the horizontal distance between the child and the crackers is 2.9 ft .
<u>Step-by-step explanation:</u>
Here we have , A child spots crackers in a cabinet at an angle of elevation of 50. The crackers are 3.5 ft above the child. We need to find what is the horizontal distance between the child and the crackers . Let's find out:
According to question , we have a right angle triangle with parameters as :
, Where x is angle of elevation !
We know that , 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
Therefore , the horizontal distance between the child and the crackers is 2.9 ft .
X| 1 | 15 | 225 |
y| 4 | ? | 900 |

therefore
[tex]\dfrac{4}{1}=4;\ \dfrac{900}{225}=4;\ \dfrac{?}{15}=4\to?=60[\tex]
Answer: ? = 60
Answer:
The answer is six
Step-by-step explanation:
12 3/5 = 63/5
2 1/10 = 21/10
63/5 divided by 21/10 is equal to 63/5 multiplied by 10/21 which is equal to 630/105 = 6 <-- answer
Answer:
D. Undefined
Step-by-step explanation:
4-4 = 0, 10-6 = 4
Change of x = 0
Change of y = 4
Since the change of x is 0 (a horizontal line), it is undefined.