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Scorpion4ik [409]
3 years ago
12

What is the area of the surface of a cookie was a diameter of 32 feet?

Mathematics
1 answer:
aliya0001 [1]3 years ago
4 0

That would be \pi(32/2)^2\approx804.25\mathrm{ft^2}.

Hope this helps.

You might be interested in
What is the constant of proportionality in the equation y=25X?
Ad libitum [116K]

Answer:

Step-by-step explanation:

y = kx

If y = 25x , then k = 25

5 0
3 years ago
Given the values of the linear functions f (x) and g(x) in the tables, where is (f – g)(x) positive?
LuckyWell [14K]

From the tables, and definitions of functions, it is found that the function (f - g)(x) is positive in the interval (–∞, –2).

--------------------

  • The function (f - g)(x) is given by: (f - g)(x) = f(x) - g(x)
  • Thus, it will be positive if: f(x) > g(x)
  • Looking at the table, and considering that it is a linear function, we have that f(x) > g(x) if x < -2.
  • Thus, (f - g)(x) is positive in the interval (–∞, –2).

A similar problem is given at brainly.com/question/24388889

7 0
2 years ago
Adds up to 13 multiplies to -48
fiasKO [112]
You only need two numbers to do that.
The numbers are  16  and  -3 .

           16  -  3  =  13

         (16) x (-3)  =  -48
7 0
3 years ago
x = c1 cos(t) + c2 sin(t) is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the seco
igomit [66]

Answer:

x=-cos(t)+2sin(t)

Step-by-step explanation:

The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:

A differential equation of the form:

a_n y^n +a_n_-_1y^{n-1}+...+a_1y'+a_oy=0

Will have a characteristic equation of the form:

a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0

Where solutions r_1,r_2...,r_n are the roots from which the general solution can be found.

For real roots the solution is given by:

y(t)=c_1e^{r_1t} +c_2e^{r_2t}

For real repeated roots the solution is given by:

y(t)=c_1e^{rt} +c_2te^{rt}

For complex roots the solution is given by:

y(t)=c_1e^{\lambda t} cos(\mu t)+c_2e^{\lambda t} sin(\mu t)

Where:

r_1_,_2=\lambda \pm \mu i

Let's find the solution for x''+x=0 using the previous information:

The characteristic equation is:

r^{2} +1=0

So, the roots are given by:

r_1_,_2=0\pm \sqrt{-1} =\pm i

Therefore, the solution is:

x(t)=c_1cos(t)+c_2sin(t)

As you can see, is the same solution provided by the problem.

Moving on, let's find the derivative of x(t) in order to find the constants c_1 and c_2:

x'(t)=-c_1sin(t)+c_2cos(t)

Evaluating the initial conditions:

x(0)=-1\\\\-1=c_1cos(0)+c_2sin(0)\\\\-1=c_1

And

x'(0)=2\\\\2=-c_1sin(0)+c_2cos(0)\\\\2=c_2

Now we have found the value of the constants, the solution of the second-order IVP is:

x=-cos(t)+2sin(t)

3 0
3 years ago
Which is the next term in the pattern 1/12, 4/15, 9/18, 16/21
photoshop1234 [79]
I’m guessing it would be 21/24
5 0
3 years ago
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