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2 years ago

Solve for x A) 0.9 B) 160 C) 10 D) 1.1

Mathematics

1 answer:

Alex777 [14]2 years ago

3 0

Answer:

A) 0.9

Step-by-step explanation:

Since these chords are perpendicular, we know that the proportion of each length is equal. Thus 3/40 = x/12, which we can solve to find x = 0.9

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A substance with a half life is decaying exponentially. If there are initially 12grams of the substance and after 70 minutes the

dybincka [34]

Answer:

Step-by-step explanation:

Use the half life formula

$N=N_0e^{kt}$ where N is the amount after decay, No is the initial amount, e is Euler's number, k is the decay constant, and t is the time in minutes. For us, the first equation looks like this:

$7=12e^{k(70)}$ and we will solve that for k and then use that value of k in the second equation to find time.

Begin by dividing 7 by 12 to get

$\frac{7}{12}=e^{k(70)}$ Now take the natural log of both sides since the natural log and that e are inverses. The e then disappears:

$ln(\frac{7}{12})=k(70)$

Plug the left side into your calculator and then set it equal to the right side, giving us:

$-.5389965007=70k$

Divide both sides by 70 to get the k value of

k = -.00769995

Now the second equation looks like this:

$2=12e^{(-.00769995)t$ where t is our only unknown now that we know k.

Begin by dividing both sides by 12 to get

$\frac{1}{6}=e^{-.00769995t$ and take the natural log of both sides again to eliminate the e:

$ln(\frac{1}{6})=-.00769995t$

Take care of the left side on your calculator to get

-1.791759469 = -.00769995t and divide both sides by -.00769995 to get

t = 232.697 minutes

4 0

2 years ago

Particle 1 of charge q1 �� 5.00q and particle 2 of charge q2 �� 2.00q are fixed to an x axis. (a) as a multiple of distance

Naya [18.7K]

Assuming that the particle is the 3rd particle, we know that it’s location must be beyond q2; it cannot be between q1 and q2 since both fields point the similar way in the between region (due to attraction). Choosing an arbitrary value of 1 for L, we get

k q1 / d^2 = - k q2 / (d-1)^2

Rearranging to calculate for d:

(d-1)^2/d^2 = -q2/q1 = 0.4 
 d^2-2d+1 = 0.4d^2 
0.6d^2-2d+1 = 0
d = 2.72075922005613
d = 0.612574113277207

We pick the value that is > q2 hence,

d = 2.72075922005613*L

d = 2.72*L

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2 years ago

A vertical cylinder is leaking water at a rate of 4m3/sec. If the cylinder has a height of 10m and a radius of 2m, at what rate

Lyrx [107]

Answer:

Therefore the rate change of height is $\frac{1}{\pi}$ m/s.