Answer:
Step-by-step explanation:
We have:
And we want to find the value of x such that the expression is positive. So, we can write this as the following inequality:
Solve for the inequality. First, we can solve for the zeros like a normal quadratic. So, pretend the inequality is with an equal sign:
Zero Product Property:
On the left, subtract 5.
On the right, add 1.
So, our zeros are:
Since our inequality is a <em>greater than</em>, our answer is an "or" inequality with our answer being all the values to the <em>left</em> of our lesser zero and all the values to the <em>right </em>of our greater zero.
So, our solution is:
And we're done!
Solution:
1) Add 80 to both sides
-np<60+80
2) Simplify 60+80 to 140
-np<140
3) Divide both sides by p
-n<\frac{140}{p}
4) Multiply both sides by -1
n>-\frac{140}{p}
Done!
Answer:
x = 2
y = 5
Step-by-step explanation:
3x + y = 11 equation (1)
-x + y = 3 multiply this equation by -1
x - y = -3 equation (2)
by adding 2 equations
4x = 8
x = 2
by sub. in Equation (2)
2 - y = -3
y = 2+3
y = 5
The answer is D. SIS idk how to do this
Ur welcome
Answer:
66 ≤ f ≤100
Explanation
Mean= ( Σ x ) / n
Mean= sum of scores/ number of subject she took
Now, she already too 3 subject which sum is 85+83+86=254
Now we need to know range of score for her to have (grade) a mark between 80 and 89
Now let take the lower limit mean=80
The lowest score she can get is
Mean = ( Σx) / n
80=(85+83+86+f)/4
80×4= 254+f
Therefore, f= 320-254=66
Therefore the minimum score she can have to have a B is 66.
Then, let take the upper limit mean 89. i.e the maximum she can have so that she don't have an A grade.
Mean = ( Σx) / n
89=( 83+85+86+f)/4
89×4= 254+f
f= 356-254
f=102.
Therefore this shows that she cannot have an A grade in the exam. The maximum score for the exam is 100.
There the range of score is 66 ≤ f ≤100 to have a B grade
66 ≤ f ≤100 answer
Since she cannot score 102 in the examination.
