Answer:
= r²−4r+3
Step-by-step explanation:
(r - 3) (r - 1)
(r x r) + (r x -1) + (-3 x r) + (-3 x -1)
r² + - r - 3r + 3
= r²−4r+3
Answer:
C
Step-by-step explanation:
it describes how much the shown graph extends left and right.
Note that 4 is included, but -2 itself not. that's why it's empty and A isn't correct.
B and D are completely inapplicable
pls brainliest
Answer:
C
Step-by-step explanation:
We have:
Where a<0.
We would like to solve for a. So, let’s divide both sides by a.
Notice that a<0. Therefore, a is negative. Hence, we must flip the sign since we are dividing by a negative. This yields:
Now, we will subtract b from both sides. Therefore, our inequality is:
Hence, our answer is C.
Answer:
part a: 5
part b: no, ABC is not an equilateral triangle
Step-by-step explanation:
part a: if you do the distance formula for points A and B, you get 5
part b: using the distance formula again, points A to B and points A to C are both 5, but points B to C is 6, therefore it's not equilateral because all the sides have to be the same length
Answer:
Cartesian
z₁= 3 +4*j
z₂= 2 +3*j
Polar
z₁=5 * e^ (0.927*j)
z₁=√13 * e^ (0.982*j)
Step-by-step explanation:
for the complex numbers z the cartesian form of is
z= x + y*j
then
1) z₁= 3 +4*j (cartesian form)
2) z₂= 2 +3*j (cartesian form)
the polar form is
z= r* e^jθ
where
r= √(x²+y²) → r₁ = √(3²+4²) = 5 , r₂ = √(2²+3²) = √13
and
θ = tan⁻¹ (y/x) → θ₁ = tan⁻¹ (4/3)= 0.927 rad , θ₂ = tan⁻¹ (3/2)= 0.982 rad
then
z₁=5 * e^ (0.927*j)
z₁=√13 * e^ (0.982*j)
