Answer:
(x+2)(x+5) or if you were looking for the x-intercepts they would be -5, and -2
Answer:

Step-by-step explanation:
We need at least two points to write the equation of a straight line.
The recursive formula that Elijah wrote is:


When we substitute n=0, we get:



The points (0,30) and (1,37) lies on this line.
The equation of this line is of the form:

where b =30 is the y-intercept and m=7 is the slope.
We plug in these values to get:

Note that the slope of the line is equal to the common difference of the Arithmetic Sequence.
You could also use the two points to find the slope:

Answer:
Step-by-step explanation:
Represent the width by W. Then, "The length of a rectangular field is 7 m less than 4 times the width" expressed symbolically is
L = 4W - 7 (dimensions in meters)
Recall that the perimeter formula in this case is P = 2L + 2W, and recognize that the perimeter value is 136 m. After substituting 4W - 7 for L, we get:
136 m = 2(4W - 7) + 2W, or
136 = 8W - 14 + 2W, or
150 = 10W These three equations are equivalent mathematical statements.
150 = 10W reduces to W = 15 (meters).
Part A: the independent variable is W, the width of the field.
Part B: The mathematical statement is 136 m = 2(4W - 7) + 2W, which after algebraic manipulation becomes 150 = 10W.
Part C: The above equation can be solved for W: W = 15 meters. This is the value of the independent variable.
Answer: x = 9
Mel's mistake was in the last step, she multiplied by 2 instead of dividing by 2
Step-by-step explanation:
2x + 1 = 19
Subtract 1 from each side
2x = 18
Divide by 2
x = 9
Answer:
Fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.
Step-by-step explanation:
Given the figure with dimensions. we have to find the area of given figure.
Area of figure=ar(1)+ar(2)+ar(3)
Area of region 1 = ar(ANGI)+ar(AIB)
![=L\times B+\frac{1}{2}\times base\times height\\\\=[1500\times (5000-2000-1500)]+\frac{1}{2}\times (3000-1500)\times (5000-2000-1500)\\\\=3375000m^2=337.5ha](https://tex.z-dn.net/?f=%3DL%5Ctimes%20B%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20base%5Ctimes%20height%5C%5C%5C%5C%3D%5B1500%5Ctimes%20%285000-2000-1500%29%5D%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%283000-1500%29%5Ctimes%20%285000-2000-1500%29%5C%5C%5C%5C%3D3375000m%5E2%3D337.5ha)
Area of region 2 = ar(DHBC)

Area of region 3 = ar(GFEH)

Hence, Area of figure=ar(1)+ar(2)+ar(3)=337.5ha+300ha+350ha
=987.5 ha
Now, we have to do straight-line fencing such that area become half and cost of fencing is minimum.
Let the fencing be done through x m downward from B which divides the two into equal area.
⇒ Area of upper part above fencing=Area of lower part below fencing
⇒
Hence, fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.