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Novosadov [1.4K]
2 years ago
15

What is the missing constant term in the perfect square that starts with x2 – 12x ?

Mathematics
1 answer:
Dima020 [189]2 years ago
8 0

Answer:

x^2 -12x+36

Step-by-step explanation:

x^2 – 12x

Take the coefficient of x

-12

Divide by 2

-12/2 = 6

Square it

6^2 = 36

We need to add 36 to make x^2 -12x a perfect square trinomial

x^2 -12x+36

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327

Step-by-step explanation:

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Solve by factorising<br><br><img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" align="abs
Paladinen [302]

Step-by-step explanation:

\underline{ \underline{ \text{Solving \: a \: quadratic \: equation \: by \: factorisation \: method}}} :

In this method , the second order of polynomial ax² + bx + c is factorised and expressed as the product of two linear factors. Then each linear factor is separately solved to get the required solutions of the equation by applying zero factor property. In zero factor property , if p•q = 0 , then either p = 0 or q = 0 . In other words , if the product of two numbers is 0 , then one or both of the numbers must be 0.

\underline{ \underline{ \text{Let's \: solve}}} :

\tt{ {x}^{2}  + 16x +  55 = 0}

Here , we have to find the two numbers that multiplies to 55 and adds to 16.

⤑ \tt{ {x}^{2}  + (11 + 5)x + 55 = 0}

⤑ \tt{ {x}^{2}  + 11x + 5x +  55 = 0}

⤑ \tt{x(x + 11) + 5(x + 11) = 0}

⤑ \tt{(x + 5)(x + 11) = 0}

Either :

\tt{x + 5 = 0 \: \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \: Or :  \:  \:  x + 11 = 0}

\sf{⟶ x  = 0 - 5} \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    \:⟶  x =  0 - 11

\tt{⟶x =  - 5 } \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:     \:  \:⟶   x =  - 11

\red{ \boxed{ \boxed{ { \tt{Our \: final \: answer :  \boxed{ \underline{  \bold{ \tt{x  =  - 5 \: , - 11}}}}}}}}}

Hope I helped ! ♡

Have a wonderful day / night ! ツ

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6 0
3 years ago
If x+3/3 = y+2/2, then x/3 = ______
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n a Gallop poll of 1012 randomly selected adults, 9% said that cloning of humans should be allowed. We are going to use a .05 si
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As per given , the appropriate set of hypothesis would be :-

H_0: p\geq0.10\\\\ H_a: p

Since H_a is left-tailed and sample size is large(n=1012) , so we perform left-tailed z-test.

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P-value for left tailed test,

P(z<-1.06)=1-P(z<1.06)        [∵P(Z<-z)=1-P(Z<z]

=1-0.8554=0.1446          (Using z-value table.)

Hence, the p-value for this test statistic= 0.1446

4 0
2 years ago
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