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3 years ago

9

Check all of the points PLEASE DUE TOMORROW

1 answer:

igor_vitrenko [27]3 years ago

7 0

The answers are E and F

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Ignore my work but solve the problem and check. Thank you!!

Free_Kalibri [48]

Yes good job you are correct

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2 years ago

<img src="https://tex.z-dn.net/?f=%20-%2020x%20-%205" id="TexFormula1" title=" - 20x - 5" alt=" - 20x - 5" align="absmiddle" cla

Aloiza [94]

It equals 4 hhnug flight

5 0

3 years ago

HELP MEE PLEASEEEE BROO !!!

Anika [276]

Answer:

5

Step-by-step explanation:

Scale factor is just the amount every side is multiplied by.

7*5=35

6 0

3 years ago

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

Translate into mathematical phrase.the weight (w) is more than 2kg

Ludmilka [50]

Answer:

w>2kg

Step-by-step explanation:

the weight is greater than 2kg.

3 0

3 years ago