 0

for y=e^{-5x}:

25e^{-5x}+5e^{-5x}-30e^{-5x}=0\\25+5-30=0

for y=e^{6x}:

36e^{6x}-6e^{6x}-30=0\\36-6+30=0

Now, to know if both function are linearly independent we calculate the Wronskian

W(f,g)=fg'-f'g

W(e^{-5x},e^{6x})=(e^{-5x})(6e^{6x})-(-5e^{-5x})(e^{6x})\neq 0

I hope this is useful for you

Best regard

7 0
3 years ago
Suppose that you're paid $63.24 for 7 hours of work. What is your hourly pay rate?
Vesnalui [34]

Answer:

$9.03

Step-by-step explanation:

Using x for pay rate

63.24=7x Divide both sides by 7

x=9.03

5 0
3 years ago
Find the solution of this system of two linear equations using substitution 11x+3y=103 and y =3x+1
Lesechka [4]
11x+3y=103 and y=3x+1. 

11x+3(3x+1)=103  -----> plug in (3x+1) for y in the first equation.  You will want to distribute the 3 to the 3x+1 to get something that looks like:

11x+9x+3=103    ------> now you want to combine like terms 

20x+3=103 ---> subtract 3 from both sides
20x=100   ----> divide both sides by 20
x=5    

y=3(5)+1     ---> I like to plug in this to the equation that already has y isolated.  3*5 is 15, add 1 and you find that y=16.

(5, 16) will be your final answer (:
8 0
3 years ago
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