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2 years ago

Which number in the monomial 215x18y3z21 needs to be changed to make it a perfect cube?

2 answers:

8 0

Remember
xyz=(x)(y)(z)
if x and y and z are all perfect cubes, xyz is also a perfect cube

remmeber
(x^n)^m=x^(mn)
so if it is perfect cube it can be factored into
(x^n)^3, such taht n is a whole number
basiclaly, see if the expoent is divisble by 3

all of them should be perfect cubes

215x^18y^3z^21

split them up to see which ones need changing

215=5*43, not a perfect cube
could be changed to 6^3, which is 216

x^18=(x^6)^3

y^3=y^3
z^21=(z^7)^3

the 215 needs to be changed

Tanya [424]2 years ago

6 0

Answer: E2020 = 215

Step-by-step explanation:

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4vir4ik [10]

<u>Answer:</u>

x = 4 (extraneous solution)

<u>Step-by-step explanation:</u>

$\frac { 4 } { x - 4 } = \frac { x } { x - 4 } - \frac { 4 } { 3 } \\ \frac { 4 } { x - 4 } - \frac { x } { x - 4 } = - \frac { 4 } { 3 } \\ \frac { 4 - x } { x - 4 } = - \frac { 4 } { 3 } \\ 3 ( 4 - x ) = - 4 ( x - 4 ) \\ 1 2 - 3 x = - 4 x + 1 6 \\ 4 x - 3 x = 1 6 - 1 2 \\ x = 4 \\$

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How many calories will Darrel burn in 1 minute while kayaking enter the answer below in the box 50/200, 100/400, 150/600, 200/80

ra1l [238]

Answer:100/400

Step-by-step explanation:

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1 year ago

Which are correct representations of the inequality -3(2x - 5) <5(2 - x)? Select two options.

Sonja [21]

Answer:

The third and fourth options.

Step-by-step explanation:

You simplify the inequality first...

-6x + 15 < 10 - 5x

-x < -5

x > 5

The first option is incorrect since it is less than.

The second option is basically -5x < 15; x > -3; that's incorrect.

The third option is basically -x < -5; x > 5; that is correct!

The fourth option is correct since it shows more than 5.

The fifth option is incorrect because it shows less than -5.

Hope this helps!

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1 year ago

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Simplify the expression. 3⁹ x 3⁶ x 3⁶

aev [14]

Answer:

I might be wrong but I think this is the answer 10460353203

Step-by-step explanation:

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1 year ago

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Find the area between y=4sinx and y=10cosx over the interval [0,π]. Sketch the curves if necessary.

FromTheMoon [43]

$\sin x\ge0$ for all $x\in[0,\pi]$ , while $\cos x>0$ for $0$ and $\cos x$ for $\dfrac\pi2$ . So you already know that $4\sin x>10\cos x$ over the second half of the interval.

In the first half, $4\sin x$ is an increasing function, from $4\sin0=0$ to $4\sin\dfrac\pi2=4$ . Meanwhile $10\cos x$ is a decreasing function, from $10\cos0=10$ to $10\cos\dfrac\pi2=0$ . Therefore there must be a point between 0 and $\dfrac\pi2$ where the two curves intersect. So we find it:

$4\sin x=10\cos x\implies \tan x=\dfrac{10}4=\dfrac52\implies x=\arctan\dfrac52\approx1.1903$

So we know that $10\cos x>4\sin x$ in $\left[0,\arctan\dfrac52\right)$ , and $4\sin x>10\cos x$ in $\left(\arctan\dfrac52,\pi\right]$ .

The area between the curves is then

$\displaystyle\int_0^{\arctan(5/2)}(10\cos x-4\sin x)\,\mathrm dx+\int_{\arctan(5/2)}^\pi(4\sin x-10\cos x)\,\mathrm dx=4\sqrt{29}$

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2 years ago