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seropon [69]
3 years ago
9

Plssss help me i'll give u 40 points evaluate: 3/5^-2

Mathematics
2 answers:
Rama09 [41]3 years ago
8 0

Answer:

75

Step-by-step explanation:

3/5^-2= 75 :)

Nikitich [7]3 years ago
5 0

Answer:

75

Step-by-step explanation:

5^-2 = 5*5 = 25/1

because it's (-) you need to flip it so 1/25

3/1 ÷ 1/25 = 3/1 × 25/1

because when you divide fraction you need to keep the first fraction as it is and change the symbol from (÷) to (×) and flip the second fraction. (KCF simple way to memorise it)

So  3/1 × 25/1 = 75/1

75/1 is same as 75

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I think the math question u are asking is y=mx+b
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3 years ago
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If each light fixture on a job requires 4 lamps and each room requires 16 fixtures, how many lamps will be required for 6 rooms?
kaheart [24]

Answer:

384 lamps

Step-by-step explanation:

This is simply a multiplication problem. From the question, we know that each fixture needs 4 lamps with a single room needing 16 fixtures.

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7 0
3 years ago
What is the answer 21 11/12 + 17 2/3 :P
jekas [21]
The answer would be 39 7/12
4 0
4 years ago
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A circle has a diameter of 4 m.
Margaret [11]
A= π r ²

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7 0
3 years ago
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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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