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brilliants [131]
3 years ago
12

Write an equation of the circle with center (-7, -4) and radius 6 .

Mathematics
1 answer:
siniylev [52]3 years ago
8 0

Answer:

(x+7)^2 + (y+4)^2 = 6^2

or

(x+7)^2 + (y+4)^2 = 36

Step-by-step explanation:

We can write the equation of a circle in the form

(x-h)^2 + (y-k)^2 = r^2

Where (h,k) is the center and r is the radius

(x--7)^2 + (y--4)^2 = 6^2

(x+7)^2 + (y+4)^2 = 6^2

or

(x+7)^2 + (y+4)^2 = 36

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Since A‘(-1,1) is on the line y=-x, reflecting it about the line doesn't change its coordinate, so A's coordinate is (-1,1). Reflecting B'(-2,1) about y=-x switches both the sign and the number of x and y, so B=(-1,2), and C'(-1,0) becomes C(0,1). Drawing out the graph with the points and the line helps.
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3 years ago
Thirteen equals one plus four s
victus00 [196]
13=1+4s
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6 0
3 years ago
The diffrence of twice a number and 3 is -21
Alika [10]

Answer:

x = - 9

Step-by-step explanation:

Step 1:

2x - 3 = - 21    Equation

Step 2:

2x = - 18     Add 3 on both sides

Step 3:

- 18 ÷ 2    Divide

Answer:

x = - 9

Hope This Helps :)

8 0
4 years ago
In parallelogram LMNP if PQ=21 find QM.
Simora [160]

Answer:

QM = 21

Step-by-step explanation:

The diagonals of a parallelogram bisect each other, then

QM = PQ = 21

8 0
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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