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nirvana33
1 month ago
14

8) Write an absolute value equation that has 5 and 15 as its solutions?

Mathematics
1 answer:
Svet_ta [12.7K]1 month ago
7 0
(I’ll use brackets to represent my absolute value bars, haha) [5 x -3] [-15] = 15.
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Two functions are shown in the table below. Function 1 2 3 4 5 6 f(x) = −x2 + 4x + 12 g(x) = −x + 6 Complete the table on your o
Svet_ta [12734]

For \fbox{\begin \\\math{x}=6\\\end{minispace}} the function f(x)=-x^{2} +4x+12 and g(x)=-x+6 both yield the same result.

Detailed breakdown:  

The functions involved are

f(x)=-x^{2}+4x+12

g(x)=-x+6

Step 1:  

Insert x=1 in f(x)=-x^{2} +4x+12 to find the value of f(1).

f(1)=-1^{2} +4(1)+12\\f(1)=-1+4+12\\f(1)=15

Insert x=1 in g(x)=-x+6 to find the value of g(1).

g(1)=-1+6\\g(1)=5

Step 2:

Insert x=2 in f(x)=-x^{2} +4x+12 to obtain the value of f(2).

f(2)=-2^{2} +4(2)+12\\f(2)=-4+8+12\\f(2)=16

Substitute x=2 into g(x)=-x+6 to find the value of g(2).

g(2)=-2+6\\g(2)=4

Step 3:

Replace x=3 in f(x)=-x^{2} +4x+12 to find the value of f(3).

f(3)=-3^{2} +4(3)+12\\f(3)=-9+12+12\\f(3)=15

Also, replace x=3 in g(x)=-x+6 to find the value of g(3).

g(3)=-3+6\\g(3)=3

Step 4:

Insert x=4 in f(x)=-x^{2} +4x+12 to find the value of f(4).

f(4)=-4^{2} +4(4)+12\\f(4)=-16+16+12\\f(4)=12

Also, replace x=4 in g(x)=-x+6 to obtain the value of g(4).

g(4)=-4+6\\g(4)=2

Step 5:

Insert x=5 in f(x)=-x^{2} +4x+12 to obtain the value of f(5).

f(5)=-5^{2} +4(5)+12\\f(5)=-25+20+12\\f(5)=7

Replace x=5 in g(x)=-x+6 to find the value of g(5).

g(5)=-5+6\\g(5)=1

Step 6:

Insert x=6 into f(x)=-x^{2} +4x+12 to find the value of f(6).

f(6)=-6^{2} +4(6)+12\\f(6)=-36+24+12\\f(6)=0

Also, substitute x=6 in g(x)=-x+6 to obtain the value of g(6).

g(6)=-6+6\\g(6)=0

Step 7:

According to the provided condition f(x)=g(x).

(a). Insert f(x)=-x^{2} +4x+12 and g(x)=-x+6 into the previously mentioned equation.

-x^{2} +4x+12=-x+6

(b). Multiply through by -1 on both sides.

x^{2} -4x-12=x-6

(c). Move the term x-6 to the left side of the equation.

x^{2} -4x-12-x+6=0\\x^{2} -5x-6=0

(d). Divide the middle term so that its sum equals 5 and the product equals 6.

x^{2} -(6-1)x-6=0\\x^{2} -6x+x-6=0\\x(x-6)+1(x-6)=0\\(x+1)(x-6)=0\\x=-1,6

From the analysis above, it is noted that for x=6 both functions f(x) and g(x) yield the same outcome.

Using a direct approach:

f(x)=g(x)\\\Leftrightarrow-x^{2} +4x+12=-x+6\\\Leftrightarrow-x^{2} +4x+12+x-6=0\\\Leftrightarrow-x^{2} +5x+6=0\\\Leftrightarrow-x^{2} +6x-x+6=0\\\Leftrightarrow x^{2} -6x+x-6=0\\\Leftrightarrow x(x-6)+1(x-6)=0\\\Leftrightarrow(x+1)(x-6)=0\\\Leftrightarrow x=6,-1

The table representing function f(x)=-x^{2} +4x+12 and g(x)=-x+6 is included below.

For more information:

1. What is the y-intercept of the quadratic function f(x) = (x – 6)(x – 2)? (0,–6) (0,12) (–8,0) (2,0)

2. Which is the graph of f(x) = (x – 1)(x + 4)?

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Response:- 0.8/8+x=5/100

Detailed explanation:

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By the seventh day, there are zero rabbits remaining.
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Part 1) The radius of the circle is r=17 units. Part 2) The points (-15,14) and (-15,-16) are situated on this circle. Step-by-step explanation: Step 1 Find the radius of the circle. We know that the distance from the center of the circle to any point on its circumference equals the radius of the circle. The formula to determine the distance between two points is equal to......we have (-7, -1) and (8, 7) substitute... Step 2 Determine the y-coordinate of point (-15,y). The standard form of the circle's equation is given by... where (h,k) represents the center, and r is the radius. Replace the values, substituting x=-15 in the equation... square root both sides... ultimately, we find two solutions: point (-15,14) and point (-15,-16) refer to the attached figure for a clearer understanding of the problem.
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