Note: Consider we need to find the vertices of the triangle A'B'C'
Given:
Triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C'.
Triangle A,B,C with vertices at A(-3, 6), B(2, 9), and C(1, 1).
To find:
The vertices of the triangle A'B'C'.
Solution:
If triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C', then
Using this rule, we get
Therefore, the vertices of A'B'C' are A'(6,3), B'(9,-2) and C'(1,-1).
The answer is cubic trinomial because the highest exponent is 3 which makes it a cubic function and there are 3 terms which makes it a trinomial
Answer:correct answer =0
Step-by-step explanation:
Step 1
5x - (x + 3)2
Putting the coefficient of x =2 in the equaton we have
5(2) - (2 + 3)2
Solving out
5(2) - (5)2
= 10- 10
=0
Step 2
Amir's mistake was in the multiplication and wrong use of the negative sign of the bolded below
5(2) - (2 + 3)2 = 10 + (-5)2 = 10 + 25 = 35
-(5)2 cannot give 25 instead 10, so him getting a 25 and wrong use of the negative and positive made his answer wrong.
The right answer as calculated in step 1 = 0
Answer:
y=4x+2 or f(x)=4x+2
Step-by-step explanation:
This is because the equation format for a line is y=mx+b, the slope is m and the y-intercept is b. So all you have to do is fill it out to get y=4x+2. Or the other way I listed the equation is right too. Hope this helps!
Answer:
b = √2
Step-by-step explanation:
Pythagorean Theorem: a² + b² = c²
<em>a</em> = Leg₁
<em>b</em> = Leg₂
<em>c</em> = Hypotenuse
Step 1: Plug in variables
3² + b² = (√11)²
Step 2: Solve
9 + b² = 11
b² = 2
b = √2
And we have our answer!
