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3 years ago

12

Select whether the triangle is inscribed in the circle, circumscribed about the circle, or neither.

1 answer:

mash [69]3 years ago

8 0

The correct answer is neither

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Evaluate f(x) = 3/6x+6 when x = -1<br><br> a) undefined <br> b) 0<br> c) 3<br> d) 3/36

Alekssandra [29.7K]

Answer:

a) Undefined.

Step-by-step explanation:

$f(x) = \dfrac 3{6x+6} = \dfrac{3}{6(x+1)} = \dfrac{1}{2(x+1)}\\\\\\f(-1) =\dfrac{1}{2(-1+1)}= \dfrac{1}{0}$

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2 years ago

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Step-by-step explanation:

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3 years ago

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Julli [10]

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Step-by-step explanation:

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3 years ago

What is the value of x in the equation ?

zheka24 [161]

Answer:

$x=2$

Step-by-step explanation:

This is our equation

$2.5\left(6x-4\right)=10+4\left(1.5+0.5x\right)$

First off we can distribute the 2.5 and the 4

Now our equation is $15x-10=10+6+2x$

subtract $2x$ from both sides

$13x-10=16$

add $10$ to both sides

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Divide both sides by $13$

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3 years ago

I need Help with calculus please, thank you! 10points

natta225 [31]

3) We have

$f(x) = \sec\left(\dfrac{\pi x}2\right) = \dfrac1{\cos\left(\frac{\pi x}2\right)}$

which has vertical asymptotes (i.e. infinite discontinuities) whenever the denominator is zero. This happens for

$\cos\left(\dfrac{\pi x}2\right) = 0$

$\implies \dfrac{\pi x}2 = \cos^{-1}(0) + 2n\pi \text{ or } \dfrac{\pi x}2 = -\cos^{-1}(0) + 2n\pi$

(where $n$ is any integer)

$\implies \dfrac{\pi x}2 = \dfrac\pi2 + 2n\pi \text{ or } \dfrac{\pi x}2 = -\dfrac\pi2 + 2n\pi$

$\implies x = 1 + 4n \text{ or } x = -1 + 4n$

So the graph of $f(x)$ has vertical asymptotes whenever $x=4n\pm1$ and $n\in\Bbb Z$ .

4) Given

$h(t) = \begin{cases} t^3+1 & \text{if } t$

we have the one-sided limits

$\displaystyle \lim_{t\to1^-} h(t) = \lim_{t\to1} (t^3+1) = 1^3+1 = 2$

and

$\displaystyle \lim_{t\to1^+} h(t) = \lim_{h\to1} \frac{t+1}2 = \frac{1+1}2 = 1$

The one-sided limits don't match, so the two-sided limit $L$ does not exist. In other words, the limit does not exist at $x=1$ because the function approaches different values from the left and right side of $x=1$ .

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2 years ago