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drek231 [11]
3 years ago
15

The volume of a cube with edge length e is modeled by the function v(e)=e3. Find a function for the edge of a cube given its vol

ume. HELP PLEASE!!!!

Mathematics
1 answer:
dsp733 years ago
4 0
V(e) = e^3.

For any value of e, (the edge length), just cube it to get the volume. Easy

V(-2) = -8
V(-1) = -1
V(0) = 0
V(1) = 1
V(2) = 8
V(3) would be 27...etc

It doesn't make sense to use a negative or zero value for e (since you can't have a length like that), so technically there should be a restriction of e > 0.

To find the edge length given the volume, we would just solve for e.

V = e³
Take the cube root of each side...
∛V = ∛e³
∛V = e

Write as e(V) = ∛V.
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What does<br> -$20 -$15 +$30 $2<br> Equal
ad-work [718]

Answer:

-3

Step-by-step explanation:

-20-15+30+2

-35+32

-3

4 0
3 years ago
-5y+8x=-18<br><br> 5y+2x=58<br><br> Solve for x and y
xxMikexx [17]

Answer:

in -5y+8x=-18 and 5y+2x=58

x = 4

in -5y+8x=-18 and 5y+2x=58

y = 10

7 0
3 years ago
Read 2 more answers
GI = 6, HI = 8, DT = 12, what is IE?​<br>a.12<br>b.16<br>c.8<br>d.4
harina [27]

Answer:

Option D

Step-by-step explanation:

By the property of intersecting chords,

If two chords are intersecting at a point inside a circle, product of lengths of the line segments on each chord is equal.

DI × EI = HI × GI

12 × EI = 8 × 6

EI = \frac{48}{12}

EI = 4

Therefore, Option D is the correct option.

5 0
3 years ago
Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.
Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

\mathbf{ 5e^{-x^2} cos (4x)  }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

Step-by-step explanation:

GIven that:

f(x) = 5e^{-x^2} cos (4x)

The Maclaurin series of cos x can be expressed as :

\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...  \ \ \ (1)}

\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0}  \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!}  -\dfrac{x^6}{3!}+... \ \ \  (2)}

From equation(1), substituting x with (4x), Then:

\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}

The first three terms of cos (4x) is:

\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}

\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}

\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}

Multiplying equation (2) with (3); we have :

\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }

\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }

\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }

\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

Finally , multiplying 5 with \mathtt{ e^{-x^2} cos (4x) } ; we have:

The first three nonzero terms in the Maclaurin series is

\mathbf{ 5e^{-x^2} cos (4x)  }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

7 0
3 years ago
On an assembly line, 3 out of every 100 light bulbs are found to be defective . If 5000 light bulbs are produced, how many would
STatiana [176]

Answer:

Step-by-step explanation:

There are 50 hundreds in 5000.

No of defective bulb expected = 50 * 3 = 150

3 0
3 years ago
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