Answer:
Step-by-step explanation:
Think of a right triangle. It may help you to draw it.
Start at a point. Draw a point and label it point A. That is where the plane takes off from. Now draw a diagonal segment tilted up to the right and stop at a point and label it B. Segment AB is the path of the airplane. Now draw a segment down vertically to a point you will label C, so that points A and C are on the same horizontal line. Connect points A and C.
You now have a right triangle. Angle C is the right angle. AB is the hypotenuse and is 5.4 km. That is the actual distance the plane traveled. BC is 1.12 km (since 1120 m = 1.12 km) and is a leg of the right triangle. The angle you are looking for is angle A.
For angle A, BC is the opposite leg. AB is the hypotenuse.
The trig ratio that relates the opposite leg to the hypotenuse is the sine.
Answer: 12.0 degrees
Answer:
17.3
Step-by-step explanation:
area of paralellogram=base x height
height=sin(60) x 4= 
area = 5 x
= 
= 17.3(to nearest tenth)
Answer:
4.625
Step-by-step explanation:
All you do is divide 15/24 and then add the four before the decimal, and BOOM there is your answer! Good Day!
The domain:The number of which the logarithm is taken must be greater than 0.
The base of the logarithm must be greater than 0 and not equal to 1.
* greater than 0:
*not equal to 1:
Sum up all the domain restrictions:
The solution:
Now if the base of the logarithm is less than 1, then you need to flip the sign when solving the inequality. If it's greater than 1, the sign remains the same.
* if the base is less than 1:
The inequality:
* if the base is greater than 1:
The inequality:
![\log_{8x^2-23x+15} (2x-2) \leq \log_{8x^2-23x+15} 1 \ \ \ \ \ \ \ |\hbox{the sign remains the same} \\ 2x-2 \leq 1 \\ 2x \leq 3 \\ x \leq \frac{3}{2} \\ x \leq 1 \frac{1}{2} \\ x \in (-\infty, 1 \frac{1}{2}] \\ \\ \hbox{including the condition that the base is greater than 1:} \\ x \in (-\infty, 1 \frac{1}{2}] \ \land \ x \in (2, \infty) \\ \Downarrow \\ x \in \emptyset](https://tex.z-dn.net/?f=%5Clog_%7B8x%5E2-23x%2B15%7D%20%282x-2%29%20%5Cleq%20%5Clog_%7B8x%5E2-23x%2B15%7D%201%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%7C%5Chbox%7Bthe%20sign%20remains%20the%20same%7D%20%5C%5C%202x-2%20%5Cleq%201%20%5C%5C%202x%20%5Cleq%203%20%5C%5C%20x%20%5Cleq%20%5Cfrac%7B3%7D%7B2%7D%20%5C%5C%20x%20%5Cleq%201%20%5Cfrac%7B1%7D%7B2%7D%20%5C%5C%20x%20%5Cin%20%28-%5Cinfty%2C%201%20%5Cfrac%7B1%7D%7B2%7D%5D%20%5C%5C%20%5C%5C%20%5Chbox%7Bincluding%20the%20condition%20that%20the%20base%20is%20greater%20than%201%3A%7D%20%5C%5C%20x%20%5Cin%20%28-%5Cinfty%2C%201%20%5Cfrac%7B1%7D%7B2%7D%5D%20%5C%20%5Cland%20%5C%20x%20%5Cin%20%282%2C%20%5Cinfty%29%20%5C%5C%20%5CDownarrow%20%5C%5C%20x%20%5Cin%20%5Cemptyset)
Sum up both solutions:
The final answer is:
Answer:
<CFE = 28
<RPQ = 53
Step-by-step explanation:
1)
<CFE + 152 = 180
<CFE = 180 - 152
<CFE = 28
2)
Complementary means the angles add up to 90 so:
<RPQ + 37 = 90
<RPQ = 90 - 37
<RPQ = 53
