Answer:
5 packs of paper.
Step-by-step explanation:
Let x be the number of packs of paper.
We have been given that a pack of paper costs $3.75 including tax. The price of x paper packs would be .
Since Mr. Cooper has a $20 budget, so the cost of x paper pack should be lees than or equal to 20.
We can represent this information in an inequality as:
Since we can purchase whole number of paper packs, therefore, Mr. Cooper can purchase 5 packs of paper with his budget.
Answer: 18 cm^2
Step-by-step explanation:
There are four equal triangle and a rectangle.
Formula of a rectangle = b * h = 2*3 = 6
Formula of triangle = but there four of them so we times them by 4, this leaves us with 2 (b * h) = 2(2*3) = 12
Area of pentagon = area of all triangles + rectangle
Area of pentagon = 12 + 6 = 18
Hope this helps.
Answer:
There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.
Step-by-step explanation:
We notice that the expression on the left of the equation is a quadratic with leading term , which means that its graph is that of a parabola with branches going up.
Therefore, there can be three different situations:
1) if its vertex is ON the x axis, there would be one unique real solution (root) to the equation.
2) if its vertex is below the x-axis, the parabola's branches are forced to cross it at two locations, giving then two real solutions (roots) to the equation.
3) if its vertex is above the x-axis, it will have NO real solutions (roots) but only non-real ones.
So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently.
We recall that the x-position of the vertex for a quadratic function of the form is given by the expression:
Since in our case and
, we get that the x-position of the vertex is:
Now we can find the y-value of the vertex by evaluating this quadratic expression for x = 3/4:
This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.
We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:
See the graph produced in the attached image.
