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3 years ago

The equation of a circle is (x + 4)2 + (y + 6)2 = 16. Determine the length of the radius.

Mathematics

1 answer:

Viefleur [7K]3 years ago

5 0

Answer:

see explanation

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

(x + 4)² + (y + 6)² = 16 is in this form

with r² = 16 ⇒ r = $\sqrt{16}$ = 4

---------------------------------------------------------------

given (h, k) = (5, 2) and r = 20, then

(x - 5)² + (y - 2)² = 400 ← represents the delivery area

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3 years ago

2 jars of spaghetti sauce $3<br> What is the constant poportion of this problem

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The constant proportion would be 1.50 for 1

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3 years ago

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Please help me with the below question.

VMariaS [17]

By letting

$y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}$

we get derivatives

$y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}$

$y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}$

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

$5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0$

Examine the lowest degree term $\left(x^{r-1}\right)$ , which gives rise to the indicial equation,

$5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0$

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients $c_k$ is

$(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}$

so that with r = 4/5, the coefficients are governed by

$c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}$

c) Starting with $c_0=1$ , we find

$c_1 = -\dfrac{c_0}5 = -\dfrac15$

$c_2 = -\dfrac{c_1}{10} = \dfrac1{50}$

so that the first three terms of the solution are

$\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}$

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2 years ago

HELPPP

Mazyrski [523]

Answer:

Hello,

Step-by-step explanation:

y-intercept: x=0 ,f(0)=-10

$x^3-8x^2+17x-10\\\\=x^3-5x^2-3x^2+15x+2x-10\\\\=x^2(x-5)-3x(x-5)+2(x-5)\\\\=(x-5)(x^2-3x+2)\\\\=(x-5)(x^2-x-2x+2)\\\\=(x-5)(x(x-1)-2(x-1))\\\\=(x-5)(x-1)(x-2)\\$

x-intercepts are: 5,1,2.

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3 years ago

2.4x10^2+7.6x10^-1 awnser quick

timurjin [86]

= 2.4 x 100 x 7.6 x 0.1
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3 years ago

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