9514 1404 393
Answer:
A. √13
Step-by-step explanation:
You can make an educated guess and come to the right conclusion.
The triangle is nearly an equilateral triangle. A triangle with two sides 3 and an angle of 60° would have a third side of 3. A triangle with two sides of 4 and an angle of 60° would have a third side of 4.
So, the third side must be between 3 and 4. Here is an evaluation of the answer choices:
__
A -- between 3 and 4, the correct choice
B -- 3, too short
C -- 1.73, too short
D -- more than 4, too long
__
The question can be answered using your triangle solver app on your calculator, or using the Law of Cosines.
c = √(a^2 +b^2 -2ab·cos(C))
c = √(3^2 +4^2 -2·3·4·(1/2)) = √(9 +16 -12)
c = √13 . . . . . length of the side opposite the 60° angle
Answer: No.
Step-by-step explanation: g should also be factored out. 10g(a - 3z).
Answer:
Step-by-step explanation:
2:3 is the correct answer
Answer:
the third option
Step-by-step explanation:
what does that mean ?
to "rationalize" it is to transform it into a rational number (that is a number that can be described as a/b, and is not an endless sequence of digits after the decimal point without a repeating pattern).
a square root of a not square number is irrational (not rational).
so, what this question is asking us to get rid of the square root part in the denominator (the bottom part).
for this we need to multiply to and bottom with the same expression (to keep the whole value of the quotient the same) that, when multiplied at the bottom, eliminates the square root.
what can I multiply a square root with to eliminate the square root ? the square root again - we are squaring the square root.
so, what works for 9 - sqrt(14) as factor ?
we cannot just square this as
(9- sqrt(14))² = 81 -2sqrt(14) + 14
we still have the square root included.
but remember the little trick :
(a+b)(a-b) = a² - b²
without any mixed elements.
so, we need to multiply (9-sqrt(14)) by (9+sqrt(14)) to get
81-14 = 67 which is a rational number.
therefore, the third answer option is correct.
