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Daniel [21]
3 years ago
7

Drew bout a chemistry book for $30. Later that book was marked down by 20%. By how much has the value been decreased

Mathematics
1 answer:
Rasek [7]3 years ago
5 0
The answer is $24

To find the answer, multiply 30 by 20% than divide by 100%. The answer is 6. So, subtract $6 from $30 which equals $24
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What is being done to the variable in the equation 3 + g = -9?
slavikrds [6]

Answer:

g=-12

Step-by-step explanation:

3+g=-9

-3     -3

g=-12

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3 years ago
Solve the compound inequality shown below. Show steps. <br> -4&lt; 3x -1&lt;5
Liono4ka [1.6K]
Add 1 to all parts.
-4+1<3x-1+1<5+1
-3x<3x<6
-3/3<3x/3<6/3
=-1hope this helps!
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Dont comment under here if you and just commenting for points.. i really need to pass this:(
Scilla [17]

Answer:

the y intercept is -9/2 or -4 1/2

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4 years ago
Which set of ordered pairs does not show y as a function of x​
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Read 2 more answers
Let C be the curve of intersection of the parabolic cylinder x^2 = 2y, and the surface 3z = xy. Find the exact length of C from
Maslowich
I've attached a plot of the intersection (highlighted in red) between the parabolic cylinder (orange) and the hyperbolic paraboloid (blue).

The arc length can be computed with a line integral, but first we'll need a parameterization for C. This is easy enough to do. First fix any one variable. For convenience, choose x.

Now, x^2=2y\implies y=\dfrac{x^2}2, and 3z=xy\implies z=\dfrac{x^3}6. The intersection is thus parameterized by the vector-valued function

\mathbf r(x)=\left\langle x,\dfrac{x^2}2,\dfrac{x^3}6\right\rangle

where 0\le x\le 4. The arc length is computed with the integral

\displaystyle\int_C\mathrm dS=\int_0^4\|\mathbf r'(x)\|\,\mathrm dx=\int_0^4\sqrt{x^2+\dfrac{x^4}4+\dfrac{x^6}{36}}\,\mathrm dx

Some rewriting:

\sqrt{x^2+\dfrac{x^4}4+\dfrac{x^6}{36}}=\sqrt{\dfrac{x^2}{36}}\sqrt{x^4+9x^2+36}=\dfrac x6\sqrt{x^4+9x^2+36}

Complete the square to get

x^4+9x^2+36=\left(x^2+\dfrac92\right)^2+\dfrac{63}4

So in the integral, you can substitute y=x^2+\dfrac92 to get

\displaystyle\frac16\int_0^4x\sqrt{\left(x^2+\frac92\right)^2+\frac{63}4}\,\mathrm dx=\frac1{12}\int_{9/2}^{41/2}\sqrt{y^2+\frac{63}4}\,\mathrm dy

Next substitute y=\dfrac{\sqrt{63}}2\tan z, so that the integral becomes

\displaystyle\frac1{12}\int_{9/2}^{41/2}\sqrt{y^2+\frac{63}4}\,\mathrm dy=\frac{21}{16}\int_{\arctan(3/\sqrt7)}^{\arctan(41/(3\sqrt7))}\sec^3z\,\mathrm dz

This is a fairly standard integral (it even has its own Wiki page, if you're not familiar with the derivation):

\displaystyle\int\sec^3z\,\mathrm dz=\frac12\sec z\tan z+\frac12\ln|\sec x+\tan x|+C

So the arc length is

\displaystyle\frac{21}{32}\left(\sec z\tan z+\ln|\sec x+\tan x|\right)\bigg|_{z=\arctan(3/\sqrt7)}^{z=\arctan(41/(3\sqrt7))}=\frac{21}{32}\ln\left(\frac{41+4\sqrt{109}}{21}\right)+\frac{41\sqrt{109}}{24}-\frac98

4 0
4 years ago
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