First set up a linear equation and using the x and y values in the table see if it solves.
It doesn't solve so we know it isn't linear. ( I won't show all those steps because they aren't needed.)
Using the quadratic formula y = ax^2 +bx +c
Build a set of 3 equations from the table:
C is the Y intercept ( when X is 0), this is shown in the table as 6
Now we have y = ax^2 + bx + 6
-2.4 =4a-2b +6
1.4 = a-b +6
Rewrite the equations
a=b/2 -2.1
1.4 = b/2-2.1 +6
b = 5
a = 5/2 -2.1 = 0.4
replace the letters to get y = 0.4x^2 + 5x +6
Problem 44
The term "bisect" means "cut in half".
Since BD bisects angle ABC, this means the smaller angles ABD and DBC are congruent.
angle ABD = angle DBC
x+15 = 4x-45
15+45 = 4x-x
60 = 3x
3x = 60
x = 60/3
x = 20
<h3>Answer: x = 20</h3>
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Problem 45
We use the same idea as the previous problem
angle ABD = angle DBC
2x+35 = 5x-22
35+22 = 5x-2x
57 = 3x
3x = 57
x = 57/3
x = 19
<h3>Answer: x = 19</h3>
Answer:
60 degrees
Step-by-step explanation:
Restructured question:
The measure of two opposite interior angles of a triangle are x−14 and x+4. The exterior angle of the triangle measures 3x-45 . Solve for the measure of the exterior angle.
First you must know that the sum of interior angle of a triangle is equal to the exterior angle
Interior angles = x−14 and x+4
Sum of interior angles = x-14 + x + 4
Sum of interior angles = 2x - 10
Exterior angle = 3x - 45
Equating both:
2x - 10 = 3x - 45
Collect like terms;
2x - 3x = -45 + 10
-x = -35
x = 35
Get the exterior angle:
Exterior angle = 3x - 45
Exterior angle = 3(35) - 45
Exterior angle = 105 - 45
Exterior angle = 60
Hence the measure of the exterior angle is 60 degrees
<em>Note that the functions of the interior and exterior angles are assumed. Same calculation can be employed for any function given</em>
Answer:
no
Step-by-step explanation:
