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Lilit [14]
3 years ago
6

F(x) = x2 - 10x + 25?​

Mathematics
1 answer:
otez555 [7]3 years ago
3 0

Answer:

hope this is correct

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Which of the following statements is true based on the data set shown below?
alisha [4.7K]

Answer:

median=mode

Step-by-step explanation:

I just did it.

8 0
3 years ago
the graph shows two functions, f(x) and g(x). if the functions are combined so that h(x) = f(x) - g(x), then the domain of the f
Sergio039 [100]

The domain of the function h(x) is x is greater than​ -1

<h3>How to determine the domain of the function h(x)?</h3>

The graphs of the functions are given as attachment

From the attachment, we have the following domains:

  • Domain of f(x): x > 2
  • Domain of g(x): x > -1

The equation of function h(x) is

h(x) = f(x) - g(x)

The domain of the function g(x) is greater than that of the function f(x)

This means that the function h(x) will assume that domain of the function g(x)

Hence, the domain of the function h(x) is x is greater than​ -1

Read more about domain at:

brainly.com/question/1770447

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5 0
1 year ago
I’m not sure what the last one is .. can u help pls
Anarel [89]

Answer:

136

Step-by-step explanation:

180 - <BAC

6 0
3 years ago
One vendor at a craft fair wants to earn $1,200 selling various items over a weekend. The following function can be used to mode
Vlada [557]

Answer:

A) The reasonable domain is limited to positive whole numbers.

Step-by-step explanation:

got it right

5 0
3 years ago
Find the volume v of the described solid s. the base of s is an elliptical region with boundary curve 4x2 + 9y2 = 36. cross-sect
Tasya [4]
4x^2+9y^2=36\iff\dfrac{x^2}9+\dfrac{y^2}4=1

defines an ellipse centered at (0,0) with semi-major axis length 3 and semi-minor axis length 2. The semi-major axis lies on the x-axis. So if cross sections are taken perpendicular to the x-axis, any such triangular section will have a base that is determined by the vertical distance between the lower and upper halves of the ellipse. That is, any cross section taken at x=x_0 will have a base of length

\dfrac{x^2}9+\dfrac{y^2}4=1\implies y=\pm\dfrac23\sqrt{9-x^2}
\implies \text{base}=\dfrac23\sqrt{9-{x_0}^2}-\left(-\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac43\sqrt{9-{x_0}^2}

I've attached a graphic of what a sample section would look like.

Any such isosceles triangle will have a hypotenuse that occurs in a \sqrt2:1 ratio with either of the remaining legs. So if the hypotenuse is \dfrac43\sqrt{9-{x_0}^2}, then either leg will have length \dfrac4{3\sqrt2}\sqrt{9-{x_0}^2}.

Now the legs form a similar triangle with the height of the triangle, where the legs of the larger triangle section are the hypotenuses and the height is one of the legs. This means the height of the triangular section is \dfrac4{3(\sqrt2)^2}\sqrt{9-{x_0}^2}=\dfrac23\sqrt{9-{x_0}^2}.

Finally, x_0 can be chosen from any value in -3\le x_0\le3. We're now ready to set up the integral to find the volume of the solid. The volume is the sum of the infinitely many triangular sections' areas, which are

\dfrac12\left(\dfrac43\sqrt{9-{x_0}^2}\right)\left(\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac49(9-{x_0}^2)

and so the volume would be

\displaystyle\int_{x=-3}^{x=3}\frac49(9-x^2)\,\mathrm dx
=\left(4x-\dfrac4{27}x^3\right)\bigg|_{x=-3}^{x=3}
=16

6 0
3 years ago
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