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vladimir2022 [97]
3 years ago
15

PLEASE HELP ME WITH THIS FOR BRAINLIEST AND A UNBURNED COOKIE

Mathematics
2 answers:
schepotkina [342]3 years ago
8 0
I think it would be 43. 8 times 5 is 40, plus 3 would equal 43. if it’s not that then it would be 55. (11 times 5)
Amiraneli [1.4K]3 years ago
3 0

Answer:

43

Step-by-step explanation:

ur replacing z in 3+8z with 5 and 8z is the same thing as 8xZ so the new equation is

3+8x5

3+40

43 is your answer

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There are 4 fewer white bunnies than gray bunnies there are 9 white bunnies how many gray bunnies are there
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Answer:

5 gray bunnies

Step-by-step explanation:

I subtracted 4 from 9 and got 5 as my answer

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Find the quotient -88, 11
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The quotient is 8 because 11*8 is 88
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A Scale model of a race car is 18.6 cm long and 6.9 cm wide each centimeter on the model represent 0.3 m on the actual car how l
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18,6 cm x 0,3m/cm = 5,58m
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3 years ago
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Help me please and thank you
vekshin1

Answer: flip the 5 and the y so you get 6x-5=y then plug in whatever number you need into x so its 6 times whatever number minus 5 then whatever number you plug in is your x and the solution is the y

Step-by-step explanation:

im doin the same thing

8 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}  

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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}  

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.  

.  

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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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