1=12-w Help solve please
1 answer:
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Answer:
1. The equation is solved by the graphed systems of equations is:
Third option: 7x + 3 = x − 3
2. The solution to the system is:
Fourth option: (3,-1)
3. The graph of the system y = −2x + 3 and 2x + 4y = 8 is:
Third option: Line through point (0, 3) and (1, 1). Line through (0, 2) and (4, 0).
4. The solution to the system is:
Second option: Infinite solutions.
5. Bill will have read the same amount of books as Mandy in:
Fourth option: December
Step-by-step explanation:
Question 1. What equation is solved by the graphed systems of equations?
Two linear equations that intersect at the point (-1, -4)=(x,y)→x=-1, y=-4
a. 7x + 3 = x + 3
Solving for x: Grouping the x's on the left side of the equation: Subtracting x both sides of the equation:
7x+3-x=x+3-x
6x+3=3
Subtracting 3 both sides of the equation:
6x+3-3=3-3
6x=0
Dividing both sides of the equation by 6:
6x/6=0/6
Dividing:
x=0 different to the intersection at x=-1, then this is not the equation.
b. 7x − 3 = x − 3
Solving for x: Grouping the x's on the left side of the equation: Subtracting x both sides of the equation:
7x-3-x=x-3-x
6x-3=-3
Adding 3 both sides of the equation:
6x-3+3=-3+3
6x=0
Dividing both sides of the equation by 6:
6x/6=0/6
Dividing:
x=0 different to the intersection at x=-1, then this is not the equation.
c. 7x + 3 = x − 3
Solving for x: Grouping the x's on the left side of the equation: Subtracting x both sides of the equation:
7x+3-x=x-3-x
6x+3=-3
Subtracting 3 both sides of the equation:
6x+3-3=-3-3
6x=-6
Dividing both sides of the equation by 6:
6x/6=-6/6
Dividing:
x=-1 equal to the intersection at x=-1, then this is the equation.
d. 7x − 3 = x + 3
Solving for x: Grouping the x's on the left side of the equation: Subtracting x both sides of the equation:
7x-3-x=x+3-x
6x-3=3
Adding 3 both sides of the equation:
6x-3+3=3+3
6x=6
Dividing both sides of the equation by 6:
6x/6=6/6
Dividing:
x=1 different to the intersection at x=-1, then this is not the equation.
Question 2 (Multiple Choice Worth 2 points) (06.03) Solve the system 2x + 3y = 3 and 3x − 2y = 11 by using graph paper or graphing technology. What is the solution to the system?
The solution to the system is (3, -1) Fourth option
Please, see the attached graph.
Question 3 (Multiple Choice Worth 2 points) (06.03) What is the graph of the system y = −2x + 3 and 2x + 4y = 8?
Third option: Line through point (0, 3) and (1, 1). Line through (0, 2) and (4, 0).
Please, see the attached graph.
Question 4 (Multiple Choice Worth 2 points) (06.03) Solve the system y = 3x + 2 and 3y = 9x + 6 by using graph paper or graphing technology. What is the solution to the system?
Second option: Infinite solutions.
Please, see the attached graph.
Question 5 (Multiple Choice Worth 2 points) (06.03) At the end of April, Mandy told Bill that she has read 16 books this year and reads 2 books each month. Bill wants to catch up to Mandy. He tracks his book reading with a table on his door. Using his table below, what month will Bill have read the same amount of books as Mandy?
Bill
Month Books
May 4
June 8
July 12
August 16
September 20
October 24
November 28
December 32
Books read by Mandy: y=16+2x
Numbers of months since april: x
May: x=1→y=16+2(1)=16+2→y=18 different to 4 (books read by Bill)
June: x=2→y=16+2(2)=16+4→y=20 different to 8 (books read by Bill)
July: x=3→y=16+2(3)=16+6→y=22 different to 12 (books read by Bill)
September: x=5→y=16+2(5)=16+10→y=26 different to 20 (books read by Bill)
October: x=6→y=16+2(6)=16+12→y=28 different to 24 (books read by Bill)
November: x=7→y=16+2(7)=16+14→y=30 different to 28 (books read by Bill)
December: x=8→y=16+2(8)=16+16→y=32 equal to 32 (books read by Bill)
Answer. Fourth option: December
The answers are shown in the attached image
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Explanation:
Set the denominator x^4-8x^3+16x^2 equal to zero and solve for x
x^4-8x^3+16x^2 = 0
x^2(x^2-8x+16) = 0
x^2(x-4)^2 = 0
x^2 = 0 or (x-4)^2 = 0
x = 0 or x-4 = 0
x = 0 or x = 4
The x values 0 and 4 make the denominator zero
These x values lead to asymptote discontinuities because the numerator 8x-24 = 8(x-3) has no common factors which cancel with the denominator factors.
