Find the velocity first, the "slope".
m=(y2-y1)/(x2-x1)
m=(95.5-53)/(4-2)
m=21.25 ? Your equations are wrong, I'll double check with another set of points...
m=(350.5-308)/(16-14)
m=21.25 LOL, yep none of your lines has the correct slope.
Anyway...
y=21.25x+b, using any point we can solve for the y-intercept, or "b", (2, 53)
53=21.25(2)+b
53=42.5+b
10.5=b so the distance to the lighthouse as a function of time is:
y=21.25x+10.5
So essentially, you should fire your teacher or burn that book you are using :)
24 sponge fingers and 420 jelly
Answer:
x<-3 or x>-1
Step by step explanation:
Start with -5x>15. To get x by itself, we have to divide each side by -5. When we do that we get x<-3. (Don't forget to flip the sign when dealing with negative · and ÷.) Now answer the second equation. x-5>-6 To get x by itself, add 5 to both sides to get x>-1. Now we have x<3 or x>-1.
The bisector of the angle at A (call it AQ) divides the segment BC into segments BQ:QC having the ratio AB:AC. Use this fact to find x.
.. 9:15 = (2x -1):3x
.. 15(2x -1) = 9*3x . . . . . the product of the means equals the product of extremes
.. 30x -15 = 27x
.. 3x = 15
.. x = 5
___
According to the value of x, the bisector AQ divides the triangle into two isosceles triangles: ABQ, ACQ.
Answer:
10
Step-by-step explanation:
Here it's given that the area of the shaded region is 48 u² and we need to find out the value of x .
The given figure is made up of two rectangles where the smaller one is inscribed into the bigger one .
- The area of shaded region will be equal to the difference of the areas of the two rectangles , that is equal to 48 u² .
- The area of rectangle is length × breadth .
- The dimensions of bigger rectangle is (x+3)(x-4) and that of smaller one is 3 × x .
So that ,
ar(bigger) - ar(smaller) = 48
[ (x+3) (x-4)] - 3(x) = 48
[ x² - 4x +3x -12 - 3x ] = 48
x² -4x -12 -48 = 0
x² - 4x - 60 = 0
x² - 10x + 4x - 60 = 0
x( x -10) +4(x -10) = 0
(x +4)(x-10) = 0
x = -4,10
- Since sides can't be negative ,
x = 10
<h3>
Hence the required answer is 10.</h3>
