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pychu [463]
3 years ago
12

331/3% of what number is 57

Mathematics
1 answer:
Alika [10]3 years ago
8 0
3 % = 3 / 100 = 0.03
let number to be found = Y
of always means multiply

so....... 331 / 0.03 x Y = 57
11033.33 x Y = 57
Y = 57 / 11033.33
Y = 5.16 x 10 power -3

HOPE IT HELPS !!
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Help plz I don’t get it
laila [671]

(I will do the first two parts of each question)

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2. 75's factors are: 1, 3, 5, 15, 25, and 75. So, there can be 75 piles of 1, 25 piles of 3, 15 piles of 5, etc. (3 more)

3a. 3*1=3, 3*2=6, 3*3=9, etc. so the first 5 multiples of 3 are: 3, 6, 9, 12, 15

3b. Same as before, so: 10, 20, 30, 40, 50

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tatuchka [14]

Answer:

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Which equation best describes the line of best fit?
Andrej [43]
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8 0
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Seema covered 3 4/13(three four by thirteen)km by bus,2 5/13(Two five by thirteen)km by metro and 1 2/13(One two by thirteen)km
Ad libitum [116K]

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Marine biologists have determined that when a shark detectsthe presence of blood in the water, it will swim in the directionin w
siniylev [52]

Solution :

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$C(x,y) = e^{-(x^2+2y^2)/10^4}$

are actually the curves

$e^{-(x^2+2y^2)/10^4}=k$

where k is a positive constant.

The equation is equivalent to

$x^2+2y^2=K$

$\Rightarrow \frac{x^2}{(\sqrt K)^2}+\frac{y^2}{(\sqrt {K/2})^2}=1, \text{ where}\ K = -10^4 \ln k$

which is a family of ellipses.

We sketch the level curves for K =1,2,3 and 4.

If the shark always swim in the direction of maximum increase of blood concentration, its direction at any point would coincide with the gradient vector.

Then we know the shark's path is perpendicular to the level curves it intersects.

b). We have :

$\triangledown C= \frac{\partial C}{\partial x}i+\frac{\partial C}{\partial y}j$

$\Rightarrow \triangledown C =-\frac{2}{10^4}e^{-(x^2+2y^2)/10^4}(xi+2yj),$ and

$\triangledown C$ points in the direction of most rapid increase in concentration, which means $\triangledown C$ is tangent to the most rapid increase curve.

$r(t)=x(t)i+y(t)j$  is a parametrization of the most $\text{rapid increase curve}$ , then

$\frac{dx}{dt}=\frac{dx}{dt}i+\frac{dy}{dt}j$ is a tangent to the curve.

So then we have that $\frac{dr}{dt}=\lambda \triangledown C$

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$\frac{dy}{y}=2\frac{dx}{x}$

$\int\frac{dy}{y}=2\int \frac{dx}{x}$

$\ln y=2 \ln x$

⇒ y = kx^2 for some constant k

but we know that $y(x_0)=y_0$

$\Rightarrow kx_0^2=y_0$

$\Rightarrow k =\frac{y_0}{x_0^2}$

∴ The path of the shark will follow is along the parabola

$y=\frac{y_0}{x_0^2}x^2$

$y=y_0\left(\frac{x}{x_0}\right)^2$

7 0
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