There are two vertical asymptotes
Let's see what happens when we plug in a value to the left of x = 0, say x = -1, we'd get
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(-1) = (8(-1)-24)/((-1)^4-8(-1)^3+16(-1)^2)
f(-1) = -1.28
So as x gets closer and closer to x = 0 from the left side, the f(x) is heading to negative infinity
Now plug in some value to the right of x = 0. I'm going to pick x = 1
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(1) = (8(1)-24)/((1)^4-8(1)^3+16(1)^2)
f(1) = -1.78 (approximate)
So as x gets closer and closer to x = 0 from the right side, the f(x) is heading to negative infinity
Overall, as x approaches 0 from either the left or right side of x = 0, the y value is heading off to negative infinity
---------------------
Repeat for values to the left and right of x = 4
We can't use x = 1 as it turns out that x = 3 is a root
But we can use something like x = 3.5 to find that...
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(3.5) = (8(3.5)-24)/((3.5)^4-8(3.5)^3+16(3.5)^2)
f(3.5) = 1.31 approx
So as x gets closer to x = 4 from the left, y is getting closer to positive infinity
Plug in x = 5 to find that
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(5) = (8(5)-24)/((5)^4-8(5)^3+16(5)^2)
f(5) = 0.64
which has the same behavior as the left side
So overall, as we approach x = 4, the y value is heading off to positive infinity
Again everything is summarized in the image attachment
Note: you could make a table of more values but they would effectively say what has already been said. It would be redundant busy work. However, its always good practice for function evaluation.
-------------------------------------------------------------------------
Explanation:
Set the denominator x^4-8x^3+16x^2 equal to zero and solve for x
x^4-8x^3+16x^2 = 0
x^2(x^2-8x+16) = 0
x^2(x-4)^2 = 0
x^2 = 0 or (x-4)^2 = 0
x = 0 or x-4 = 0
x = 0 or x = 4
The x values 0 and 4 make the denominator zero
These x values lead to asymptote discontinuities because the numerator 8x-24 = 8(x-3) has no common factors which cancel with the denominator factors.
There are two vertical asymptotes
Let's see what happens when we plug in a value to the left of x = 0, say x = -1, we'd get
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(-1) = (8(-1)-24)/((-1)^4-8(-1)^3+16(-1)^2)
f(-1) = -1.28
So as x gets closer and closer to x = 0 from the left side, the f(x) is heading to negative infinity
Now plug in some value to the right of x = 0. I'm going to pick x = 1
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(1) = (8(1)-24)/((1)^4-8(1)^3+16(1)^2)
f(1) = -1.78 (approximate)
So as x gets closer and closer to x = 0 from the right side, the f(x) is heading to negative infinity
Overall, as x approaches 0 from either the left or right side of x = 0, the y value is heading off to negative infinity
---------------------
Repeat for values to the left and right of x = 4
We can't use x = 1 as it turns out that x = 3 is a root
But we can use something like x = 3.5 to find that...
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(3.5) = (8(3.5)-24)/((3.5)^4-8(3.5)^3+16(3.5)^2)
f(3.5) = 1.31 approx
So as x gets closer to x = 4 from the left, y is getting closer to positive infinity
Plug in x = 5 to find that
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(5) = (8(5)-24)/((5)^4-8(5)^3+16(5)^2)
f(5) = 0.64
which has the same behavior as the left side
So overall, as we approach x = 4, the y value is heading off to positive infinity
Again everything is summarized in the image attachment
Note: you could make a table of more values but they would effectively say what has already been said. It would be redundant busy work. However, its always good practice for function evaluation.