If M ⊥ N and L ∥ M, then _____

Mathematics

1 answer:

joja [24]3 years ago

5 0

Answer:

L ⊥ N

Step-by-step explanation:

Since M and N are perpendicular, and L is parallel to M, anything that's perpendicular to M is also perpendicular to L. In fact, since we have parallel lines, we now have many sets of congruent angles, but the only ones we know the actual measurements of are the right angles from the perpendicular lines.

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In which sentence is the literary device litotes used?

Lilit [14]

It’s letter D. “Litotes is a figure of speech and a form of understatement in which a sentiment is expressed ironically by negating its contrary.” Hope this helps

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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

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

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3 years ago

A data set lists earthquake depths. The summary statistics are

irinina [24]

Answer:

Step-by-step explanation:

The summary of the given statistics data include:

sample size n = 400

sample mean $\overline x$ = 6.86

standard deviation = 4.37

Level of significance ∝ = 0.01

Population Mean $\mu$ = 6.00

Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.

To start with the hypothesis;

The null and the alternative hypothesis can be computed as :

$H_o: \mu = 6.00 \\ \\ H_1 : \mu \neq 6.00$