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2 years ago

The radius of the circle whose equation is (x + 4)² + (y - 2)² = 36 is

Mathematics

1 answer:

bija089 [108]2 years ago

8 0

Answer: 6

The equation given to you is in the form (x-h)^2 + (y-k)^2 = r^2

The right hand side 36 is in the place of r^2, so r^2 = 36 leading to r = 6 after you apply the square root to both sides.

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2 years ago

(1 point) consider the function f(t)=⎧⎩⎨⎪⎪⎪⎪0,−5,−6,6,t<00≤t<11≤t<7t≥7;f(t)={0,t<0−5,0≤t<1−6,1≤t<76,t≥7; 1. wr

sergij07 [2.7K]

$f(t)=\begin{cases}0&\text{for }t$

Recall that

$u(t)=\begin{cases}0&\text{for }t$

Take it one piece at a time. For $t\ge0$ , we can scale $u(t)$ by -5:

$-5u(t)=\begin{cases}0&\text{for }t$

If we shift the argument by 1 and scale by -5, we have

$-5u(t-1)=\begin{cases}0&\text{for }t$

so if we subtract this from $-5u(t)$ , we'll end up with

$-5u(t)+5u(t-1)=\begin{cases}0&\text{for }t$

For the next piece, we can add another scaled and shifted step like

$-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t$

so that

$-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t$

For the last piece, we add one more term:

$6u(t-7)=\begin{cases}0&\text{for }t$

and so putting everything together, we get $f(t)$ :

$f(t)\equiv-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)+6u(t-7)$
$f(t)\equiv-5u(t)-u(t-1)+12u(t-7)$

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2 years ago