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4 years ago

Find the circumference of the object. Use 3.14 or

Mathematics

2 answers:

Maru [420]4 years ago

6 0

Answer:

3.14

Step-by-step explanation:

34kurt4 years ago

4 0

Answer:

C = 3.14 ft

Step-by-step explanation:

C = 2πr

Step 1: Define

r = 0.5 ft

Step 2: Substitute and Evaluate

C = 2π(0.5 ft)

C = 1π ft

C = π ft

C = 3.14 ft

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Help pls and explain!!

quester [9]

Answer:

100

Step-by-step explanation:

$(100 \times 3) - (100 \times 2) = 100$

4 0

3 years ago

Identify Variables

loris [4]

Step-by-step explanation:

Substitute x = 12 and y = 4 into the equations:

1.

$12 + 4 = 12$

Incorrect.

2.

$12 + 4 = 4$

Incorrect.

3.

$12 - 4 = 12$

Incorrect.

4.

$12 - 4 = 4$

Incorrect.

All equations are incorrect when x and y are substituted.

7 0

2 years ago

Find the absolute maximum and minimum values of f on the set D. f(x, y) = x3 − 3x − y3 + 12y + 1, D is quadrilateral whose verti

mojhsa [17]

$D$ is the set of points,

$\left\{(x,y)\mid-2\le x\le2,x\le y\le3\right\}$

Check for critical points:

$f(x,y)=x^3-3x-y^3+12y+1\implies\begin{cases}f_x=3x^2-3=0\implies x=\pm1\\f_y=-3y^2+12=0\implies y=\pm2\end{cases}$

Of these 4 points, only 2 belong to $D$ , (-1, 2) and (1, 2), for which we have

$\begin{cases}f(-1,2)=19\\f(1,2)=15\end{cases}$

Now look for extrema along the boundary.

If $x=-2$ , then

$f(-2,y)=-y^3+12y-1\implies f'(-2,y)=-3y^2+12=0\implies y\pm2$

We have $f'(-2,y)>0$ for $-2$ and $f'(-2,y)$ for $2$ , which indicates a local maximum at $y=2$ and minima at the endpoints of this boundary. So

$\begin{cases}f(-2,2)=15\\f(-2,-2)=-17\\f(-2,3)=8\end{cases}$

If $x=2$ , then

$f(2,y)=y^3+12y+3\implies f'(2,y)=-3y^2+12=0\implies y=\pm2$

We have $f'(2,y)$ for $2$ , so we have extrema at the endpoints of this boundary.

$\begin{cases}f(2,2)=19\\f(2,3)=12\end{cases}$

If $y=x$ , then

$f(x,x)=9x+1\implies f'(x,x)=9>0$

which tells us $f$ is strictly increasing on this boundary, giving the extrema we already know about,

$\begin{cases}f(-2,-2)=-17\\f(2,2)=19\end{cases}$

If $y=3$ , then

$f(x,3)=x^3-3x+10\implies f'(x,3)=3x^2-3=0\implies x=\pm1$

We have $f'(x,3)>0$ for $-2$ and $1$ , and $f'(x,3)$ for $-1$ . This indicates a maximum at $x=-1$ and a minimum at $x=1$ , with

$\begin{cases}f(-2,3)=8\\f(-1,3)=12\\f(1,3)=8\\f(2,3)=12\end{cases}$

From this analysis, we find that $f$ attains an absolute maximum of 19 at (-1, 2) and (2, 2), and an absolute minimum of -17 at (-2, -2).

4 0

3 years ago

What makes the equation true -3 y = 18. A. y = 54 B. y = -54 C. y = 6 D y = -6

Zolol [24]

Answer:

Step-by-step explanation:

-3y=18

divide each side by -3

y=-6

I hope this is correct and have a great day

7 0

3 years ago

Read 2 more answers

CAN SOMEONE PLEASE HELP ME WITH THIS FUNCTION PROBLEM

ycow [4]

Answer:

Choose f(x) = 11x + 1

Step-by-step explanation:

Note that we will simply plug the value of 2 into the equation for February:

f(2) = 11(2) + 1 = 23

And plug the value of 6 into the equation for June:

f(6) = 11(6) + 1 = 67

Note how the points on the graph seem to match up with these values. If we evaluate following the same style for each:

f(3) = 11(3) + 1 = 34

f(4) = 11(4) + 1 = 45

f(5) = 11(5) + 1 = 56

Note, these values seems to be very close in approximation to the graph points for each month.

The other three functions return values that are just to far away from what are represented in the graph.

Cheers.

8 0

4 years ago