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3 years ago

What is the form of the two squares identity?

Mathematics

1 answer:

zvonat [6]3 years ago

6 0

Answer:

Step-by-step explanation:

Consider all options:

A. False, because

$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\ \\(ab-cd)^2+(ac+bd)^2=a^2b^2-2abcd+c^2d^2+a^2c^2+2abcd+b^2d^2=\\ =a^2b^2+c^2d^2+a^2c^2+b^2d^2\\ \\a^2c^2+a^2d^2+b^2c^2+b^2d^2\neq a^2b^2+c^2d^2+a^2c^2+b^2d^2$

B. False, because

$(a^2-b^2)(c^2+d^2)=a^2c^2+a^2d^2-b^2c^2-b^2d^2\\ \\(ac-bd)^2-(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=\\=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ \\a^2c^2+a^2d^2-b^2c^2-b^2d^2\neq a^2c^2+b^2d^2+a^2d^2+b^2c^2$

C. False, because

$(a^2+b^2)(c^2-d^2)=a^2c^2-a^2d^2+b^2c^2-b^2d^2\\ \\(ac+bd)^2-(ad+bc)^2=a^2c^2+2abcd+b^2d^2-a^2d^2-2abcd-b^2c^2=\\=a^2c^2+b^2d^2-a^2d^2-b^2c^2\\ \\a^2c^2-a^2d^2+b^2c^2-b^2d^2\neq a^2c^2+b^2d^2-a^2d^2-b^2c^2$

D. True, because

$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\ \\(ac-bd)^2-(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=\\=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ \\a^2c^2+a^2d^2+b^2c^2+b^2d^2= a^2c^2+b^2d^2+a^2d^2+b^2c^2$

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Program the VR Robot to drive forward 6 grid squares, turn around, and drive back 6 grid squares. How many degrees did you have

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Answer::3

Step-by-step explanation:

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3 years ago

WILL MARK BRAINLIEST FEELING LAZY WORTH 15 POINTS

Zinaida [17]

The irrational numbers are: √8, √10 and √15

Step-by-step explanation:

A rational number is a number that can be written in the form p/q where p&q are integers and q≠0.

"All the numbers whose square root is not a whole number and has an infinite number of digits after decimal, are irrational numbers"

So in the given options

$\sqrt{4} = 2$

Which can be written in the required form so √4 is a rational number

$\sqrt{8}=2.828427124746190097603.....$

√8 has an infinite expansion hence it cannot be written in the p/q form, so it is an irrational number

$\sqrt{10}=3.16227766016837933....$

√10 has an infinite expansion hence it cannot be written in the p/q form, so it is an irrational number

$\sqrt{15}=3.87298334620.....$

√15 has an infinite expansion hence it cannot be written in the p/q form, so it is an irrational number

$\sqrt{36} = 6$

Which can be written in the required form so √36 is a rational number

Hence,

The irrational numbers are: √8, √10 and √15

Keywords: Rational numbers, Irrational numbers

Learn more about rational numbers at:

#LearnwithBrainly

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3 years ago

In the expression below, a is an integer. 12^3+ax-20 if 3x+4 is a factor of the expession above, what is the value of a

SVEN [57.7K]

Our function is:

$f(x)=12^3+ax-20$

Now, if $3x+4$ is a factor of $f(x)$ then $x=\frac{-4}{3}$ must satisfy the equation:

So, $f(\frac{-4}{3} )=0$

Putting $x=\frac{-4}{3}$

We get,

$f(\frac{-4}{3} )=12^3+a(\frac{-4}{3} )-20=0$

$12^3+\frac{-4}{3} a-20=0$

$\frac{-4}{3} a=20-12^3=20-1708$

$a=-1708 \times \frac{-3}{4} =1281$

So, the value of 'a' for the given expression is 1281.

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3 years ago

What is the square root of 5 divided by the square root of 15 and simplify and in fraction form

ozzi

Answer:

$\sqrt{\frac{1}{3} }$

$\frac{\sqrt{3} }{3}$

Step-by-step explanation:

The expression $\frac{\sqrt{5} }{\sqrt{15} }$ can be simplified by first writing the fraction under one single radical instead of two.

$\frac{\sqrt{5} }{\sqrt{15} } = \sqrt{\frac{5}{15} }$

5/15 simplifies because both share the same factor 5.

It becomes $\sqrt{\frac{5}{15} } = \sqrt{\frac{1}{3} }$

This can simplify further by breaking apart the radical.

$\sqrt{\frac{1}{3} } = \frac{\sqrt{1} }{\sqrt{3} } = \frac{1}{\sqrt{3} }$

A radical cannot be left in the denominator, so rationalize it by multiplying by √3 to numerator and denominator.

$\frac{1}{\sqrt{3} } *\frac{\sqrt{3} }{\sqrt{3} } =\frac{\sqrt{3} }{3}$

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3 years ago

Jack looks at a clock tower from a distance and determines that the angle of elevation of the top of the tower is 40°. John, who

zmey [24]

Answer:

18.7939 m

Step-by-step explanation:

-Let x be the distance between John and clock tower.

-Let y be the vertical distance from the eyes of the two men standing to the top of the clock tower.

#Taking the right triangle ACD:

$\Tan \ theta=\frac{Perpendicular \ Height}{Base}\\\\Tan \ 60\textdegree=\frac{y+1.5}{x}\\\\y=x \ Tan \ 60\textdegree -1.5$

#Taking the right triangle ABD:

$\Tan \ theta=\frac{Perpendicular \ Height}{Base}\\\\Tan \ 40\textdegree=\frac{y+1.5}{x+20}\\\\y=(x+20)\ Tan \ 40\textdegree -1.5$

#We equate the two yo solve for x and y;

$(x+20)\ Tan \ 40\textdegree -1.5=x\ Tan \ 60\textdegree -1.5\\\\(x+20)\ Tan \ 40\textdegree=x\ Tan \ 60\textdegree\\\\0.8391x+16.7820=1.7321x\\\\x=18.7939$

Hence, John's distance from the tower's base is 18.7939 m

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3 years ago