Two students are 1 foot apart on a straight path. Each student starts walking away from the other at the same moment. The first
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Answer:
<em>Interval notation: [0, 2.236]</em>
<em>Set Builder notation: </em>
Step-by-step explanation:
Given that:
Equation of height of the ball dropped from a height of 10 foot, as:
Where is the time since the ball was dropped.
To find:
The domain of the function in Interval and set builder notation.
Solution:
<em>Domain of a function </em>is defined as the set of valid input values that can be given to the function for which the function is defined.
Here, input is time.
We can not have negative values for time.
Therefore, starting value for time will be <em>0 seconds</em>.
And the value of height can not be lesser than that of 0 ft.
Maximum value for time can be 2.236 seconds.
Therefore the domain is:
<em>Interval notation: [0, 2.236]</em>
<em>Set Builder notation: </em>
Pictures always help; see the one I've attached for reference.
Call the first vector (from HQ to the supply drop) u and the second vector (from supply drop to medics) v. We then want to find w, the vector from the medics to HQ, which corresponds to the vector -(u + v). (This is because u + v is the vector pointing from HQ to the medics; we're talking about the one in the opposite direction.)
Write the vectors in horizontal/vertical component form:
u = (125 km) (cos 235º x + sin 235º y) = (-71.70 x - 102.39 y) km
v = (75 km) (cos 110º x + sin 110º y) = (-25.65 x + 70.48 y) km
Why these angles?
- "55 degrees south of west" is 235º; due west is 180º from the positive horizontal axis, and you add 55º to this
- "20 degrees west of north" is 110º; due north is 90º, so add 20º to this
Add the vectors:
u + v = (-97.35 x - 31.92 y) km
Multiply by -1 to get the vector w:
w = -(u + v) = (97.35 x + 31.92 y) km
The distance covered by this vector is equal to its magnitude:
||w|| = √((97.35 km)^2 + (31.92 km)^2) = 102.45 km
The direction <em>θ</em> is given by
tan<em>θ</em> = (31.92 km)/(97.35 km) ==> <em>θ</em> = 18.15º
Answer:
<h2><DEF = 40</h2><h2><EBF = <EDF = 56</h2><h2><DCF = <DEF =40</h2><h2><CAB = 84</h2>Step-by-step explanation:
In triangle DEF, we have:
<u>Given</u>:
<EDF=56
<EFD=84
So, <DEF =180 - 56 - 84 =40 (sum of triangle angles is 180)
____________
DE is a midsegment of triangle ACB
( since CD=DA(given)=>D is midpoint of [CD]
and BE = EA => E midpoint of [BA] )
According to midsegment Theorem,
(DE) // (CB) "//"means parallel
and DE = CB/2 = FB =CF
___________
DEBF is a parm /parallelogram.
<u>Proof</u>: (DE) // (FB) ( (DE) // (CB))
AND DE = FB
Then, <EBF = <EDF = 56
___________
DEFC is parm.
<u>Proof</u>: (DE) // (CF) ((DE) // (CB))
And DE = CF
Therefore, <DCF = <DEF =40
___________
In triangle ACB, we have:
<CAB =180 - <ACB - <ABC =180 - 40 - 56 =84(sum of triangle angles is 180)