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Salsk061 [2.6K]
1 year ago
7

Which number property is illustrated in the Identy ?

Mathematics
1 answer:
zhuklara [117]1 year ago
4 0
The property of this equation is associative property
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4(2x+3)=x(3+5)+12<br> Help
Karo-lina-s [1.5K]

Answer:

Step-by-step explanation:

8x + 12 = 3x + 5x + 12

8x + 12 = 8x + 12

Bringing like terms on one side

8x - 8x = 12 - 12

x = 0

7 0
3 years ago
Read 2 more answers
Use the properties of exponents to rewrite y=e^-0.75t in the form of a(1-r)^t
Ivahew [28]

Answer: y = (1 - 0.527)^t

Step-by-step explanation:

y=e^(-0.75t)


y=(e^-0.75)^t


y= 0.47236655^t


1 - 0.47236655 = 0.52763345

y = (1 - 0.527)^t

5 0
2 years ago
Factor f(x) = 15x^3 - 15x^2 - 90x completely and determine the exact value(s) of the zero(s) and enter them as a comma separated
Illusion [34]

Answer:

x=-2,0,3

Step-by-step explanation:

We have been given a function f(x)=15x^3-15x^2-90x. We are asked to find the zeros of our given function.

To find the zeros of our given function, we will equate our given function by 0 as shown below:

15x^3-15x^2-90x=0

Now, we will factor our equation. We can see that all terms of our equation a common factor that is 15x.

Upon factoring out 15x, we will get:

15x(x^2-x-6)=0

Now, we will split the middle term of our equation into parts, whose sum is -1 and whose product is -6. We know such two numbers are -3\text{ and }2.

15x(x^2-3x+2x-6)=0

15x((x^2-3x)+(2x-6))=0

15x(x(x-3)+2(x-3))=0

15x(x-3)(x+2)=0

Now, we will use zero product property to find the zeros of our given function.

15x=0\text{ (or) }(x-3)=0\text{ (or) }(x+2)=0

15x=0\text{ (or) }x-3=0\text{ (or) }x+2=0

\frac{15x}{15}=\frac{0}{15}\text{ (or) }x-3=0\text{ (or) }x+2=0

x=0\text{ (or) }x=3\text{ (or) }x=-2

Therefore, the zeros of our given function are x=-2,0,3.

7 0
3 years ago
Which of the following statements is true for the logistic differential equation?
solong [7]

Answer:

All of the above

Step-by-step explanation:

dy/dt = y/3 (18 − y)

0 = y/3 (18 − y)

y = 0 or 18

d²y/dt² = y/3 (-dy/dt) + (1/3 dy/dt) (18 − y)

d²y/dt² = dy/dt (-y/3 + 6 − y/3)

d²y/dt² = dy/dt (6 − 2y/3)

d²y/dt² = y/3 (18 − y) (6 − 2y/3)

0 = y/3 (18 − y) (6 − 2y/3)

y = 0, 9, 18

y" = 0 at y = 9 and changes signs from + to -, so y' is a maximum at y = 9.

y' and y" = 0 at y = 0 and y = 18, so those are both asymptotes / limiting values.

8 0
3 years ago
Which line of symmetry for the parabola (x-1)^2= 4(y-1)^?
jonny [76]
This is a vertical parabola, because  (x-1)².
Vertex of the parabola (1,1).
So line symmetry is x=1.
4 0
3 years ago
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