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2 years ago

12

Which statement describes whether a right triangle can be formed using one side length from each of these squares?

1 answer:

balu736 [363]2 years ago

4 0

Answer:

Step-by-step explanation:

The area of the given squares are

Area = 64 in

Area = 225 in

Area = 289 in

The length of each side of a square is determined by finding the square root of its area. For the first square, the length of its side is √64 = 8inches. For the second square, the length of its side is √225 = 15inches. For the third square, the length of its side is √289 = 17inches.

For a right angle triangle to be formed, Pythagorean theorem must be obeyed. Sum of the square of the smaller sides must equal the square of the longer side. Therefore,

8² + 15² = 17²

289 = 289

Therefore, the correct statement is

Yes, a right triangle can be formed because the sum of the areas of the two smaller squares equals the area of the largest square

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F(x) = x^2 - 3x+ 5 g(x) = 2x^2 - 4x - 11 what is h(x) = f(x) + g(x)

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Hey there! :)

Answer:

h(x) = 3x² - 7x - 6

Step-by-step explanation:

Calculate h(x) by adding the two polynomials:

h(x) = f(x) + g(x):

h(x) = x² - 3x + 5 + 2x² - 4x - 11

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h(x) = 3x² - 7x - 6

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Consider the differential equation x2y′′ − 9xy′ + 24y = 0; x4, x6, (0, [infinity]). Verify that the given functions form a funda

pantera1 [17]

Answer:

The functions satisfy the differential equation and linearly independent since W(x)≠0

Therefore the general solution is

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Step-by-step explanation:

Given equation is

$x^2y'' - 9xy+24y=0$

This Euler Cauchy type differential equation.

So, we can let

$y=x^m$

Differentiate with respect to x

$y'= mx^{m-1}$

Again differentiate with respect to x

$y''= m(m-1)x^{m-2}$

Putting the value of y, y' and y'' in the differential equation

$x^2m(m-1) x^{m-2} - 9 x m x^{m-1}+24x^m=0$

$\Rightarrow m(m-1)x^m-9mx^m+24x^m=0$

$\Rightarrow m^2-m-9m+24=0$

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⇒m²-6m -4m+24=0

⇒m(m-6)-4(m-6)=0

⇒(m-6)(m-4)=0

⇒m = 6,4

Therefore the auxiliary equation has two distinct and unequal root.

The general solution of this equation is

$y_1(x)=x^4$

and

$y_2(x)=x^6$

First we compute the Wronskian

$W(x)= \left|\begin{array}{cc}y_1(x)&y_2(x)\\y'_1(x)&y'_2(x)\end{array}\right|$

$= \left|\begin{array}{cc}x^4&x^6\\4x^3&6x^5\end{array}\right|$

=x⁴×6x⁵- x⁶×4x³

=6x⁹-4x⁹

=2x⁹

≠0

The functions satisfy the differential equation and linearly independent since W(x)≠0

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