Distribute 
to every term inside the parentheses.
Simplify with multiplication.
You can't simplify further by adding, so leave the answer as-is.
Answer:
6
Step-by-step explanation:
Answer:
The ship is located at (3,5)
Explanation:
In the first test, the equation of the position was:
5x² - y² = 20 ...........> equation I
In the second test, the equation of the position was:
y² - 2x² = 7 ..............> equation II
This equation can be rewritten as:
y² = 2x² + 7 ............> equation III
Since the ship did not move in the duration between the two tests, therefore, the position of the ship is the same in the two tests which means that:
equation I = equation II
To get the position of the ship, we will simply need to solve equation I and equation II simultaneously and get their solution.
Substitute with equation III in equation I to solve for x as follows:
5x²-y² = 20
5x² - (2x²+7) = 20
5x² - 2y² - 7 = 20
3x² = 27
x² = 9
x = <span>± </span>√9
We are given that the ship lies in the first quadrant. This means that both its x and y coordinates are positive. This means that:
x = √9 = 3
Substitute with x in equation III to get y as follows:
y² = 2x² + 7
y² = 2(3)² + 7
y = 18 + 7
y = 25
y = +√25
y = 5
Based on the above, the position of the ship is (3,5).
Hope this helps :)
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.
5 Steps to Solving the Problems With Your Problem Solving
Step 1: Pin the Problem. Clearly define the issue at hand. ...
Step 2: Identify the Issues. Start breaking down the problem into subcomponents. ...
Step 3: Generate Hypotheses and Prioritize Proving Them. ...
Step 4: Conduct Your Analysis. ...
Step 5: Advance Your Answer.
