Remark
This is a very interesting question. Draw a line from the origin to where the upper right vertex of the square touches the line. That line has the property that the its equation is y = x. So the "solution" to the point of intersection is the solution of the two equations.
y = x (1)
3x + 4y = 12 (2)
Put x in for y in equation 2
3x + 4x = 12
7x = 12
x = 12/7
x = 1.714
y = 1.714
Problem A
<em><u>x intercept</u></em>
The x intercept occurs when y = 0
3x + 4(0) = 12
3x = 12 Divide by 3
x = 12/3
x = 4
the x intercept = (4,0)
<em><u>y intercept</u></em>
The y intercept occurs when x =0
3(0) + 4y = 12
4y = 12
y = 12/4
y = 3
y intercept = (0,3)
Problem B
x and y both equal 1.714 so they are also the length of the square's side.
Problem C
See solution above. x =y is the key fact.
x = y = 1.714
12x > 18x - 27 - 15
12x > 18x - 42
12x - 18x > -42
-6x > -42
x < -42/-6
two negatives make a positive so now its x < 42/6
Lastly divide 42/6 so it can equal 7
Answer: X < 7.
Answer:
4
Step-by-step explanation:
Class width is said to be the difference between the upper class limit and the lower class limit consecutive classes of a grouped data. To calculate class width, this formula can be used:
CW = UCL - LCL
Where,
CW= Class width
UCL= Upper class limit
LCL= Lower class limit
From the table above:
For class 1, CW = 64 - 60 = 4
For class 2, CW = 69 - 65 = 4
For class 3, CW = 74 - 70 = 4
For class 4, CW = 79 - 75 = 4
For class 5, CW = 84 - 80 = 4
Therefore, the class width of the grouped data = 4
Answer:
the answer is 34
Step-by-step explanation:
first we put the equation together (4x^2+4x+1)by(2x^2+x-2)
so we multiply the formula together so we (4x^2+4x+1) (2x^2+-2) and that's how you get 34
the set of x values are called the domain and the corresponding values of y the range
