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Alenkasestr [34]
1 year ago
8

align="absmiddle" class="latex-formula">can you help me solve this problem please? I also need help knowing how to work it out.

Mathematics
1 answer:
icang [17]1 year ago
8 0

We are given the following expression:

\sqrt[3]{875}

We can factor 875 as:

\sqrt[3]{125\times7}

Now we will use the following property:

\sqrt[x]{a\times b}=\sqrt[x]{a}\times\sqrt[x]{b}

Applying the property:

\sqrt[3]{125}\times\sqrt[3]{7}

Solving the cubic root on the left:

5\sqrt[3]{7}

Since we get simplify any further this is the answer.

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3 0
3 years ago
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NEED HELP PLEASE I NEED TO GET THIS ANSWER RIGHT!!! AND I NEED TO KNOW HOW YOU DID IT.
creativ13 [48]
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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
What is the midpoint of the segment
Solnce55 [7]

(-3,4)(-6,-1)

The easy way

look at the y's = 4 and -1 the difference is 5 now /2 = 2.5

so the y midpoint is 2.5 from each point or 1.5

look at the x's = -3 and -6 the differebce is 3 now /2 = 1.5

so its 1.5 from each point or -4.5

so midpoint is (-4.5, 1.5) or (-9/2, 3/2)  or C


the formula way:

for x

x1+x2/2

-3+-6/2

-9/2 (which is -4.5)

for y

y1+y2/2

4+-1/2

3/2 (which is 1.5)



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3 years ago
9. Ella bought a $379 tablet for 15% off. The
Stels [109]
If she would have waited a day she would have saved D) $64.43 more. This is because the original sale price comes to $322.15 and then we find the answer by taking this number and multiplying it by the additional sale and get $64.43.
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