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2 years ago

Need this answered quick

Mathematics

1 answer:

melisa1 [442]2 years ago

7 0

95.7942857 pie form hope this is right got it off the internet so yeah

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Select the vertical asymptote(s) of the function <img src="https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7B%28x%2B6%29%28x-1%29%7D%

Bess [88]

Answer:

As the described function, we want to find vertical asymtotes, we find the value of x so that the denominator is equal to 0.

Here, (x - 2)(x + 6) = 0

=> x = 2, x =-6

=> Option C is correct.

Hope this helps!

4 0

2 years ago

Read 2 more answers

PLEASE HELP !! I'LL GIVE YOU BRAINLIEST !!

GuDViN [60]

Answer:

Step-by-step explanation:

tan45° = 10/x

=> 1 = 10/x

=> x = 10

=> y^2 = 10^2 + 10^2 = 200

=> y = 100*sqrt(2)

=> b

5 0

2 years ago

Please help !!! I’m stuck

Kruka [31]

Answer:

2/3 ≈ 0.6667

Step-by-step explanation:

apply soh cah toa

sine=opposite/hypo

sine=4/6

sine=2/3

4 0

2 years ago

The proceeds of a $7,500.00, 10 percent simple discounted note for 85 days is

SIZIF [17.4K]

Answer:

7325.34

Step-by-step explanation:

5 0

2 years ago

The amount A of the radioactive element radium in a sample decays at a rate proportional to the amount of radium present. Given

slavikrds [6]

Answer:

a) $\frac{dm}{dt} = -k\cdot m$ , b) $m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$ , c) $m(t) = 10\cdot e^{-\frac{t}{2438.155} }$ , d) $m(300) \approx 8.842\,g$

Step-by-step explanation:

a) Let assume an initial mass m decaying at a constant rate k throughout time, the differential equation is:

$\frac{dm}{dt} = -k\cdot m$

b) The general solution is found after separating variables and integrating each sides:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $\tau$ is the time constant and $k = \frac{1}{\tau}$

c) The time constant is:

$\tau = \frac{1690\,yr}{\ln 2}$

$\tau = 2438.155\,yr$

The particular solution of the differential equation is:

$m(t) = 10\cdot e^{-\frac{t}{2438.155} }$

d) The amount of radium after 300 years is:

$m(300) \approx 8.842\,g$

4 0

3 years ago

Read 2 more answers