<em>The question doesn't show any graph, but I have included it to help you with the problem.</em>
Answer:
<em>The new graph will be shifted up by 3 units</em>
Step-by-step explanation:
<u>Graph of a Real Function</u>
The total cost of belonging to the book club depends on one fixed cost, the initial joining fee, and a variable cost, depending on the number of books purchased. Call C the total cost and x the number of purchased books, then
By giving x some values, we get the corresponding values of C and then we can plot the graph as a blue line, in the image provided below.
If the joining fee goes up by $3, the new function is
The graph is also shown in the image, but with a red line. We can see the graph is shifted up by 3 units in the y-axis, but the slope remains the same, as the cost per book hasn't changed
Let, amount of 14% solution required is x ounces.
So, amount of 8% solution will be ( 28 - x ) ounces.
Now, mathematical equation is :
Therefore, amount of 14% solution and 8% solution required is 4 ounces and 24 ounces respectively.
Hence, this is the required solution.
(2 x-3 y) (3 x+2 y) if it is wrong just look on math way
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The coordinate of the midpoint of line segment LT is determined as: -12.
<h3>How to Find the Coordinate of the Midpoint of a Line Segment?</h3>
The midpoint of a line segment is the point where the distance between the endpoints of the line segment are equidistant. The distance from that midpoint to each endpoint is the same.
Given the following:
- Coordinate of point L is: -35
- Coordinate of point T is: 11
Distance from point L to T = |-35 - 11| = 46 units.
Half of 46 units would be: 46/2 = 23 units.
This means that, both point L and point T are 23 units from the midpoint of segment LT.
Thus, the coordinate of the midpoint would be 23 units from -35 = -35 + 23 = -12
Or 23 units from the midpoint to point T = 11 - 23 = -12
Therefore, the coordinate of the midpoint of line segment LT is determined as: -12.
Learn more about the midpoint of a segment on:
brainly.com/question/19149725
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A. All of the members of the PTA.
