Number 19 you are comparing one measurement to another. Since it says 1/2 inch equals 4 ft, we want to find out how many more inches are needed if the given scale was 2/3 = 4 ft. Now lets find a common denominator for both scales stated in inches. We have 2/3 inch and 1/2 inch. Our denominator are the bottom parts of the fraction where we need to find a common factor for the denominator so we can add or subtract fractions. We have a 3 and a 2. You may always use the multiplication between two denominators to find a common factor such as 3 times 2 which equals 6 for both denominators. Now we multiplied the 3 by 2 to get 6 so the top part (numerator needs to be multiplied the the 2 because we changed the bottom part by 2 as well. You should notice that when you reduce your fraction now 4/6 is 2/3. Just a self check example there. As for 1/2 we multiplied a 3 to get 6 for the denominator so we need to multiply the numerator by 3 as well. You now should have 4/6 and 3/6. Since the question asks for how many more inches we need to subtract 4/6 from 3/6 and we get 1/6 inch for our answer.
Answer:

Step-by-step explanation:
We want to combine alike terms so
-2/3c+14c & -9/5+3/10
14c-2/3

Multiply 14/1 by 3/3 to get the denominators the same

Subtract

simplify

now

Multiply 9/5 by 2/2

Subtract

Simplify

put both together

Hope this helps! If you have any questions on how I got my answer feel free to ask. Stay safe!
Answer:
huh????
Step-by-step explanation:
what did you try to say
Answer:
Hence the ground distance is about 889m between stadium and Tennis court.
Step-by-step explanation:
Given:
A pilot at height of 125 m(flying the advertising blimp)
angle of depression=8 degrees.
To find :
The ground distance stadium and tennis court.
Solution :
Using Trigonometric Function we can solve it ,
(Refer the attachment for fig)
Considering above data we get a triangle(ABC)
with point A represent pilot position and C court position and B as stadium.
The height AB=125 m
angle of depression= 8 degrees.
Using
<em>tan∅ =AB/BC</em>
tan(8)=125/BC
BC=125/tan(8)
=125/0.14054
=889.2
=889 m
Hence the ground distance is about 889m between stadium and Tennis court.