Answer:
Option C is correct.
Ratio of longer leg to hypotenuse is; 
Step-by-step explanation:
This is the special right angle triangle 30°-60°-90° as shown below in the figure.
- The side opposite the 30° angle is always the shortest because 30 degrees is the smallest angle.
- The side opposite the 60° angle will be the longer leg, because 60 degrees is the mid-sized degree angle in this triangle.
- Finally , the side opposite the 90° angle will always be the largest side(Hypotenuse) because 90 degrees is the largest angle.
In 30°−60°−90° right triangle,
- the length of the hypotenuse is twice the length of the shorter leg,also
- the length of the longer leg is
times the length of the shorter leg.
Then:
the sides are in proportion i.e, 
Therefore, the ratio of the length of the longer leg to the length of its hypotenuse is: 
Answer:
The equations represent the line that is parallel to 3x - 4y = 7 and pass through the point (-4,-2) are:
Step-by-step explanation:
The slope-intercept form of the line equation
where
Given the line
3x - 4y = 7
writing in the slope-intercept form
4y = 3x - 7
dividing both sides by 4
4y/4 = 3/4x - 7/4
y = 3/4x - 7/4
Now, comparing with the slope-intercept form of the line equation
y = 3/4x - 7/4
The slope of the line m = 3/4
We know that parallel lines have the same slopes.
Therefore, the slope of the parallel line is: 3/4
now we have,
The point (-4, -2)
The slope m of parallel line = 3/4
Given the point-slope form of the line equation
where m is the slope of the line and (x₁, y₁) is the point
substituting (-4, -2) and m = 3/4 in the point-slope form of line equation


Thus, the equation in the point-slope form of the line equation is:

Simplifying the equation

Subtract 3 from both sides


Multiplying the equation by 4


Therefore, the equations represent the line that is parallel to 3x - 4y = 7 and pass through the point (-4,-2) are:
Answer:
12?
Step-by-step explanation:
you didnt mention how many cups of flour to make one bath but you need 12 cups of sugar for 4 batches
There's an app called photomath that would be useful for these kinds of problems!