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LUCKY_DIMON [66]
3 years ago
13

Change each fraction or mixed number to a percent. a. 2⁄5 b. 7⁄20 c. 111⁄16 d. 141⁄4

Mathematics
1 answer:
tino4ka555 [31]3 years ago
6 0

a. 2⁄5 = 0.4 = 40%

b. 7⁄20 = 0.35 = 35%

c. 11 1⁄16 = 6.9375  ≈694%

d. 141⁄4 = 35.25 = 3525%

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A rock was thrown from the top of a cliff such that its distance above sea level was given by s(t)= at² + bt + c, where t is the
frutty [35]

Answer:

<h2>a) a = -3, b = 18, c = 48; </h2><h2>s(t) = -3t²+18t+48</h2><h2>b) 48m</h2><h2>c) 8secs</h2>

Step-by-step explanation:

The question is incomplete. Here is the complete question.

'A rock was thrown from the top of a cliff such that its distance above sea level was given by s(t)=at²+bt+c, where t is the time in seconds after the rock was released. After 1 second the rock was 63 m above sea level, after 2 seconds 72 m, and after 7 seconds 27 m. a. Find a, b and c and hence an expression for s(t). b. Find the height of the cliff. c. Find the time taken for the rock to reach sea level.'

Given the equation of the distance modelled as s(t)=at²+bt+c

If after 1 second the rock was 63 m, then 63 = a+b+c

If after 2 seconds, the distance was 72 m then 72 = 4a+2b+c

Also if after 7 seconds, the distance is 27 m, then 27 = 49a+7b+c

i) Solving the 3 equations simultaneously to get a, b and c we have;

a+b+c = 63 ... 1

4a+2b+c = 72 ...2

49a+7b+c = 27 ...3

Subtracting 2 from 1 and 3 from 2 we will generate 2 new equations as shown;

eqn 2- eqn1: 3a + b = 9...4

eqn 3- eqn 2: 45a + 5b = -45

eqn 3- eqn 2: 9a+b = -9 ... 5

solving 4 and 5 simultaneously

3a + b = 9 ...4

9a+b = -9 ... 5  

Taking the difference of 4 and 5 we have

6a - -9-9

6a = -18

a = -3

substituting a = -3 into equation 4 to get b we have;

3(-3)+b = 9

-9 + b = 9

b = 9+9

b = 18

substituting a = -3 and b = 18 into equation 1 to get c we have;

-3+18+c = 63

15+c = 63

c = 48

a = -3, b= 18 and c = 48

The distance function will be s(t) = -3t²+18t+48

ii) If the height of the cliff is modelled by the equation  s(t)=at²+bt+c

The height of the cliff is at when t = 0

s(0) = -3(0)²+18(0)+48

s(0) = 48m

The height of the cliff is 48m

iii) At the sea level, the height of the rock will be 0m, substituting this into the modeled equation for the height to get the time we have;

s(t)=at²+bt+c

0 = -3t²+18t+48

3t²-18t-48 = 0

t² - 6t - 16 =0

t² - 8t+2t - 16 = 0

t(t-8)+2(t-8) = 0

(t+2)(t-8) = 0

t = -2 or 8

Taking the positive value of the time, t = 8secs

Time taken for the rock to reach sea level is 8secs

7 0
4 years ago
Circle O has a circumference of approximately 44 in.
sleet_krkn [62]

Answer:

44

Step-by-step explanation:

because yea

4 0
3 years ago
The quotient of a number and 2 increased by 9 is -15
Agata [3.3K]

Answer:

x/2+9=-15

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
A graphing calculator is recommended.
kogti [31]

Answer:

Q(3\,min) = 0.349\,pd

Step-by-step explanation:

The expression for the salt in the barrel is:

Q(t) = 6\cdot (1-e^{-0.02\cdot t})

The quantity of salt in the barrel after 3 minutes is:

Q(3\,min) = 6\cdot [1-e^{-0.02\cdot (3\,min)}]

Q(3\,min) = 0.349\,pd

5 0
4 years ago
A scientist kept 100 grams of a radioactive material in a container. After 280 days, he observed that 25 grams of the radioactiv
Wittaler [7]

Answer:

B: About 50 grams of the material was left in the container after 140 days.

Step-by-step explanation:

Radioactive materials decay at a rate called "half-lives." After a certain amount of time, the grams of radioactive material will be reduced to half. After that same amount of time again, the material will be reduced to half of the half.

1/4 is equal to 1/2 times 1/2. So, when 1/4 of the material remains, two half-lives have elapsed.

The question states that after 280 days, 25 grams of the material was left in the container. 25 grams is 1/4 of the original 100 grams that the scientist started with at time = 0 days.

When 25 grams of material remain, two half-lives have elapsed. 280 days equals two half-lives for this material. So 140 days (half of 280 days) equals one half-life for this material.

After 280 days (two half-lives), one fourth of the material (25 grams) will remain.

After 140 days (one half-life), half of the material (50 grams) will remain.

5 0
3 years ago
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