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3 years ago

What's the quadratic equation?

2 answers:

6 0

The standard form of the quadratic equation is ax^2 + bx + c. In vertex form, it's y = a(x – h)2 + k.
The graph of a quadratic equation is a parabola, which looks like a u. If <em>a</em> is negative then it's an upside down u.

o-na [289]3 years ago

4 0

in 8th grade my algebra teacher made us memorize the quadratic equation song (to the rhythm of pop goes the weasel):

x equals negative b plus or minus the square root of b squared minus 4ac all over 2a

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Find the side of a square whose diagonal is of the given measure.<br> Given = 15.2 cm

Neko [114]

Answer:

15cm

Step-by-step explanation:

First, a square's diagonal is basically the hypotenuse of a 45-45-90 triangle. a 45-45-90 triangle has a really special relationship, where the side length is x, and the diagonal is x $\sqrt{2}$ . So, the side length is 15.

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(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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3 years ago

Use models to show the sum of 120+205

maksim [4K]

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2 years ago

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