Answer:
Part 1) In the procedure
Part 2) In the procedure
Step-by-step explanation:
Part 1) Observing the graph
The roots are x=-1/2 and x=4
Remember that, the roots are the x-values when the value of y is equal to zero (the x-intercepts)
That's why Sheldon immediately realizes Howard's solution is incorrect
Part 2) we have
The formula to solve a quadratic equation of the form 
is equal to
in this problem we have
equate the equation to zero
so
substitute
Howard's error is in this step, is wrong with the sign of the number 7, is positive instead of negative
Work the information to set inequalities that represent each condition or restriction.
2) Name the
variables.
c: number of color copies
b: number of black-and-white copies
3)
Model each restriction:
i) <span>It
takes 3 minutes to print a color copy and 1 minute to print a
black-and-white copy.
</span><span>
</span><span>
3c + b</span><span>
</span><span>
</span><span>ii) He needs to print
at least 6 copies ⇒
c + b ≥ 6</span><span>
</span><span>
</span><span>iv) And must have
the copies completed in
no more than 12 minutes ⇒</span>
3c + b ≤ 12<span />
4) Additional restrictions are
c ≥ 0, and
b ≥ 0 (i.e.
only positive values for the number of each kind of copies are acceptable)
5) This is how you
graph that:
i) 3c + b ≤ 12: draw the line 3c + b = 12 and shade the region up and to the right of the line.
ii) c + b ≥ 6: draw the line c + b = 6 and shade the region down and to the left of the line.
iii) since c ≥ 0 and b ≥ 0, the region is in the
first quadrant.
iv) The final region is the
intersection of the above mentioned shaded regions.v) You can see such graph in the attached figure.
Answer:
11, 13, 17, 19
Step-by-step explanation:
If you look at the prime numbers chart between 10 and 20, there are:
11, 13, 17, 19
Those are all of the prime numbers that are between 10 and 20.
Answer:
what is the question?
Step-by-step explanation:
uhh ok
Answer:
x = {nπ -π/4, (4nπ -π)/16}
Step-by-step explanation:
It can be helpful to make use of the identities for angle sums and differences to rewrite the sum:
cos(3x) +sin(5x) = cos(4x -x) +sin(4x +x)
= cos(4x)cos(x) +sin(4x)sin(x) +sin(4x)cos(x) +cos(4x)sin(x)
= sin(x)(sin(4x) +cos(4x)) +cos(x)(sin(4x) +cos(4x))
= (sin(x) +cos(x))·(sin(4x) +cos(4x))
Each of the sums in this product is of the same form, so each can be simplified using the identity ...
sin(x) +cos(x) = √2·sin(x +π/4)
Then the given equation can be rewritten as ...
cos(3x) +sin(5x) = 0
2·sin(x +π/4)·sin(4x +π/4) = 0
Of course sin(x) = 0 for x = n·π, so these factors are zero when ...
sin(x +π/4) = 0 ⇒ x = nπ -π/4
sin(4x +π/4) = 0 ⇒ x = (nπ -π/4)/4 = (4nπ -π)/16
The solutions are ...
x ∈ {(n-1)π/4, (4n-1)π/16} . . . . . for any integer n
