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3 years ago

11 is the quotient of a number y and b

Mathematics

1 answer:

ICE Princess25 [194]3 years ago

5 0

Y/b=11 that’s a turned around phase question

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Two stacks of quarters are shown. Both stacks have the same amount of quarters.

Sophie [7]

Answer:

in geometry nation its in two solids of equal altitude, parallel to and at the same distance.

Step-by-step explanation:

just got it wrong :(

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3 years ago

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An equation is shown.<br> 18+ 5(2x + 5) = 12<br> Solve the equation for x.

andriy [413]

Step-by-step explanation:

18+ 5(2x + 5) = 12

18 + 10x + 25 = 12

43 + 10x = 12

10x = 12 -43

10x = -31

x = -3.1

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3 years ago

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AysviL [449]

Wow! So so so cool lol

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3 years ago

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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

13-2x = 4x-5<br> I need it explaining to me if so

worty [1.4K]

Answer:

x=3

Step-by-step explanation:

13-2x = 4x-5

-6x=-18

x=3

7 0

3 years ago

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