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alexira [117]
3 years ago
9

X/7=3/5 so x=??????????????????

Mathematics
2 answers:
g100num [7]3 years ago
6 0
X = 3 --->  3*7 = 21 so 5x = 21   
7    5

5x = 21   this equals 4.2
5       5     

frosja888 [35]3 years ago
5 0
By using cross multiplication x= 3×7÷5=4.2
the answer is 4.2
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He has 2 $5 bills and 10 $10 bills

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Betty is making a cake that call for 4/6 cups of milk. Which of the following is equivalent to 4/6
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Suppose $1750 is put into an account that pays an annual rate of 4.5%
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Answer:

The amount in the account after six years is $2,288.98

Step-by-step explanation:

In this question, we are asked to calculate the amount that will be in an account that has a principal that is compounded quarterly.

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A = P(1+r/n)^nt

Where P is the amount deposited which is $1,750

r is the rate which is 4.5% = 4.5/100 = 0.045

t is the number of years which is 6 years

n is the number of times per year, the interest is compounded which is 4(quarterly means every 3 months)

we plug these values into the equation

A = 1750( 1 + 0.045/4)^(4 * 6)

A = 1750( 1 + 0.01125)^24

A = 1750( 1.01125)^24

A = 2,288.98

The amount in the account after 6 years is $2,288.98

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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
3 years ago
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