What are the coordinates of point B?

What's 2/7 of 21? I would appreciate an explanation but it's okay if you can't.

Svetlanka [38]

<h2>Answer:6</h2>

Step-by-step explanation:

Here's a little tip for you. Any number can be converted to fraction if you use 1 as the denominator:

21

1

So now that we've converted 21 into a fraction, to work out the answer, we put the fraction 2/7 side by side with our new fraction, 21/1 so that we can multiply those two fractions.

That's right, all you need to do is convert the whole number to a fraction and then multiply the numerators and denominators. Let's take a look:

2 x 21

7 x 1

=

42

7

In this case, our new fraction can actually be simplified down further. To do that, we need to find the greatest common factor of both numbers.

You can use our handy GCF calculator to work this out yourself if you want to. We already did that, and the GCF of 42 and 7 is 7.

We can now divide both the new numerator and the denominator by 7 to simplify this fraction down to its lowest terms.

42/7 = 6

7/7 = 1

When we put that together, we can see that our complete answer is:

6

1

The complete and simplified answer to the question what is 2/7 of 21 is:

6

7 0

3 years ago

Read 2 more answers

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

It cost $7 for a matinee and $8 for an evening movie. With $56, would you be able to buy morematinee tickets or evening tickets?

aniked [119]

To find how many matinee tickets you could buy, divide $56 by $7, which is the price of 1 ticket.

56 ÷ 7 = 8

Thus, you could buy 8 matinee tickets.

To find how many evening tickets you could buy, divide $56 by $8, which is the price of 1 ticket.

56 ÷ 8 = 7

Thus, you could buy 7 evening tickets. In conclusion, you could buy more matinee tickets than evening tickets with the same amount of money.

Let me know if you need any clarifications, thanks!

5 0

3 years ago

Question 13 of 30

Ivan

Answer:

608.33 cm

Step-by-step explanation:

that is a simple subtraction :

703.33 - 95 =

in our heads : let's assume we initially subtract only 93.

then it is easy

703.33 - 93 = 610.33

and now we subtract the remaining 2 (to subtract in total 95), and we get

608.33 cm

4 0

2 years ago

Least common denominator of 7/2 and1/2

nataly862011 [7]

The least common denominator of 7/2 and 1/3 is two (2)

8 0

4 years ago