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4 years ago

PLZ HELP! IM GIVING EXTRA POINTS!

Mathematics

2 answers:

horsena [70]4 years ago

8 0

3 (4r+y) if it dont help let me know

zhenek [66]4 years ago

7 0

I hope this helps you

3 and (4r+y)

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A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 29 ft/s. Its height

Ilia_Sergeevich [38]

Answer:

$\overline{v}_{@\Delta t=0.01s}=-15.22ft/s, \overline{v}_{@\Delta t=0.005s}=-15.11ft/s, \overline{v}_{@\Delta t=0.002s}=-15.044ft/s, \overline{v}_{@\Delta t=0.001s}=-15.022ft/s$

Step-by-step explanation:

Now, in order to solve this problem, we need to use the average velocity formula:

$\overline{v}=\frac{y_{f}-y_{0}}{t_{f}-t_{0}}$

From this point on, you have two possibilities, either you find each individual $y_{f}, y_{0}, t_{f}, t_{0}$ and input them into the formula, or you find a formula you can use to directly input the change of times. I'll take the second approach.

We know that:

$t_{f}-t_{0}=\Delta t$

and we also know that:

$t_{f}=t_{0}+\Delta t$

in order to find the final position, we can substitute this final time into the function, so we get:

$y_{f}=29(t_{0}+\Delta t)-22(t_{0}+\Delta t)^{2}$

so we can rewrite our formula as:

$\overline{v}=\frac{29(t_{0}+\Delta t)-22(t_{0}+\Delta t)^{2}-y_{0}}{\Delta t}$

$y_{0}$ will always be the same, so we can start by calculating that, we take the provided function ans evaluate it for t=1s, so we get:

$y_{0}=29t-22t^{2}$

$y_{0}=29(1)-22(1)^{2}$

$y_{0}=7ft$

we can substitute it into our average velocity equation:

$\overline{v}=\frac{29(t_{0}+\Delta t)-22(t_{0}+\Delta t)^{2}-7}{\Delta t}$

and we also know that the initil time will always be 1, so we can substitute it as well.

$\overline{v}=\frac{29(1+\Delta t)-22(1+\Delta t)^{2}-7}{\Delta t}$

so we can now simplify our formula by expanding the numerator:

$\overline{v}=\frac{29+29\Delta t-22(1+2\Delta t+\Delta t^{2})-7}{\Delta t}$

$\overline{v}=\frac{29+29\Delta t-22-44\Delta t-22\Delta t^{2}-7}{\Delta t}$

we can now simplify this to:

$\overline{v}=\frac{-15\Delta t-22\Delta t^{2}}{\Delta t}$

Now we can factor Δt to get:

$\overline{v}=\frac{\Delta t(-15-22\Delta t)}{\Delta t}$

and simplify

$\overline{v}=-15-22\Delta t$

Which is the equation that will represent the average speed of the ball. So now we can substitute each period into our equation so we get:

$\overline{v}_{@\Delta t=0.01s}=-15-22(0.01)=-15.22ft/s$

$\overline{v}_{@\Delta t=0.005s}=-15-22(0.005)=-15.11ft/s$

$\overline{v}_{@\Delta t=0.002s}=-15-22(0.002)=-15.044ft/s$

$\overline{v}_{@\Delta t=0.001s}=-15-22(0.001)=-15.022ft/s$

5 0

3 years ago

Find the number of real number solutions for the equation x^2-3+8=0

svlad2 [7]

Find value of determinant.
The determinant is a term that is inside a square root and part of the quadratic formula used for solving quadratic equations.
Let determinant be 'd'.
If d >0, Then there are 2 real solutions
If d = 0, Then there is only 1 real solutions
If d < 0, Then there are 0 real solutions but 2 imaginary solutions

d = b^2 - 4ac

For this problem, the coefficients are:
a = 1, b = -3, c = 8

d = (-3)^2 - 4(1)(8)

d = 9 -32 = -23

d is less than 0, therefore there are 0 real solutions and 2 imaginary solutions.

This is true because you cannot take square root of a negative number.

5 0

3 years ago

Hai are yall real:/?

sammy [17]

Answer:

yes surprisingly

Step-by-step explanation:

4 0

3 years ago

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X^2 + x - 56 = 0<br> Solve for x

NISA [10]