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2 years ago

The lengths of the legs of a right triangle are 4.5 inches and 6 inches.

Mathematics

2 answers:

MAVERICK [17]2 years ago

5 0

Answer:

$\huge\boxed{\sf H = 7.5\ inches}$

Step-by-step explanation:

Let,

Base = 4.5 inches

Perpendicular = 6 inches

We can find the hypotenuse by Pythgoras theorem.

$\sf (Hypotenuse)^2=(Base)^2+(Perpendicular)^2$

Put the given values

(H)² = (4.5)² + (6)²

(H)² = 20.25 + 36

(H)² = 56.25

Take sqrt on both sides

H = √56.25

H = 7.5 inches

$\rule[225]{225}{2}$

OLga [1]2 years ago

4 0

7.5
Using the Pythagorean theorem you can determine this answer! Let me know if I’m wrong!

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The average and standard deviation of a measurement is 15.0 ± 5.0. if 100 students measured the value, how many would you expect

AURORKA [14]

Solution: We are given:

$\mu=15, \sigma=5$

Using the empirical rule, we have:

$\mu \pm \sigma$ covers 68% of data.

Also the percentage of values below mean = Percentage of values above mean = 50%

Now, let's find the z score for x=20

$z=\frac{20-15}{5}=1$

Therefore, the percentage of values greater than 1 standard deviation above mean $50\% - \frac{68\%}{2} =50\%-34\%=16%$

Expected number of students = 16% of 100 = 16

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3 years ago

Write the name of each angle arm<br> A to O to C

tester [92]

Answer:

∠AOC

Step-by-step explanation:

The statement "A to O to C" is talking about each of the points that form an angle. Each point is given a letter as its name.

The symbol ∠ <u>means angle</u>.

You need two lines to form an angle, and you need two points to form a line. By reusing one of the points, you need three points to form an angle. The point that is being reused is always the point in the middle, in this case, that point is called 'O'.

8 0

3 years ago

Find the surface area of x^2+y^2+z^2=9 that lies above the cone z= sqrt(x^@+y^2)

Mashcka [7]

The cone equation gives

$z=\sqrt{x^2+y^2}\implies z^2=x^2+y^2$

which means that the intersection of the cone and sphere occurs at

$x^2+y^2+(x^2+y^2)=9\implies x^2+y^2=\dfrac92$

i.e. along the vertical cylinder of radius $\dfrac3{\sqrt2}$ when $z=\dfrac3{\sqrt2}$ .

We can parameterize the spherical cap in spherical coordinates by

$\mathbf r(\theta,\varphi)=\langle3\cos\theta\sin\varphi,3\sin\theta\sin\varphi,3\cos\varphi\right\rangle$

where $0\le\theta\le2\pi$ and $0\le\varphi\le\dfrac\pi4$ , which follows from the fact that the radius of the sphere is 3 and the height at which the sphere and cone intersect is $\dfrac3{\sqrt2}$ . So the angle between the vertical line through the origin and any line through the origin normal to the sphere along the cone's surface is

$\varphi=\cos^{-1}\left(\dfrac{\frac3{\sqrt2}}3\right)=\cos^{-1}\left(\dfrac1{\sqrt2}\right)=\dfrac\pi4$

Now the surface area of the cap is given by the surface integral,

$\displaystyle\iint_{\text{cap}}\mathrm dS=\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dv\,\mathrm du$
$=\displaystyle\int_{u=0}^{u=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}9\sin v\,\mathrm dv\,\mathrm du$
$=-18\pi\cos v\bigg|_{v=0}^{v=\pi/4}$
$=18\pi\left(1-\dfrac1{\sqrt2}\right)$
$=9(2-\sqrt2)\pi$