Simplify 10^4 to 10000
8.5x10000-3.0x10^3
Simplify 8.5x*10000 to 85000x
85000x-3.0*10^3
Simplify 3.0x10^3 to 3*10^3
Answer 85000x-3*10^3
You need to understand that you're solving for the average, which you already know: 90. Since you know the values of the first three exams, and you know what your final value needs to be, just set up the problem like you would any time you're averaging something.
Solving for the average is simple:
Add up all of the exam scores and divide that number by the number of exams you took.
(87 + 88 + 92) / 3 = your average if you didn't count that fourth exam.
Since you know you have that fourth exam, just substitute it into the total value as an unknown, X:
(87 + 88 + 92 + X) / 4 = 90
Now you need to solve for X, the unknown:
87
+
88
+
92
+
X
4
(4) = 90 (4)
Multiplying for four on each side cancels out the fraction.
So now you have:
87 + 88 + 92 + X = 360
This can be simplified as:
267 + X = 360
Negating the 267 on each side will isolate the X value, and give you your final answer:
X = 93
Now that you have an answer, ask yourself, "does it make sense?"
I say that it does, because there were two tests that were below average, and one that was just slightly above average. So, it makes sense that you'd want to have a higher-ish test score on the fourth exam.
Answer:
x=9 and y=115
Step-by-step explanation:
Solve y=9x+34;y=16x−29
Steps:
I will solve your system by substitution.
y=9x+34;y=16x−29
Step: Solvey=9x+34for y:
Step: Substitute9x+34foryiny=16x−29:
y=16x−29
9x+34=16x−29
9x+34+−16x=16x−29+−16x(Add -16x to both sides)
−7x+34=−29
−7x+34+−34=−29+−34(Add -34 to both sides)
−7x=−63
−7x
−7
=
−63
−7
(Divide both sides by -7)
x=9
Step: Substitute9forxiny=9x+34:
y=9x+34
y=(9)(9)+34
y=115(Simplify both sides of the equation)
Hope this helps :)
Answer:
-3.92 N
Step-by-step explanation:
We are given that
Mass of ball=0.4 kg
Acceleration due to gravity=
Because the ball flies against to gravity
We have to find the net force acting on the ball while it is in motion.
We know that when there is no air resistance then the net force acts on the ball is equal to weight of ball.
Net force,F =mg
Substitute the values then we get
Hence, the net force acting on the ball while it is on motion=-3.92 N
