Answer:
Is it 2?
Step-by-step explanation:
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 numbers after the decimal place are represented as 10ths, hundredths, thousandths, and so on. 0.6 is said as "Zero point six tenths." Or "Zero and six tenths."
0.06 is said as "Zero point zero six hundredths." Or "Zero and six hundredths."
Answer:
B.
Step-by-step explanation:
factored form actually contains the x intercepts right in it. Basically whatever sets the whole equation equal to zero.
On the graph you can see that it passes through both positive and negative 4. so (x+4)(x-4) is the answer because plugging in 4 for x will make it zero. Same for plugging in negative 4
Hope this helps!
If you mean "factor over the rational numbers", then this cannot be factored.
Here's why:
The given expression is in the form ax^2+bx+c. We have
a = 3
b = 19
c = 15
Computing the discriminant gives us
d = b^2 - 4ac
d = 19^2 - 4*3*15
d = 181
Note how this discriminant d value is not a perfect square
This directly leads to the original expression not factorable
We can say that the quadratic is prime
If you were to use the quadratic formula, then you should find that the equation 3x^2+19x+15 = 0 leads to two different roots such that each root is not a rational number. This is another path to show that the original quadratic cannot be factored over the rational numbers.
