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luda_lava [24]
2 years ago
13

Variable y varies directly with x2, and y = 96 when x = 4.

Mathematics
1 answer:
lakkis [162]2 years ago
4 0
Direct variation is y = kx, where k is the constant of variation.
 But now it says y varies directly with x2 (or 2x), so now the x in the equation is 2x.

The equation is y = k(2x)
 Now you find k.
y = 96 when x = 4.
(96) = k(2*4)
96 = k(8)
k = 12
 
The equation is now y = 12(2x)
To find the value of y when x=2, plug 2 into the equation you made. y = 12(2*2) y = 48
_________________
Now it's with a "quadratic variation," which is the same thing except x is squared.
The equation is y = kx^2
 
But y varies directly with x2 (same thing as 2x), so now it's y = k(2x)^2.
 
Now you find k by substituting y and x values that were given.
y = 180 when x = 6
(180) = k(2*6)^2
180 = k(12)^2
180 = k(144)
k = 1.25
k, 1.25, is the constant of variation.
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P=3w+6000000w^{-1}\\\\
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\textit{so the critical points are at }
\begin{cases}
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solve for "w", to see what critical points you get, and then run a first-derivative test on them, for the minimum

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8 0
3 years ago
Point B is a point of tangency. Find the radius r of OC.<br> А 9<br> B<br> r<br> n
insens350 [35]

Answer:

Radius, r, is 3.75

Step-by-step explanation:

From the given diagram,

AB = 9

AC = 6 + r

BC = r

Thus applying the Pythagoras theorem, we have;

/AC/^{2} = /AB/^{2} + /BC/^{2}

(6 + r)^{2} = 9^{2} + r^{2}

36 + 12r + r^{2} = 81 + r^{2}

collecting like terms,

12r = 81 - 36

12r = 45

r = \frac{25}{12}

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r = 3.75

The radius is 3.75.

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2 years ago
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If f(x) = x+9 what is f(#)?
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3 years ago
In the derivation of Newton’s method, to determine the formula for xi+1, the function f(x) is approximated using a first-order T
dimaraw [331]

Answer:

Part A.

Let f(x) = 0;

suppose x= a+h

such that f(x) =f(a+h) = 0

By second order Taylor approximation, we get

f(a) + hf'±(a) + \frac{h^{2} }{2!}f''(a) = 0

h = \frac{-f'(a) }{f''(a)} ± \frac{\sqrt[]{(f'(a))^{2}-2f(a)f''(a) } }{f''(a)}

So, we get the succeeding equation for Newton's method as

x_{i+1} = x_{i} + \frac{1}{f''x_{i}}  [-f'(x_{i}) ± \sqrt{f(x_{i})^{2}-2fx_{i}f''x_{i} } ]

Part B.

It is evident that Newton's method fails in two cases, as:

1.  if f''(x) = 0

2. if f'(x)² is less than 2f(x)f''(x)    

Part C.

In case  x_{i+1} is close to x_{i}, the choice that shouldbe made instead of ± in part A is:

f'(x) = \sqrt{f'(x)^{2} - 2f(x)f''(x)}  ⇔ x_{i+1} = x_{i}

Part D.

As given x_{i+1} = x_{i} = h

or                 h = x_{i+1} - x_{i}

We get,

f(a) + hf'(a) +(h²/2)f''(a) = 0

or h² = -hf(a)/f'(a)

Also,             (x_{i+1}-x_{i})² = -(x_{i+1}-x_{i})(f(x_{i})/f'(x_{i}))

So,                f(a) + hf'(a) - (f''(a)/2)(hf(a)/f'(a)) = 0

It becomes   h = -f(a)/f'(a) + (h/2)[f''(a)f(a)/(f(a))²]

Also,             x_{i+1} = x_{i} -f(x_{i})/f'(x_{i}) + [(x_{i+1} - x_{i})f''(x_{i})f(x_{i})]/[2(f'(x_{i}))²]

6 0
3 years ago
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