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4 years ago

!!30 points!! Are points A, B, and E collinear or noncollinear?

Mathematics

1 answer:

NeX [460]4 years ago

7 0

A. Noncolinear

ABE although they are in the same side, but they don't lie on the same line.

so its a noncolinear.

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What is the value of x

Sloan [31]

Answer:53

Step-by-step explanation:

6 0

4 years ago

Can a triangle with the following sides exist? 7, 8, 1

11Alexandr11 [23.1K]

Answer:

Step-by-step explanation:

Can a triangle with the following sides exist? 7, 8, 1

Answer is NO

Because the sum of any two sides has to be greater than the third side but here we have:

8 = 7+1

3 0

3 years ago

1:Under what condition will the line px+py+r=0 mat be a normal to the circke x²+y²+2gx+2fy+c=0

ahrayia [7]

Answer:

The normal overlaps with the diameter, so it passes through the center.

<u>Let's find the center of the circle:</u>

x² + y² + 2gx + 2fy + c = 0
(x + g)² + (y + f)² = c + g² + f²

<u>The center is:</u>

(-g, -f)

<u>Since the line passes through (-g, -f) the equation of the line becomes:</u>

p(-g) + p(-f) + r = 0
r = p(g + f)

This is the required condition

Rewrite equations and find centers and radius of both circles.

<u>Circle 1</u>

x² + y² + 2ax + c² = 0
(x + a)² + y² = a² - c²
The center is (-a, 0) and radius is √(a² - c²)

<u>Circle 2</u>

x² + y² + 2by + c² = 0
x² + (y + b)² = b² - c²
The center is (0, -b) and radius is √(b² - c²)

<u>The distance between two centers is same as sum of the radius of them:</u>

d = √(a² + b²)

<u>Sum of radiuses:</u>

√(a² - c²) + √(b² - c²)

<u>Since they are same we have:</u>

√(a² + b²) = √(a² - c²) + √(b² - c²)

<u>Square both sides:</u>

a² + b² = a² - c² + b² - c² + 2√(a² - c²)(b² - c²)
2c² = 2√(a² - c²)(b² - c²)

<u>Square both sides:</u>

c⁴ = (a² - c²)(b² - c²)
c⁴ = a²b² - a²c² - b²c² + c⁴
a²c² + b²c² = a²b²

<u>Divide both sides by a²b²c²:</u>

1/a² + 1/b² = 1/c²

Proved

6 0

3 years ago

Read 2 more answers

Given the functionf ( x ) = x^2 + 7 x + 10/ x^2 + 9 x + 20

vladimir1956 [14]

<em>x = -4 is a vertical asymptote for the function.</em>

<h2>Explanation:</h2>

The graph of $y=f(x)$ is a vertical has an asymptote at $x=a$ if at least one of the following statements is true:

$1) \ \underset{x\rightarrow a^{-}}{lim}f(x)=\infty\\ \\ 2) \ \underset{x\rightarrow a^{-}}{lim}f(x)=-\infty \\ \\ 3) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty \\ \\ 4) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty$

The function is:

$f(x)=\frac{x^2+7x+10}{x^2+9x+20}$

First of all, let't factor out:

$f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20} \\ \\ f(x)=\frac{x(x+5)+2(x+5)}{x(x+5)+4(x+5)} \\ \\ f(x)=\frac{(x+5)(x+2)}{(x+5)(x+4)} \\ \\ f(x)=\frac{(x+2)}{(x+4)}, \ x\neq 5$

From here:

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ right: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(-4^{+}+2)}{(-4^{+}+4)} \\ \\ \\ The \ numerator \ is \ negative \ and \ the \ denominator \\ is \ a \ small \ positive \ number. \ So: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=-\infty$

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ left: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(-4^{-}+2)}{(-4^{-}+4)} \\ \\ \\ The \ numerator \ is \ a \ negative \ and \ the \ denominator \\ is \ a \ small \ negative \ number \ too. \ So: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=+\infty$