y=f(x) is odd (f(-x)=-f(x)) and f(x>0) = x^2-5x-12, x is not = to 0. What points does the line f(x)=12 go through.
1 answer:
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Answer:
- parallel: y = 4x -6
- perpendicular: y = -1/4x +27/4
Step-by-step explanation:
If we want the new line to be written in slope-intercept form, we need to find the new value of the y-intercept. The equation of the line is ...
y = mx +b . . . . . . . for slope m and y-intercept b
Solving for b gives ...
b = y -mx . . . . . . . subtract mx from both sides.
The values of x and y come from the point we want the line to pass through. The value of m will be the same for the parallel line as for the given line: 4. For the perpendicular line, it will be the opposite reciprocal of this: -1/4.
<u>Parallel line</u>
b = 6 -4(3) = 6 -12 = -6
y = 4x -6
Perpendicular line
b = 6 -(-1/4)(3) = 6 +3/4 = 27/4
y = -1/4x +27/4
7) Certainly there is a typo in the statement, just see that the expression of item (ii) is different from that of item (i). Probably the correct expression is: 
. With this consideration, we can continue.
(i) Let E the expression that we are analyzing:
Since (x-1)² is a perfect square, it is a positive number. So, E is a result of a sum of two positive numbers, 2(x-1)² and 3. Hence, E is a positive number, too.
(ii) Manipulating the expression:
So, it's the case when E=0. However, E is always a positive number. Then, there is no real number x that satisfies the expression.
8) Let E the expression that we want to calculate:
Multiplying by (2-1) in the both sides:
Repeating the process, we obtain:
(i) Let E the expression that we are analyzing:
Since (x-1)² is a perfect square, it is a positive number. So, E is a result of a sum of two positive numbers, 2(x-1)² and 3. Hence, E is a positive number, too.
(ii) Manipulating the expression:
So, it's the case when E=0. However, E is always a positive number. Then, there is no real number x that satisfies the expression.
8) Let E the expression that we want to calculate:
Multiplying by (2-1) in the both sides:
Answer:
The correct options are;
EFGH has 4 congruent sides
Diagonal FH bisects angles EFG and EHG
Angle FEH is congruent to angle FGH
Step-by-step explanation:
1) Given that for a reflection, we have;
The distance of the reflected preimage from the line of reflection = The distance of the reflected image from the line of reflection
Therefore;
The distance of the point E from the line HF = The distance of the point G from the line HF
Also the reflection of an preimage (x, y) about the x-axis, gives an image (x, -y)
We can show that from the length of a line given by the equation, that the length EH ≅ GH and EF ≅ GF
Therefore since we are given that EH = EF, we have;
EH = GH = GF = EF by the definition of congruency, which gives 4 congruent sides
2) Given that EH = GH = GF = EF and HF = FH by reflective property, we have;
ΔEHF ≅ ΔGHF
∴ ∠GHF ≅ ∠EHF by Congruent Parts of Congruent Triangles are Congruent
Similarly, ∠GFH ≅ ∠EFH
Therefore, ∠GFH = ∠EFH and ∠GHF = ∠EHF
Therefore, diagonal FH bisects angles EFG and EHG
3) Given that ΔEHF ≅ ΔGHF, we have;
Angle FEH is congruent to angle FGH, by Congruent Parts of Congruent Triangles are Congruent