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yan [13]
3 years ago
12

Spencer ran the first leg of a relay in 13.56 seconds. Michael ran the second leg. The total time it took both boys to run the r

ace was 26.49 seconds. How long did it take Michael to run the second leg of the race?
Mathematics
1 answer:
zalisa [80]3 years ago
3 0

Answer:

12.93

Step-by-step explanation:

26.49-13.56= 12.93

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The width of a deck is 5 feet shorter than its length. The equation that represents this situation is y = x – 5. What does x rep
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X represents the length of a deck. There are three parts to this equations. The 1. width, 2. the length, and 3. the 5 feet shorter. Using the key words, y = the width of a deck (uses the term is which means =) So that takes the width part out of the equation. Then, 5 feet shorter means -5 than a number. That takes away the 5 feet shorter part of the equation. That leaves x to equal the length of the deck.

8 0
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Hose 1 in June took 45 mins to fill a pool of height of 20 inches and length 110 inches. Hose 2 in July took 15 mins to fill the
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Answer:

there isnt enough evidence to colclude an answer

Step-by-step explanation:

6 0
3 years ago
Which statement is true about the sum of (5y/y^2)+(y-4/5y) ? A: The denominator of the simplified sum is y2 + 5y. B: The denomin
lora16 [44]

Answer:

Dude the answer is B

Step-by-step explanation:

7 0
3 years ago
A quadrilateral has angle measures of 120°, 37°, and 48°. What is the measure of the fourth angle?
White raven [17]
155 degrees because the sum of the angles in a quadrilateral is 360 degrees.
x + 120 + 37 + 48 = 360
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5 0
3 years ago
Read 2 more answers
A population of bees is decreasing. The population in a particular region this year is 1,250. After 1 year, it is estimated tha
8_murik_8 [283]
To model this situation, we are going to use the decay formula: A=Pe^{rt}
where 
A is the final pupolation
P is the initial population 
e is the Euler's constant
r is the decay rate 
t is the time in years

A. We know for our problem that the initial population is 1,250, so P=1250; we also know that after a year the population is 1000, so A=1000 and t=1. Lets replace those values in our formula to find r:
A=Pe^{rt}
1000=1250e^{r}
e^{r}= \frac{1000}{1250}
e^{r}= \frac{4}{5}
ln(e^{r})=ln( \frac{4}{5} )
r=ln( \frac{4}{5} )
r=-02231

Now that we have r, we can write a function to model this scenario:
A(t)=1250e^{-0.2231t}.

B. Here we are going to use a graphing utility to graph the function we derived in the previous point. Please check the attached image.

C. 
- The function is decreasing
- The function doe snot have a x-intercept 
- The function has a y-intercept at (0,1250)
- Since the function is decaying, it will have a maximum at t=0: 
A(0)=1250e^{(-0.2231)(0)
A_{0}=1250e^{0}
A_{0}=1250
- Over the interval [0,10], the function will have a minimum at t=10:
A(10)=1250e^{(-0.2231)(10)
A_{10}=134.28

D. To find the rate of change of the function over the interval [0,10], we are going to use the formula: m= \frac{A(0)-A(10)}{10-0}
where 
m is the rate of change 
A(10) is the function evaluated at 10
A(0) is the function evaluated at 0
We know from previous calculations that A(10)=134.28 and A(0)=1250, so lets replace those values in our formula to find m:
m= \frac{134.28-1250}{10-0}
m= \frac{-1115.72}{10}
m=-111.572
We can conclude that the rate of change of the function over the interval [0,10] is -111.572.

7 0
3 years ago
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