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3 years ago

Can the set of lengths be the side lengths of a right triangle?

Mathematics

2 answers:

ElenaW [278]3 years ago

7 0

Answer: Hello there!

We know that in a right triangle, the sum of the squares of the cathetus is equal to the square of the hypotenuse.

the smaller sides given are:

7ft and 12ft, and the longest is 17ft

so if this triangle is a right triangle, we should have that:

7^2 + 12^2 = 17^2

17^2 = 289

7^2 + 12^2 = 193

The terms are different, which means that this is not a right triangle, then the correct answer is B.

andreyandreev [35.5K]3 years ago

4 0

Answer:

B. No.

Step-by-step explanation:

We have been given 3 side lengths 7 ft, 12 ft, 17 ft. We are asked to determine, whether the given set of lengths can form a right triangle or not.

We will use Pythagoras theorem to solve our given problem, which states that the square of hypotenuse of a right triangle is equal to the sum of squares of two legs of right triangle.

$17^2=12^2+7^2$

$289=144+49$

$289>193$

Since the sum of squares of both legs is less than square of hypotenuse, therefore, the given set of lengths can not be the side lengths of a right triangle.

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2 years ago

Explain how we do decimals plus Fractions

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Assume z = x + iy, then find a complex number z satisfying the given equation. d. 2z8 – 2z4 + 1 = 0

kodGreya [7K]

Answer: complex equations has n number of solutions, been n the equation degree. In this case:

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i11,25°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i101,25°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i191,25°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i281,25°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i78,75°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i168,75°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i258,75°}$

$Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i348,75°}$

Step-by-step explanation:

I start with a variable substitution:

$Z^{4} = X$

Then:

$2X^{2}-2X+1=0$

Solving the quadratic equation:

$X_{1} =\frac{2+\sqrt{4-4*2*1} }{2*2} \\X_{2} =\frac{2-\sqrt{4-4*2*1} }{2*2}$

$X=\left \{ {{0,5+0,5i} \atop {0,5-0,5i}} \right.$

Replacing for the original variable:

$Z=\sqrt[4]{0,5+0,5i}$

or $Z=\sqrt[4]{0,5-0,5i}$

Remembering that complex numbers can be written as:

$Z=a+ib=|Z|e^{ic}$

Using this:

$Z=\left \{ {{{\frac{\sqrt{2}}{2} e^{i45°} } \atop {{\frac{\sqrt{2}}{2} e^{i-45°} }} \right.$

Solving for the modulus and the angle:

$Z=\left \{ {{\sqrt[4]{\frac{\sqrt{2}}{2} e^{i45}} = \sqrt[4]{\frac{\sqrt{2}}{2} } \sqrt[4]{e^{i45}} } \atop {\sqrt[4]{\frac{\sqrt{2}}{2} e^{i-45}} = \sqrt[4]{\frac{\sqrt{2}}{2} } \sqrt[4]{e^{i-45}} }} \right.$

The possible angle respond to:

$RAng_{12...n} =\frac{Ang +360*(i-1)}{n}$

Been "RAng" the resultant angle, "Ang" the original angle, "n" the degree of the root and "i" a value between 1 and "n"

In this case n=4 with 2 different angles: Ang = 45º and Ang = 315º

Obtaining 8 different angles, therefore 8 different solutions.

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2 years ago

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barxatty [35]

Answer:

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2 years ago