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3 years ago

Help plzzzSolution set of 25x2 – 1 = 0 is ____________.

Mathematics

1 answer:

kari74 [83]3 years ago

8 0

Answer:

x = 1/5 , and x = - 1/5

Step-by-step explanation:

25x² – 1 = 0

5²x² - 1² = 0

(5x - 1)(5x + 1) = 0

5x -1 = 0, 5x + 1 = 0

x = 1/5 , x = - 1/5

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Answer:

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3 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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3 years ago

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Mademuasel [1]

Answer: 32,000

Step-by-step explanation:

the thousands place (1000) is the fourth number to the left of the decimal, so round that one either up or down based on the number just one spot closer to the decimal, in this case, the hundreds place (100).

so we have 32,420

we can ignore the numbers that are going to become zeros as a result of the rounding . .

32,400

let's round 32,XXX either up or down based on the value in the hundreds place. if that value is 5 or greater, than we round up. if it is less than 5, than we round down.

4 is in the hundreds place, so we round down

32,000 is the answer

Hope this helps!!! Good luck!!! ;)

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11Alexandr11 [23.1K]

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2 years ago

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Answer:

Step-by-step explanation:

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2 years ago

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