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jekas [21]
3 years ago
13

Two numbers are in the ratio 3:2. if 5 subtracted to each

Mathematics
1 answer:
eduard3 years ago
4 0

Answer:

3 and 2

Step-by-step explanation:

The ratio of the 2 numbers = 3 : 2 = 3x : 2x ( x is a multiple )

When 5 is subtracted from both , that is

3x - 5 : 2x - 5 = 2 : 3

Expressing the ratio in fractional form

\frac{3x-5}{2x-5} = \frac{2}{3} ( cross- multiply )

3(3x - 5) = 2(2x - 5) ← distribute both sides

9x - 15 = 4x - 10 ( subtract 4x from both sides )

5x - 15 = - 10 ( add 15 to both sides )

5x = 5 ( divide both sides by 5 )

x = 1

Thus the 2 numbers are

3x = 3(1) = 3 and 2x = 2(1) = 2

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What are the coordinates of points x,j,y, and w
Dmitry_Shevchenko [17]
X: 1, 2

J: 1, 5

Y: -3, 5

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3 years ago
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1. (a) Use the integral test to show that P[infinity] n=1 1/n4 converges. (b) Find the 10th partial sum, s10, of the series P[in
scZoUnD [109]

Answer:

Step-by-step explanation:

a) \int\limits^{\infty} _1 {\frac{1}{n^4} } \, dn\\ =\frac{n^{-3} }{-3}

Substitute limits to get

= \frac{1}{3}

Thus converges.

b) 10th partial sum =

\int\limits^{10} _1 {\frac{1}{n^4} } \, dn\\ =\frac{n^{-3} }{-3}

=\frac{-1}{3} (0.001-1)\\= 0.333

c) Z [infinity] n+1 1 /x ^4 dx ≤ s − sn ≤ Z [infinity] n 1 /x^ 4 dx, (1)

where s is the sum of P[infinity] n=1 1/n4 and sn is the nth partial sum of P[infinity] n=1 1/n4 .

(question is not clear)

3 0
3 years ago
Find the absolute extrema for f(x,y)=4-x^2-y^4+1/2y^2 over the closed disk D:x^2+y^2 is less than or equal to 1
algol [13]

Find the critical points of f(x,y):

\dfrac{\partial f}{\partial x}=-2x=0\implies x=0

\dfrac{\partial f}{\partial y}=y-4y^3=y(1-4y^2)=0\implies y=0\text{ or }y=\pm\dfrac12

All three points lie within D, and f takes on values of

\begin{cases}f(0,0)=4\\f\left(0,-\frac12\right)=\frac{65}{16}\\f\left(0,\frac12\right)=\frac{65}{16}\end{cases}

Now check for extrema on the boundary of D. Convert to polar coordinates:

f(x,y)=f(\cos t,\sin t)=g(t)=4-\cos^2-\sin^4t+\dfrac12\sin^2t=3+\dfrac32\sin^2t-\sin^4t

Find the critical points of g(t):

\dfrac{\mathrm dg}{\mathrm dt}=3\sin t\cos t-4\sin^3t\cos t=\sin t\cos t(3-4\sin^2t)=0

\implies\sin t=0\text{ or }\cos t=0\text{ or }\sin t=\pm\dfrac{\sqrt3}2

\implies t=n\pi\text{ or }t=\dfrac{(2n+1)\pi}2\text{ or }\pm\dfrac\pi3+2n\pi

where n is any integer. There are some redundant critical points, so we'll just consider 0\le t< 2\pi, which gives

t=0\text{ or }t=\dfrac\pi3\text{ or }t=\dfrac\pi2\text{ or }t=\pi\text{ or }t=\dfrac{3\pi}2\text{ or }t=\dfrac{5\pi}3

which gives values of

\begin{cases}g(0)=3\\g\left(\frac\pi3\right)=\frac{57}{16}\\g\left(\frac\pi2\right)=\frac72\\g(\pi)=3\\g\left(\frac{3\pi}2\right)=\frac72\\g\left(\frac{5\pi}3\right)=\frac{57}{16}\end{cases}

So altogether, f(x,y) has an absolute maximum of 65/16 at the points (0, -1/2) and (0, 1/2), and an absolute minimum of 3 at (-1, 0).

5 0
3 years ago
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