The correct answer is: [D]: " 7.2 units" .
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Explanation:
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Use the Pythagorean theorem:
a² + b² = c² ;
in which: "6 units" and "4 units" equal the lengths of the right angle (formed by the rectangle); and "c" is the length of the diagonal of the rectangle, or the "hypotenuse", of the right triangle formed by the rectangle; We wish to solve for "c" ;
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6² + 4² = c² ; Solve for "c" ;
↔ c² = 6² + 4² ;
= (6*6) + (4*4) ;
= 36 + 16 ;
= 52 ;
c² = 52 ;
Take the "positive square root" of each side of the equation; to isolate "c" on one side of the equation; and to solve for "c" ;
√(c²) = √52 ;
c = √52 ;
At this point, we know the 7² = 49 ; 8² = 64 ; so, the answer is somewhere between "7" and "8" ; yet closer to "7" ; so among the answer choices given;
The correct answer is: [D]: " 7.2 units" .
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However, let use a calculator:
c = √52 = 7.2111025509279786 ; which rounds to "7.2" ;
which corresponds to:
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Answer choice: [C]: " 7.2 units" .
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Simplify the expression.
Exact Form:
8671/34
Decimal Form:
255.02941176
…
Mixed Number Form:
255 1/34
Solve x+4y = 13 for x by subtracting 4y from both sides.
We end up with x = -4y+13.
Since x and -4y+13 are the same, we can replace every x in the first equation with -4y+13
2x-3y = -29
2( x ) - 3y = -29
2( -4y+13 ) - 3y = -29 ... x is replaced with -4y+13
The answer is choice B
![{c}^{2} - 64 = 0](https://tex.z-dn.net/?f=%20%7Bc%7D%5E%7B2%7D%20-%2064%20%3D%200%20)
This can also be represented as:
![{c}^{2} - {8}^{2} = 0](https://tex.z-dn.net/?f=%20%7Bc%7D%5E%7B2%7D%20-%20%7B8%7D%5E%7B2%7D%20%3D%200)
Using difference of two squares;
![(c - 8)(c + 8) = 0](https://tex.z-dn.net/?f=%28c%20-%208%29%28c%20%2B%208%29%20%3D%200)
![c - 8 = 0 \: or \: c + 8 = 0 \\ \\ c = 8 \: or \: c = - 8](https://tex.z-dn.net/?f=c%20-%208%20%3D%200%20%5C%3A%20or%20%5C%3A%20c%20%2B%208%20%3D%200%20%5C%5C%20%5C%5C%20c%20%3D%208%20%5C%3A%20or%20%5C%3A%20c%20%3D%20-%208)
Hope I helped? ☺️
Answer:
Step-by-step explanation:
The graph passes through the origin. It could either open up or open down. The quadratic has two zeros, both at (0, 0).