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kherson [118]
3 years ago
6

Find the product mentally. [(x + 1)(x - 1)]^2 pls help. Be quick if you can =)

Mathematics
1 answer:
BartSMP [9]3 years ago
6 0

Answer:

x^{4} -2 x^{2} +1

Step-by-step explanation:

<u>Step 1</u>:-

Given problem is [(x + 1)(x - 1)]^2

by using formula in side square bracket is

(a+b)(a-b)=(a^{2}-b^{2})

[x^{2} -1]^{2}

<u>Step 2</u>:- again we will use formula

(a-b)^{2} = a^{2} -2 a b+b^{2}

[tex](x^2-1)^{2} = (x^{2}) ^{2} -2 x^{2}  (1)+1^{2}[/tex]

on simplification we get

x^{4}-2 x^{2}  +1

<u>Final answer:-</u>

x^{4}-2 x^{2}  +1

<u />

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Please please help please please ASAP please I'm begging you please I have 10 minutes please y'all help please any ACE that can
Andreyy89

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Angle S

Step-by-step explanation:

The two triangles are similar, with the reason of (corr. angles, similar triangles), we get that angles S = angle N

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3 years ago
A group of 199 people is going on a boat tour. If each boat holds 8 people. how many boats will they need?
umka21 [38]
Basically what your gonna do is do 199 divided by 8 so you would need 25 boats, because it would be 24.875 rounded its 25
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3 years ago
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3x?=90 Thank you have a good day
nikdorinn [45]
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6 0
3 years ago
Solve the simultaneous equations<br> y = 9 - X<br> y = 2x2 + 4x + 6
kenny6666 [7]

Answer:

\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}

\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}

Step-by-step explanation:

Given the simultaneous equations

y=9-x

y\:=\:2x^2\:+\:4x\:+\:6

Subtract the equations

y=9-x

-

\underline{y=2x^2+4x+6}

y-y=9-x-\left(2x^2+4x+6\right)

\mathrm{Refine}

x\left(2x+5\right)=3

\mathrm{Solve\:}\:x\left(2x+5\right)=3

2x^2+5x=3        ∵ \mathrm{Expand\:}x\left(2x+5\right):\quad 2x^2+5x

\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}

2x^2+5x-3=3-3

\mathrm{Solve\:with\:the\:quadratic\:formula}

\mathrm{Quadratic\:Equation\:Formula:}

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=2,\:b=5,\:c=-3:\quad x_{1,\:2}=\frac{-5\pm \sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}v\\

x=\frac{-5+\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}

  =\frac{-5+\sqrt{5^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}

  =\frac{-5+\sqrt{49}}{2\cdot \:2}

  =\frac{-5+\sqrt{49}}{4}

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  =\frac{2}{4}

  =\frac{1}{2}

Similarly,

x=\frac{-5-\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}:\quad -3

\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}

x=\frac{1}{2},\:x=-3

\mathrm{Plug\:the\:solutions\:}x=\frac{1}{2},\:x=-3\mathrm{\:into\:}y=9-x

\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}\frac{1}{2}:\quad y=\frac{17}{2}

\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}-3:\quad y=12

\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}

\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}

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4 years ago
Please help me ASAP and I have more so PLEASE
nevsk [136]

The answer is B part

As the requirement for the student is either a senior or one who studies maths or both.

So, just add the values within the two venn circles divide it by total no of values.

=> 240+40+110/240+40+110+840

=> 360/1200

8 0
3 years ago
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