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2 years ago

PLEASE PLEASE PLEASE HELP ME IM DESPERATE

Mathematics

1 answer:

artcher [175]2 years ago

3 0

I think it might be the first answer dom f = {-4, 3, 5}
Hope it helps ;))

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Janice wants to create a test containing 20 questions worth 50 points. If Janice creates questions worth either __two points or

Helga [31]

15 two points and 5 two points

8 0

3 years ago

Read 2 more answers

3x-9=3(x-3) how many solutions

Colt1911 [192]

Answer:

All Real Numbers

Step-by-step explanation:

Step 1:

3x - 9 = 3 ( x - 3 ) Equation

Step 2:

3x - 9 = 3x - 9 Multiply

Step 3:

- 9 = - 9 Subtract 3x on both sides

Answer:

All Real Numbers

Hope This Helps :)

4 0

2 years ago

How do you solve 6) n2+7n+15=5 by factoring?

charle [14.2K]

First, you need to set the equation equal to zero:

n^2 + 7n + 10 = 0

Now we factor. We need to find two numbers that add up to 7 and multiply to 10.

2 + 5 = 7
2 * 5 = 10

Now, we just need to write this as a polynomial:

(n + 2) (n + 5)
is our answer.

Hope this helps!

8 0

3 years ago

2. The water level in a reservoir is now 52 meters. Which equation can be used to find the initial depth, d, if this is the wate

Maru [420]

Answer:

1.23d = 52

Explanation:

If 52 meters is the water level after a 23% increase, then we can say that the initial depth d added to the 23% of d is equal to 52 meters. So:

d + 23%d = 52 meters

Since 23% is equivalent to 0.23, we get:

d + 0.23d = 52

Finally, adding the like terms, we get:

(1 + 0.23)d = 52

1.23d = 52

So, the equation is:

1.23d = 52

8 0

7 months 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

3 years ago