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Natalija [7]
3 years ago
6

Which statement could be used to explain why the function h(x) = x3 has an inverse relation that is also a function?

Mathematics
1 answer:
eimsori [14]3 years ago
3 0

Answer:

h(x) = x^3

A function has an inverse that is also a function is it is a one-to-one function, a function is one-to-one if each value in the domain is linked to only one value in the range, and if each value in the range is linked to only one value in the domain.

Then, a function that is monotonous growing is always one to one, and a function is monotonous growing if the derivative is positive for all the values of x

The derivative of x^3 is:

h'(x) = 3*x^2

and as you know, x^2 is always equal or greater than zero, so h(x) = x^3 is monotonous growing, then it has a inverse that is also a function.

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Please help I'll give brainliest​
SIZIF [17.4K]

Answer:

y + 1 = 4(x - 2)

y = 4x + 11

8x - 2y = 6

Step-by-step explanation:

Following equations are parallel to the graph 4x - y = 6, because their slopes are equal (4).

y + 1 = 4(x - 2)

y = 4x + 11

8x - 2y = 6

8 0
3 years ago
How many cups are in 25 gallons
dmitriy555 [2]
There are 400 cups in 25 Gallons !!!
8 0
3 years ago
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Two long jumpers competed in a world-class track meet. The first athlete jumped a distance of 28.65 feet, and the second athlete
Alja [10]

Answer:

4.4 feet.

Step-by-step explanation:

Just subtract the lower from the higher jump.

28.65 - 24.25

= 4.4 feet.

5 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
3 years ago
What is the square root of -372
wel

Answer:

19.2873015 (Done by a calculator) <---- could be wrong

Step-by-step explanation:

And fun fact:

Negative numbers don't have real square roots since a square is either positive or 0. The square roots of numbers that are not perfect square are members of irrational numbers. This means that they can't be written as the quotient of two integers.

Hope it helps!

4 0
3 years ago
Read 2 more answers
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