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Veseljchak [2.6K]
2 years ago
7

What is another word for a dependent variable in science?.

Mathematics
1 answer:
blondinia [14]2 years ago
5 0

it is sometimes called as response variable

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What is 3/13 of 91 ?
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▹ Answer

<em>21</em>

▹ Step-by-Step Explanation

3/13 x 91/1

91/13 = 7

3/1 = 3

3 * 7 = 21

Hope this helps!

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Find a rational zero of the polynomial function and use it to find all the zeros of the function. f(x) = x4 + 3x3 - 5x2 - 9x - 2
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Polynomials in the fourth degree are called quartic equations. In solving the roots of polynomials, there are techniques available. For quadratic equations, you use the quadratic formula. For cubic equations, you use the scientific calculator. But for quartic equations and higher, it is very complex. The method is very lengthy and can get very messy because you introduce a lot variables. So, I suggest you do the easiest method to estimate the roots.

Graph the equation by plotting arbitrary points. The graph looks like that in the figure. The points at which the curve passes the x-axis are the solution which are encircled in red.In approximation, the rational roots or zero's are -3.73, -1, -0.28 and 2.

3 0
3 years ago
For a boat to float in a tidal bay, the water must be at least 2.5 meters deep. The depth of water around the boat, ????(????),
Effectus [21]

Answer:

a. Period, T = 12.57 hours. b. Latest time = 294.16 hours, 6 a.m 12 days later

Step-by-step explanation:

For a boat to float in a tidal bay, the water must be at least 2.5 meters deep. The depth of water around the boat, d(t), in meters, where t is measured in hours since midnight, is d(t) = 5 + 4.6sin(0.5t). (a) What is the period of the tides in hours? (b) If the boat leaves the bay at midday, what is the latest time it can return before the water becomes too shallow?

a. The period, T of the tides is gotten from ω = 2π/T, where ω is the angular frequency of the wave and T its period. Comparing the sine part of equation of the tides with the general equqtion of a sine wave, Asinωt, thus Asinωt = 4.6sin(0.5t),

so ω = 0.5 rad/s.

Equating this to 2π/T implies

0.5 = 2π/T

therefore, T = 2π/0.5 = 12.566 hours ≈ 12.57 hours

b. The tide becomes too low if it is less than 2.5 m i.e d(t) < 2.5 m. So, equating d(t) as 2.5 in the tidal equation, we have d(t) = 5 + 4.6sin(0.5t).

2.5< 5+ 4.6sin(0.5t)

2.5-5 < 4.6sin(0.5t)

-2.5 < 4.6sin(0.5t)

-2.5/4.6 < sin(0.5t)

-0.543 < sin(0.5t)

sin⁻¹(-0.543)<0.5t

-32.92<0.5t. Since we cannot have negative time, we add 180 to -32.92 to give 147.08

147.08<0.5t

t>147.08/0.5=294.16 hours

So, if t is greater than 294.16 hours, the water will be shallow. That is, below 2.5 m. which is 12 days 6hours 8 min 38 s which is 6 a.m 12 days later

3 0
3 years ago
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