Answer:
a) Yes
b) Not a right triangle
Step-by-step explanation:
<h3><u>Part (a)</u></h3>
The sum of the lengths of any 2 sides of a triangle is greater than the length of the 3rd side.
7 + 8 = 15 > 9
7 + 9 = 16 > 8
8 + 9 = 17 > 7
Therefore, a triangle with side lengths 7, 8 and 9 <u>can</u> be made.
<h3><u>Part (b)</u></h3>
To determine if the triangle is a right triangle, use Pythagoras' Theorem:
a² + b² = c² (where a and b are the legs, and c is the hypotenuse)
The hypotenuse is the longest side, therefore:
Substituting these values into the formula:
⇒ a² + b² = c²
⇒ 7² + 8² ≠ 9²
⇒ 113 ≠ 81
As 7² + 8² ≠ 9² then the triangle is <u>not</u> a right triangle.
Answer:
8 x^5 y^7
Step-by-step explanation:
Simplify the following:
4×2 x^2 y^3 x^3 y^4
Hint: | Combine products of like terms.
4 x^2 y^3×2 x^3 y^4 = 4 x^(2 + 3) y^(3 + 4)×2:
4×2 x^(2 + 3) y^(3 + 4)
Hint: | Evaluate 3 + 4.
3 + 4 = 7:
4×2 x^(2 + 3) y^7
Hint: | Evaluate 2 + 3.
2 + 3 = 5:
4×2 x^5 y^7
Hint: | Multiply 4 and 2 together.
4×2 = 8:
Answer: 8 x^5 y^7
5>0-3
Step By Step- 5 is greater than 0-3 because 0-3 would equal a negative number, and all positive numbers are greater than negative numbers
then use the common difference "d", to get the 2nd and 3rd terms
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None of the alternatives are correct. The domain's lower bound is 6, however it contains 6, therefore a bracket is required rather than a parenthesis. [6 (infinity)] Because infinity is not included, a bracket for infinity is invalid. The range is from 0 to infinity. assuming you meant 1/sqr (x-6). However, none of the options are right if you meant 1/(sqr(x) - 6).
or did you mean 1/sqrx - 6? However, none of them are correct.
