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velikii [3]
2 years ago
14

3k^2=11+6k : complete the square

Mathematics
1 answer:
WARRIOR [948]2 years ago
4 0

Answer:

3(k - 1)² = 12

Step-by-step explanation:

To complete the square, write the equation standard form and make the a term's coefficient 1.

3k² = 11 + 6k

3k² - 6k - 11 = 0

3(k² - 2k) = 11

Take the middle term -2k and divide the coefficient in 2 and square it. Here it is -2/2 = -1. -1² = 1. Add 1 to both sides.

3(k² - 2k + 1) = 11 + 1

3(k - 1)² = 12

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What is 9/25?<br><br> A. 1/2<br> B. 81/25<br> C. 18/25<br> D. 3/5
Paha777 [63]

Answer:

3/5 would be your answer.

Step-by-step explanation:

Multiplying both numbers by 3 & 5.

= 3*3/5*5

= 9/25

hope it helps!

5 0
3 years ago
Read 2 more answers
-2x+5-40-9x - 13 help me please​
Naddika [18.5K]

Answer:

-48/11

Step-by-step explanation:

-2x+5-40-9x-13=0

-11x=40-5+13

-11x=48

x=-48/11

7 0
3 years ago
if a person paid rs 32 as VAT to buy an article of cost Rs 2750 then how much money he /she pay? find it​
bekas [8.4K]

The total amount paid for the article is Rs2782

<h3>How to determine the amount paid?</h3>

The given parameters are:

Article = Rs 2750

VAT = Rs 32

The amount paid is calculated using:

Amount = Article + VAT

This gives

Amount = 2750 + 32

Evaluate

Amount = 2782

Hence, the total amount paid for the article is Rs2782

Read more about VAT at:

brainly.com/question/20392481

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7 0
1 year ago
The Laplace Transform of a function f(t), which is defined for all t &gt; 0, is denoted by L{f(t)} and is defined by the imprope
lesya692 [45]

(1) D

L_s\left\{t\right\} = \displaystyle\int_0^\infty te^{-st}\,\mathrm dt

Integrate by parts, taking

u = t \implies \mathrm du=\mathrm dt

\mathrm dv = e^{-st}\,\mathrm dt \implies v=-\dfrac1se^{-st}

Then

L_s\left\{t\right\} = \displaystyle \left[-\frac1ste^{-st}\right]\bigg|_{t=0}^{t\to\infty}+\frac1s\int_0^\infty e^{-st}\,\mathrm dt

L_s\left\{t\right\} = \displaystyle \frac1s\int_0^\infty e^{-st}\,\mathrm dt

L_s\left\{t\right\} = \displaystyle -\frac1{s^2}e^{-st}\bigg|_{t=0}^{t\to\infty}

L_s\left\{t\right\} = \displaystyle \boxed{\frac1{s^2}}

(2) A

L_s\left\{1\right\} = \displaystyle\int_0^\infty e^{-st}\,\mathrm dt

L_s\left\{1\right\} = \displaystyle\left[-\frac1se^{-st}\right]\bigg|_{t=0}^{t\to\infty}

L_s\left\{1\right\} = \displaystyle\boxed{\frac1s}

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3 years ago
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