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oksano4ka [1.4K]
3 years ago
13

Match each word problem to the equation that models it. tiles the sum of the squares of two positive integers is 394. if one int

eger is 2 less than the other and the larger integer is x, find the integers.

Mathematics
1 answer:
Dovator [93]3 years ago
6 0
Let
x--------> the larger integer
y-------> the smaller integer

we know that
x²+y²=394-----> equation 1
x=y+2-----> equation 2
substitute equation 2 in equation 1
[y+2]²+y²=394-----> y²+4y+4+y²=394
2y²+4y-390=0

using a graph tool----> to resolve the second order equation
see the attached figure

the solution is
y=13
x=y+2---> x=13+2---> x=15

the answer is
the numbers are 15 and 13

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3 0
4 years ago
Please help me I am having a hard time with this activity
zysi [14]

Answer:

Step-by-step explanation:

b.

\frac{a^2+6a+9}{a^2-9} *\frac{3a-9}{a+3} \\=\frac{a^2+2*a*3+3^2}{a^2-3^2} *\frac{3(a-3)}{a+3} \\=\frac{(a+3)^2}{(a+3)(a-3)} *\frac{3(a-3)}{a+3} \\=\frac{3(a+3)^2}{(a+3)^2} \\=3

d.

\frac{3x^2-6x}{3x+1} *\frac{x+3x^2}{x^2-4x+4} \\=\frac{3x(x-2)}{3x+1} *\frac{x(1+3x)}{x^2-2x-2x+4} \\=\frac{3x(x-2)}{3x+1} *\frac{x(1+3x)}{x(x-2)-2(x-2)} \\=\frac{3x(x-2)}{3x+1} *\frac{x(1+3x)}{(x-2)(x-2) } \\=\frac{3x^2}{x-2)}

e.

\frac{2x^2-10x+12}{x^2-4} *\frac{2+x}{3-x} \\=\frac{2[x^2-5x+6]}{x^2-2^2} *\frac{2+x}{-(-3+x)} \\=\frac{2[x^2-2x-3x+6]}{(x+2)(x-2)} *\frac{x+2}{-(x-3)} \\=\frac{2[x(x-2)-3(x-2)]}{(x+2)(x-2)} *\frac{x+2}{-(x-3)} \\=\frac{2(x-2)(x-3)}{(x+2)(x-2)} *\frac{x+2}{-(x-3)} \\=\frac{2}{-1} \\=-2

k.

\frac{6x^2-11x-10}{6x^2-5x-6} *\frac{6-4x}{25-20x+4x^2} \\=\frac{6x^2-15x+4x-10}{6x^2-9x+4x-6} *\frac{-4x+6}{4x^2-10x-10x+25} \\=\frac{3x(2x-5)+2(2x-5)}{3x(2x-3)+2(2x-3) } *\frac{-2(2x-3)}{2x(2x-5)-5(2x-5)} \\=\frac{(2x-5)(3x+2)}{(2x-3)(3x+2)} *\frac{-2(2x-3)}{(2x-5)(2x-5)} \\=\frac{-2}{2x-5}

7 0
3 years ago
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