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3 years ago

The side lengths of a triangle are 5, 8, and 12. Is this a right triangle?

Mathematics

2 answers:

MariettaO [177]3 years ago

8 0

Answer:

Step-by-step explanation:

5²+8²=25+64=89

12²=144

5²+8²≠12²

so it is not aright angled triangle.

MrRa [10]3 years ago

7 0

Answer:

Not a right triangle

Step-by-step explanation:

A triangle is right if the sides of the triangle satisfies the Pythagoras theorem.

That is square of the largest sides = sum of the sqaures of the sides .

That is ,

$12^2 = 5^2 + 8^2 \\\\144 = 25 + 64 \\\\144 \neq 89$

It did not satisfy the Pythagoras theorem.

Hence the given triangle is not a right triangle.

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Svetllana [295]

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Step-by-step explanation:

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Write an equation in standard form of the line that passes through the given point and the given slope. (-9, -4); m = -1/3

tigry1 [53]

<h2>Answer:</h2>

First, we must determine <u><em>b</em></u> using <u><em>slope-intercept form*</em></u>, the given point, and the given slope.

$-4 = -\frac{1}{3}(-9) + b\\\\-4 = -3 + b\\\\-1 = b$

Second, we need to set up the equation of the line in slope-intercept form.

$y = -\frac{1}{3}x - 1$

Third, we convert the above equation to <em>STANDARD FORM**</em>.

$y = -\frac{1}{3}x - 1\\\\\frac{1}{3}x + y = -1$

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3 years ago

The function f(x)f(x) is a quartic function and the zeros of f(x)f(x) are -6−6, -5−5, -2−2 and 11. Assume the leading coefficien

Luden [163]

<h2>Answer: </h2>

$\boxed{f(x) = 11x^4 + 22x^3 - 1001x^2 - 5632x - 7260}$

<h2>Explanation: </h2>

A quartic function is a function given by the the following equation in standard form:

$f(x)=a_{4}x^4+a_{3}x^3+a_{2}x^2+a_{1}x+a_{0} \\ \\ \\ Where: \\ \\ a_{4},a_{3},a_{2},a_{1},a_{0} \ \text{are constant} \\ \\ \text{and} \ a_{4} \ \text{is the leading coefficient}$

From the statement we must assume the leading coefficient of f(x) is 11, so:

$a_{4}=11$

The zeros are:

$x=-6 \\ \\ x=-5 \\ \\ x=-2 \\ \\ x=11$

So, we can write:

$f(x)=11(x-(-6))(x-(-5))(x-(-2))(x-11) \\ \\ \\ So: \\ \\ \mathrm{Rule}:-\left(-a\right)=a \\ \\ \\ Then: \\ \\ f(x)=11\left(x+6\right)\left(x+5\right)\left(x+2\right)\left(x-11\right)$

$Expand:\left(x+6\right)\left(x+5\right):\ x^2+11x+30: \ \text{Distributive Property} \\ \\ \\ Then: \\ \\ f(x)=11\left(x^2+11x+30\right)\left(x+2\right)\left(x-11\right) \\ \\ \\ Expand:\left(x^2+11x+30\right)\left(x+2\right):\ x^3+13x^2+52x+60: \ \text{Distributive Property} \\ \\ \\ Then: \\ \\ f(x)=11\left(x^3+13x^2+52x+60\right)\left(x-11\right)$

$Expand: \left(x^3+13x^2+52x+60\right)\left(x-11\right):\ x^4+2x^3-91x^2-512x-660 \\ (Distributive \ property)$

Finally:

$f(x) = 11(x^4 + 2x^3 - 91x^2 - 512x - 660) \\ \\ \boxed{f(x) = 11x^4 + 22x^3 - 1001x^2 - 5632x - 7260}$

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