Answer:
$87,461
Step-by-step explanation:
Given that the dimensions or sides of lengths of the triangle are 119, 147, and 190 ft
where S is the semi perimeter of the triangle, that is, s = (a + b + c)/2.
S = (119 + 147 + 190) / 2 = 456/ 2 = 228
Using Heron's formula which gives the area in terms of the three sides of the triangle
= √s(s – a)(s – b)(s – c)
Therefore we have = √228 (228 - 119)(228 - 147)(228 - 190)
=> √228 (109)(81)(38)
= √228(335502)
=√76494456
= 8746.1109071 * $10
= 87461.109071
≈$87,461
Hence, the value of a triangular lot with sides of lengths 119, 147, and 190 ft is $87,461.
Answer:
74
Step-by-step explanation:
Say that arc JL going through M is arc E and JL going the other way is arc D
For the angle formed by two tangents, K=(1/2)(E-D)
64=E-D
Furthermore, angle K and central angle JCL (facing toward K) are supplementary, so 180-K=JCL=180-32=148
Thus, as the angles around angle C add up to 360, angle JCL (facing toward M) is 360-148=32+180=212
E is then 212
64=212-D
212-64=D=148
Thus, as JML is an inscribed angle, M=1/2(D)=1/2(148)=74
Answer:
Trapezoid and Quadrilateral
Step-by-step explanation:
Trapezoid because it has four sides that are not equal You may not be familiar with this figure because most trapezoids don't look like this. It is a quadrilateral because it is a figure that has 4 sides.
Answer:

Step-by-step explanation:
Slope Formula: 
Simply plug in 2 coordinates into the slope formula to find slope <em>m</em>:



So if you want to fit the y-intercepts or "b", on the y-axis you should go by 25's [0, 25, 50, 75, 100...]
If the x-axis <u>does not have to</u> follow the same pattern (25's), you should go by 5's [0, 5, 10, 15, 20...]
y = 7x + 50
y = 2x + 175
First I would plot the y-intercepts for each equation, then plot a few points with x = 5, 10, 15 Then draw a straight line.
The point where the two lines meet/cross paths is your solution. It should be (25, 225) The x-axis is the number of miles, and the y-axis is the total cost. So Truck driver A and B would arrive/be at the same place/meet in 25 miles at the same cost of $225