H(10)=8(10)+10
h(x)=90
hope this helps
For a single 20 sided dice, a chance to get a certain number would be 1/20. For 2 numbers it will be a 2/20 chance, which can be simplified to a 1/10 chance that you will roll a 15 or a 17.
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.
Answer:
50
Step-by-step explanation:
The figure represents a 30-60-90 special triangle
in this special right triangle the measure of sides lengths is represented by
a, a, and 2a
The side length that sees angle measure 30 is represented by a and here we see a = 50m
The height of the building, which sees angle measure 60, is equal to
50
Answer: 225
Step-by-step explanation: If the whole area is 900 square inches you divide by the sides of the square 4 and divide by the total area 900. 900/4 = 225
