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3 years ago

Find the measure of CP

Mathematics

1 answer:

dmitriy555 [2]3 years ago

6 0

Given:

A figure ABCP.

To find:

The measure of CP.

Solution:

In triangles ABP and CBP,

$\angle PAB\cong \angle PCB$ (Given right angles)

$\angle ABP\cong \angle CBP$ (Given)

$BP=BP$ (Common side)

Now,

$\Delta ABP\cong \Delta CBP$ (By AAS property of congruence)

We know that the corresponding parts of congruent triangles are congruent.

$AP=CP$

$8=CP$

Therefore, the measure of CP is 8 units.

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14. Which of the following pairs of equation are equivalent

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Answer:

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Dada la ecuacion 25x2 + 4y2 = 100, determina las coordenadas de los vertices, focos, las longitudes de los respectivos ejes mayo

Likurg_2 [28]

Answer:

The given equation is

$25x^{2} +4y^{2}=100$

Which represents an elipse.

To find its elements, we need to divide the equation by 100

$\frac{25x^{2} +4y^{2} }{100} =\frac{100}{100} \\\frac{x^{2} }{4} +\frac{y^{2} }{25} =1$

Where $a^{2} =25$ and $b^{2}=4$ . Remember that the greatest denominator is $a$ , and the least is $b$ . So, we extract the square root on each equation.

$a=5$ and $b=2$ .

In a elipse, we have a major axis and a minor axis. In this case, the major axis is vertical and the minor axis is horizontal, that means this is a vertical elipse.

The length of the major axis is $2a=2(5)=10$ .

The length of the minor axis is $2b=2(2)=4$ .

The vertices are $(0,5);(0,-5)$ and $(2,0);(-2,0)$ .

Now, the main parameters of an elipse are related by

$a^{2}=b^{2} +c^{2}$ , which we are gonna use to find $c$ , the parameter of the focus.

$c=\sqrt{a^{2}-b^{2} }=\sqrt{25-4}=\sqrt{21}$

So, the coordinates of each focus are $(0,\sqrt{21})$ and $(0,-\sqrt{21})$

The eccentricity of a elipse is defined

$e=\frac{c}{a}=\frac{\sqrt{21} }{5} \approx 0.92$

The latus rectum is defined

$L=\frac{2b^{2} }{a}=\frac{2(4)}{5} =\frac{8}{5} \approx 1.6$

Finally, the graph of the elipse is attached.

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3 years ago

1. The temperature was 80 ℉ in Arizona and then fell 20 ℉ during the night.

notsponge [240]

Answer:

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Step-by-step explanation:

simple math

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3 years ago

Find constants a and b such that the function y = a sin(x) + b cos(x) satisfies the differential equation y'' + y' − 5y = sin(x)

vichka [17]

Answers:

a = -6/37

b = -1/37

============================================================

Explanation:

Let's start things off by computing the derivatives we'll need

$y = a\sin(x) + b\cos(x)\\\\y' = a\cos(x) - b\sin(x)\\\\y'' = -a\sin(x) - b\cos(x)\\\\$

Apply substitution to get

$y'' + y' - 5y = \sin(x)\\\\\left(-a\sin(x) - b\cos(x)\right) + \left(a\cos(x) - b\sin(x)\right) - 5\left(a\sin(x) + b\cos(x)\right) = \sin(x)\\\\-a\sin(x) - b\cos(x) + a\cos(x) - b\sin(x) - 5a\sin(x) - 5b\cos(x) = \sin(x)\\\\\left(-a\sin(x) - b\sin(x) - 5a\sin(x)\right) + \left(- b\cos(x) + a\cos(x) - 5b\cos(x)\right) = \sin(x)\\\\\left(-a - b - 5a\right)\sin(x) + \left(- b + a - 5b\right)\cos(x) = \sin(x)\\\\\left(-6a - b\right)\sin(x) + \left(a - 6b\right)\cos(x) = \sin(x)\\\\$

I've factored things in such a way that we have something in the form Msin(x) + Ncos(x), where M and N are coefficients based on the constants a,b.

The right hand side is simply sin(x). So we want that cos(x) term to go away. To do so, we need the coefficient (a-6b) in front of that cosine to be zero

a-6b = 0

a = 6b

At the same time, we want the (-6a-b)sin(x) term to have its coefficient be 1. That way we simplify the left hand side to sin(x)

-6a -b = 1

-6(6b) - b = 1 .... plug in a = 6b

-36b - b = 1

-37b = 1

b = -1/37

Use this to find 'a'

a = 6b

a = 6(-1/37)

a = -6/37

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2 years ago