Please please please help fast

Between the time Iko woke up and lunchtime, the temperature rose by 11°. Then by the time he went to bed, the temperature droppe

Tanzania [10]

Answer: -3 degrees

Step-by-step explanation:

Lets say that the original temperature was "x."

The temperature rose by 11 degrees so it is now at 11+x.

Then the temperature fell by 14 degrees so you subtract. 11+x-14

11-14= -3 so you subtract 3 degrees from the original temperature and there fore the relative temperature is -3

I hope this helps!!!

4 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

Can someone please help me out :))

ki77a [65]

Answer:

sin(T)

cos(C)

Step-by-step explanation:

sine goes up or down for angles in the circle.

cosine goes left or right for angles in the circle.

consider a circle with is center at T, and it goes through C (so, the radius is 25 = TC).

then 7 = sin(T)×25

24 = cos(T)×25

so, sin(T) = 7/25

then, sin(T) = cos(90-T)

and the angle at C = 90-T.

therefore, cos(C)=sin(T)=7/25

tan(x) = sin(x)/cos(x)

and that would lead here always to a 7/24 or 24/7 ratio.

so, no tan function is right.

5 0

3 years ago

HELP DFASSSSSSSSSTTTTTTT PLZZZZZ

lana66690 [7]

A. <em>distributive property </em>

i hope this helps

4 0

3 years ago

Read 2 more answers

A puppy is tied to a leash in a back yard. His leash is 3 meters long, and he runs around in circles pulling the leash as far as

zlopas [31]

The puppy runs a distance of D) 18 - 19 meters in one lap.

Step-by-step explanation:

Step 1; It is given that the leash measures 3 meters long. He runs pulling the leash as much as he can creating circles. Even though he keeps pulling, the leash will remain at a constant length of 3 meters. So we need to calculate the perimeter of the circle.

Step 2; The perimeter of any given circle is 2π times the radius. The radius of this circle is equal to 3 meters.

Perimeter of the circle = 2π × r = 2 × 3.1415 × 3 meters = 18.85 meters.

So the puppy runs a distance of 18.85 meters in a lap.

6 0

3 years ago