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3 years ago

Find the area to this figure please help

Mathematics

1 answer:

elena55 [62]3 years ago

4 0

Answer:

the area of the circle is 25π or about 78.54 yards

Step-by-step explanation:

Area of circle=πr^2, r being radius

r=5

5^2=25

25π≈78.54 yards

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3 years ago

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A school has 200 students and spends $40 on supplies for each student. The principal expects the number of students to increase

Xelga [282]

Answer:

$\mathbf{S(t)=200(\frac{105}{100})^{x}}$

$\mathbf{A(t)=40(\frac{98}{100})^{x}}$

$\mathbf{E(t)=S(t) \cdot A(t)=200(\frac{105}{100})^{x} \cdot 40(\frac{98}{100})^{x}=8000(\frac{10290}{10000})^{x}}$

Step-by-step explanation:

<h3>The predicted number of students over time, S(t) </h3>

Rate of increment is 5% per year.

A function 'S(t)' which gives the number of students in school after 't' years.

S(0) means the initial year when the number of students is 200.

S(0) = 200

S(1) means the number of students in school after one year when the number increased by 5% than previous year which is 200.

S(1) = 200 + 5% of 200 = $200+\frac{5}{100}\time200$ = $200(1+\frac{5}{100})$ = $200(\frac{105}{100})$

S(2) means the number of students in school after two year when the number increased by 5% than previous year which is S(1)

S(2) = S(1) + 5% of S(1) = $\textrm{S}(1)(\frac{105}{100})$ = $200(\frac{105}{100})(\frac{105}{100})$ = $200(\frac{105}{100})^{2}$

Similarly $\mathbf{S(x)=200(\frac{105}{100})^{x}}$

<h3>The predicted amount spent per student over time, A(t) </h3>

Rate of decrements is 2% per year.

A function 'A(t)' which gives the amount spend on each student in school after 't' years.

A(0) means the initial year when the number of students is 40.

A(0) = 40

A(1) means the amount spend on each student in school after one year when the amount decreased by 2% than previous year which is 40.

A(1) = 40 + 2% of 40 = $40-\frac{2}{100}\time40$ = $40(1-\frac{2}{100})$ = $40(\frac{98}{100})$

A(2) means the amount spend on each student in school after two year when the amount decreased by 2% than previous year which is A(1)

A(2) = A(1) + 2% of A(1) = $\textrm{A}(1)(\frac{98}{100})$ = $40(\frac{98}{100})(\frac{98}{100})$ = $40(\frac{98}{100})^{2}$

Similarly $\mathbf{A(x)=40(\frac{98}{100})^{x}}$

<h3>The predicted total expense for supplies each year over time, E(t)</h3>

Total expense = (number of students) × (amount spend on each student)

E(t) = S(t) × A(t)

$\mathbf{E(t)=S(t) \cdot A(t)=200(\frac{105}{100})^{x} \cdot 40(\frac{98}{100})^{x}=8000(\frac{10290}{10000})^{x}}$

$\mathbf{E(t)=8000(\frac{10290}{10000})^{x}}$

(NOTE : The value of x in all the above equation is between zero(0) to ten(10).)

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Answer:

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Step-by-step explanation:

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