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Dahasolnce [82]
3 years ago
10

Drag the tiles to the correct boxes to complete the pairs.

Mathematics
2 answers:
ra1l [238]3 years ago
4 0

Step-by-step explanation:

lara [203]3 years ago
3 0
For the first problem, the integers are 8 and 11.
The second problem has missing information.  So does the third.  They cannot be worked without the information that's missing.

Explanation:
Let x be the larger integer and y be the smaller integer.  We know that the smaller integer, y, is 3 less than the larger integer, x, so:
y=x-3

The sum of their squares is 185.  This means:
x²+y²=185
x²+(x-3)²=185
x²+(x-3)(x-3)=185

Multiplying the binomials we get:
x²+x*x-3*x-3*x-3(-3)=185
x²+x²-3x-3x+9=185

Combining like terms we get:
2x²-6x+9=185

When solving quadratics, we want the function to equal 0; subtract 185 from both sides:
2x²-6x+9-185=185-185
2x²-6x-176=0

Use the quadratic formula:
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
\\
\\=\frac{--6\pm \sqrt{(-6)^2-4(2)(-176)}}{2(2)}=\frac{6\pm \sqrt{36--1408}}{4}
\\
\\=\frac{6\pm \sqrt{1444}}{4}=\frac{6\pm 38}{4}=\frac{6-38}{4}\text{ or }\frac{6+38}{4}=\frac{-32}{4}\text{ or }\frac{44}{4}=-8\text{ or }11

Since both integers must be positive, x = 11.  This means that y=11-3=8.
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Hello!

After reviewing the problem you have provided I have come up with the correct solution:

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