What is three tenths times 8
1 answer:
You might be interested in
Answer:
15°.
Step-by-step explanation:
1. Angles ADC and CDB are supplementary, thus
m∠ADC+m∠CDB=180°.
Since m∠ADC=115°, you have that m∠CDB=180°-115°=65°.
2. Triangle BCD is isosceles triangle, because it has two congruent sides CB and CD. The base of this triangle is segment BD. Angles that are adjacent to the base of isosceles triangle are congruent, then
m∠CDB=m∠CBD=65°.
The sum of the measures of interior angles of triangle is 180°, therefore,
m∠CDB+m∠CBD+m∠BCD=180° and
m∠BCD=180°-65°-65°=50°.
3. Triangle ABC is isosceles, with base BC. Then
m∠ABC=m∠ACB.
From the previous you have that m∠ABC=65° (angle ABC is exactly angle CBD). So
m∠ACB=65°.
4. Angles BCD and DCA together form angle ACB. This gives you
m∠ACB=m∠ACD+m∠BCD,
m∠ACD=65°-50°=15°.
Have a good Day!
Answer:
5 minutes 4.8 seconds
Step-by-step explanation:
We can work this problem several ways. One is to determine the component of the distance from JBLM to the mountain that is in the direction the plane is flying. We can do that by finding the dot product of the vector to the mountain with the normalized direction vector of the airplane:
JA = JR•(JP/|JP|) . . . . . . where JR = (56, -40) and JP = (112, -115)
= (56, -40)•(112, -115)/√(112² +115²)
JA = 10872/√25769 ≈ 67.7268 . . . . km
__
At a speed of 800 km/h, this distance will be covered in ...
time = distance/speed
time = (67.7268 km)/(800 km/h) = 0.0846585 h
That's about 5 minutes, 4.8 seconds.
_____
In the attached diagram, J represents JBLM, R represents Mt. Rainier, A represents the point of closest approach, and P would represent the location of the airplane after 1 hour of flying time, (112, -115). In the above, we have used the names of these line segments/vectors.
