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Naily [24]
3 years ago
6

If 2x + 5 =8 ×, then 12x=

Mathematics
2 answers:
nadya68 [22]3 years ago
5 0
<span>2x + 5 =8x
6x = 5
x = 5/6

12x = 12(5/6) = 10</span>
nikitadnepr [17]3 years ago
5 0
2x+5=8x
Then Subtract 2x from both sides
2x+5=8x
-2x. -2x

5=6x
Then divide each side by 6

X=5/6

12x= 12(5/6) = 9.996
If you need round to 10
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Use the Triangle Inequality Property to determine whether a triangle can be formed with the given side lengths. 4 m, 8 m, 9 m *
Illusion [34]

Answer:

Therefore a triangle can be formed with side lengths of 4 m, 8 m, 9 m

Step-by-step explanation:

A triangle is a polygon with three sides and three angles. There are different types of triangles such as scalene, isosceles, equilateral and so on.

The triangle inequality property states that the sum of any two sides of a triangle must be greater than the third side. If a, b and c are the sides of a triangle then:

a + b > c;   a + c > b;   b + c > a

Given a triangle with side length 4 m, 8 m, 9 m:

4m + 8m = 12m > 9m

4m + 9m = 13m > 8m

8m + 9m = 17m > 4m

Therefore a triangle can be formed with side lengths of 4 m, 8 m, 9 m.

7 0
3 years ago
The graph of y=x^3+x^2-6x is shown....
Bad White [126]

Answer:

its 0 and 2 and -3(im pretty sure)

Step-by-step explanation:

4 0
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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