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3 years ago

CAN SOMEONE HELP ME PLEASE?

Mathematics

2 answers:

rjkz [21]3 years ago

8 0

Answer:

kk!

Step-by-step explanation:

Tu = ml

km = us

lk = sr

m = t

t = m

ust = lkm

Tus = mlk

AysviL [449]3 years ago

5 0

I HAVE SOLVED ALL THE QUESTIONS ,SEE THE IMAGE....

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Helppp me Please!!!!! :)

irina [24]

Answer:

$x = 30 \\ y = 10$

Step-by-step explanation:

$8x + 6y = 300 \\ x + y = 40 \\ 6x + 6y = 240 \\ 2x = 60 \\ = > x = 30 \\ y = 10$

7 0

3 years ago

Amy's music collection includes 17 hiphop albums,9 R&B,12country and 8 Pop. She wants to keep 10 CDs for herself and share t

ella [17]

The answer I believe is 6

6 0

3 years ago

What is the range of the function graphed?

victus00 [196]

The correct answer is B)
-3 < y ≤ 3

You can tell this by looking at the graph. The lowest that it goes is -3. You'll also note that the -3 has an open circle, which means that it isn't includes and thus needs a <.

The highest it goes it 3. You'll note that it has a closed circle, which means it is included and needs a ≤.

4 0

4 years ago

The school that Scott goes to is selling tickets to the annual dance

zalisa [80]

Senior tickets (x)
Child tickets (y)

First day: 3x + 5y = 70

second day: 12x + 12y = 216

Solve the system of equations (use elimination)

Multiply first equation by -4 -4(3x + 5y = 70), which makes it

-12x - 20y = -280
(+)12x + 12y = 216 add to second equation

-8y = -64

divide by -8. y = 8

Plugin the y value to either equation ( I will choose first equation)

3x + 5(8) =70

3x+ 40 = 70

3x = 30

x = 10

Senior tickets are $10, child tickets are $8

4 0

3 years ago

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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