Answer:
Every quadrilateral's 4 angles add up to 360 degrees.
The four angles would be 100° + 7x +37 + 100° + 7x +37 = 360°
200° + 14x + 74° = 360°
14x = 360° - 274
14x = 86
x = 6.1428571429
Step-by-step explanation:
Hi,
First we should calculate within the parentheses to get 343^3
then we calculate the exponent to get 40,353,607.
But the simplified version is 343^3
Have a great day!
Answer: The fire is 3.5 miles from tower B
Step-by-step explanation: Please refer to the attached diagram. The triangle in the attached diagram illustrates the clues given in the question. Both rangers are standing at points A and B respectively with a distance of 5 miles between them, which is line AB. Also, one ranger spots a fire from a tower at an angle of 42 degrees, which is point A. Another ranger spots the same fire from another tower, but from an angle of 64 degrees, which is point B. The fire is at point C on the triangle. Now we have a triangle with only one side known (5 miles) and three angles known (the third angle is computed as 180 - {64+42} which equals 74) which are 64 degrees, 42 degrees and 74 degrees.
The distance from the fire to tower B is calculated using the law of sines. (Note that this is not a right angled triangle, hence we cannot use trigonometric ratios). The law of sines is expressed as follows;
a/SinA = b/SinB or
a/SinA = c/SinC
Depending on the sides and angles we are given and the ones we are to calculate.
The distance from the fire to tower B is line BC, labeled as a in our diagram. Using the law of sines
a/SinA = c/SinC
(Note also that a is directly facing angle A, c is directly facing angle C, and so on)
a/SinA = c/SinC
a/Sin 42 = 5/Sin 74
By cross multiplication we now have
a (Sin 74) = 5 (Sin 42)
Divide both sides of the equation by Sin 74 and we now arrive at
a = 5 (Sin 42)/Sin 74
a = 5 (0.6691)/0.9613
a = 3.3455/0.9613
a = 3.4802
{rounded to the nearest tenth of a mile, a equals 3.5}
Therefore the distance from tower B to the fire is approximately 3.5 miles
Answer:
Based on the conditions we see that the function never pases through x=2 so then we know that x=2 is not on the domain of f. And we can create the following function who satisfy the conditions:
And we can see that we have a vertical asymptote in x =2 and is negative always on the interval (-∞, 2) and always positive on the interval (2, ∞).
And we can see the function in the figure attached.
Step-by-step explanation:
For this case we need a function that satisfy that is negative on the interval (-∞, 2) and positive on the interval (2, ∞).
Based on the conditions we see that the function never pases through x=2 so then we know that x=2 is not on the domain of f. And we can create the following function who satisfy the conditions:
And we can see that we have a vertical asymptote in x =2 and is negative always on the interval (-∞, 2) and always positive on the interval (2, ∞).
And we can see the function in the figure attached.