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yanalaym [24]
3 years ago
11

Describe the inverse operation that will undo the given operation. Subtracting by 12

Mathematics
1 answer:
san4es73 [151]3 years ago
3 0

Answer:

adding 12

the inverse is the opposite

addition is the opposite side of subtract

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Suppose the function F(c) represents the earnings of a volleyball team from selling c cupcakes. They are selling each cupcake fo
navik [9.2K]

Answer:

The variable c represents the domain as it is the independent variable.

The domain of the function F(c) is given by c ≥ 0.

So, only positive values for the input make sense.

The upper limit of the domain is +∞ and lower limit is 0.

It is not possible for the team to earn $50.50 as it will be only multiple of 2.

Step-by-step explanation:

If F(c) represents the earning of a volleyball team from selling c cupcakes and each cupcake costs $2 each, then the equation that models the situation is  

F(c) = 2c ..... (1)

The variable c represents the domain as it is the independent variable. (Answer)

The domain of the function F(c) is given by c ≥ 0. (Answer)

So, only positive values for the input make sense. (Answer)

The upper limit of the domain is +∞ and the lower limit is 0. (Answer)

It is not possible for the team to earn $50.50 as it will be only multiple of 2. (Answer)

5 0
3 years ago
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Use the method of reduction of order to find a second solution to t^2y' + 3ty' – 3y = 0, t&gt; 0 Given yı(t) = t y2(t) = Preview
Anestetic [448]

Let y_2(t)=tv(t). Then

{y_2}'=tv'+v

{y_2}''=tv''+2v'

and substituting these into the ODE gives

t^2(tv''+2v')+3t(tv'+v)-3tv=0

t^3v''+5t^2v'=0

tv''+5v'=0

Let u(t)=v'(t), so that u'(t)=v''(t). Then the ODE is linear in u, with

tu'+5u=0

Multiply both sides by t^4, so that the left side can be condensed as the derivative of a product:

t^5u'+5t^4u=(t^5u)'=0

Integrating both sides and solving for u(t) gives

t^5u=C\implies u=Ct^{-5}

Integrate again to solve for v(t):

v=C_1t^{-6}+C_2

and finally, solve for y_2(t) by multiplying both sides by t:

tv=y_2=C_1t^{-5}+C_2t

y_1(t)=t already accounts for the t term in this solution, so the other independent solution is y_2(t)=t^{-5}.

6 0
3 years ago
The value -2 is a lower bound for the zeros of the function shown below f(x)=4x^3-12x^2-x+15
NISA [10]

We are given

The value -2 is a lower bound for the zeros of the function shown below f(x)=4x^3-12x^2-x+15

So, firstly, we will find value of f(x) at x=-2

we will plug x=-2 into f(x)

we get

f(x)=4x^3-12x^2-x+15

we will plug x=-2

f(-2)=4(-2)^3-12(-2)^2-(-2)+15

now, we can simplify it

f(-2)=-32-12\cdot \:4-\left(-2\right)+15

f(-2)=-63

so,

f(-2)=-63

we know that any zeros always lies between positive and negative value

Here , we got f(-3)=-63 ...which is negative value

so, this is lower bound

Hence , this is TRUE

6 0
3 years ago
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