Answer:
c = 105 degres
Step-by-step explanation:
When two lines cross like that and form an x, the opposite sides are equal (so the angle below the 75 degree one is also 75 degees). Using this, you can figure out the rest.
If both the top an bottom are equal, you know that 150 degres are taken (75+75) and you know that there can only be 360 degres in total here, so you subtract 150 from 360 and get 210. Now you know that the last two sides together are 210, and since they are equal, you divide it by two to get 105 degrees.
Answer:
0.4
Step-by-step explanation:
15 = 5p - 13
-13 = 5p -13
---------------- ( subtract 13 on both sides )
<u> 2 = 5p</u>
5 = 5p
---------------- ( divide 2 by 5 )
p = 0.4
( p equals 0.4 )
C. 427
The median is the middle number of all the data
Since there is an even number of data points the median is in between the middle two data points (434 and 420)
To find this: 434-420=14
14/2=7
434-7=427
420+7=427
The question is asking to write and equation by using the two variable in the standard form that ca be used to describe the velocity of thecar at different times and base on my calculation, I would say that the answer would be v=4t+39. I hope you are satisfied with my answer and feel free to ask for more
Find an equation of the plane that contains the points p(5,−1,1),q(9,1,5),and r(8,−6,0)p(5,−1,1),q(9,1,5),and r(8,−6,0).
topjm [15]
Given plane passes through:
p(5,-1,1), q(9,1,5), r(8,-6,0)
We need to find a plane that is parallel to the plane through all three points, we form the vectors of any two sides of the triangle pqr:
pq=p-q=<5-9,-1-1,1-5>=<-4,-2,-4>
pr=p-r=<5-8,-1-6,1-0>=<-3,5,1>
The vector product pq x pr gives a vector perpendicular to both pq and pr. This vector is the normal vector of a plane passing through all three points
pq x pr
=
i j k
-4 -2 -4
-3 5 1
=<-2+20,12+4,-20-6>
=<18,16,-26>
Since the length of the normal vector does not change the direction, we simplify the normal vector as
N = <9,8,-13>
The required plane must pass through all three points.
We know that the normal vector is perpendicular to the plane through the three points, so we just need to make sure the plane passes through one of the three points, say q(9,1,5).
The equation of the required plane is therefore
Π : 9(x-9)+8(y-1)-13(z-5)=0
expand and simplify, we get the equation
Π : 9x+8y-13z=24
Check to see that the plane passes through all three points:
at p: 9(5)+8(-1)-13(1)=45-8-13=24
at q: 9(9)+8(1)-13(5)=81+9-65=24
at r: 9(8)+8(-6)-13(0)=72-48-0=24
So plane passes through all three points, as required.
