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vladimir1956 [14]
2 years ago
6

HELP PLSS! The endpoints of CD are C( -3,5) and D(4,-2)

Mathematics
1 answer:
Nonamiya [84]2 years ago
4 0

Answer:

\left(\frac{1}{2}, \frac{3}{2} \right)

Step-by-step explanation:

\left(\frac{-3+4}{2}, \frac{5-2}{2} \right)=\left(\frac{1}{2}, \frac{3}{2} \right)

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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).)

6 0
3 years ago
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Pls help it is due today
Margarita [4]

Answer:

D. How many students enrolled in seventh grade at the start of this year

3 0
3 years ago
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Help picture below problem 5
Anna35 [415]

\angle F+ \angle E = 180^{\circ}\\\\ \implies \angle F + 164^{\circ} = 180^\circ\\\\\implies \angle F = 180^\circ - 164^\circ\\\\\implies \angle F= 16^\circ

7 0
2 years ago
Solve the equation.<br> 6W + 5y = 2z solve for y<br> Pls help!!!
Butoxors [25]

Answer:

Y = 2/5z-6/5w

Step-by-step explanation:

6W + 5y = 2z

1. Subtract 6W from both sides

2. 5y = 2z-6W

3. Divide both sides by 5

4. y = 2/5z-6/5W

6 0
3 years ago
The given measurements may or may not determine a triangle. If not, then state that no triangle is formed. If a triangle is form
kolbaska11 [484]

h=(a)(sinC)=19(sin37)

That means h ≈11.43

Since c<h, There is no triangle

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3 years ago
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