I need help on this answer and don't know. Do you know how to do it?????

What the perimeter is ?

tigry1 [53]

Answer:

the perimeter is the total length of that plane figure

5 0

3 years ago

Solve for brainliest

Evgesh-ka [11]

Answer:

7) 9.135x10^10

8) 3.428x10^-2

9) 2.5x10^-7

10) 4x10^10

Step-by-step explanation:

You’re only supposed to have one number to the left of the decimal point in scientific notation.

7) 9.135x10^10

8) 3.428x10^-2

9) 2.5x10^-7

10) 4x10^10

3 0

3 years ago

Vinny is serving punch to guests at a party. He has 2/3 of a gallon of punch, and he pours 3/64 of a gallon into each guest's cu

Nata [24]

Divide 2/3 by 3/64 and you will get the number of cups of punch. It doesn't come out evenly, but that's the way to do this.there is a total of 2/3 gallon of punch. Each guests has a cup of 3/64 gallon of punch. Now many 3/64 gallon of punch can you fit in 2/3 gallon of punch, that will tell you how many guests there are. so naturally we want to divide ( 2/3 ) / ( 3/64).

7 0

3 years ago

Read 2 more answers

What is the additive identity of rational number

Burka [1]

Answer:

zero(0)

Step-by-step explanation:

The additive identity of a set of number is a number such that the its sum with any of the numbers in the set would give a result that is equal to the number in that set.

In other words, say for example the set of numbers is rational, the additive identity of rational numbers is 0. This is because, given any rational number say x, adding zero to the number x gives the same number x. i.e

x + 0 = x

If x is say 2, then we have;

2 + 0 = 2

Since adding zero to rational numbers gives has no effect on the numbers, then zero (0) is the additive identity of rational numbers.

3 0

3 years ago

Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)

leonid [27]

Let $P=(x,y,z)$ be an arbitrary point on the surface. The distance between $P$ and the given point $(7,11,0)$ is given by the function

$d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}$

Note that $f(x)$ and $f(x)^2$ attain their extrema, if they have any, at the same values of $x$ . This allows us to consider the modified distance function,

$d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2$

So now you're minimizing $d^*(x,y,z)$ subject to the constraint $z^2=xy$ . This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

$\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)$

which has partial derivatives

$\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}$

Setting all four equation equal to 0, you find from the third equation that either $z=0$ or $\lambda=-1$ . In the first case, you arrive at a possible critical point of $(0,0,0)$ . In the second, plugging $\lambda=-1$ into the first two equations gives

$\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10$

and plugging these into the last equation gives

$z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5$

So you have three potential points to check: $(0,0,0)$ , $(2,10,2\sqrt5)$ , and $(2,10,-2\sqrt5)$ . Evaluating either distance function (I use $d^*$ ), you find that

$d^*(0,0,0)=170$
$d^*(2,10,2\sqrt5)=46$
$d^*(2,10,-2\sqrt5)=46$

So the two points on the surface $z^2=xy$ closest to the point $(7,11,0)$ are $(2,10,\pm2\sqrt5)$ .

5 0

4 years ago