Simplify: (-2i) (8i)
When multipling i's together, they fall out so your answer is 16
Oh! that makes sense, thank you cx
Take note that \[i=\sqrt{-1}\] Therefore by using distribution, you get \[-16i^2=-16(\sqrt{-1})^2=-16(-1)=16\]
True or false: Any number in the form of a+b, where a and b are real numbers and b=0 is considered a pure imaginary number.
Also what @Hoslos said.
Ohhh
False
Real number is that without the i constant, whereas an imaginary number has i constant. Therefore if a and b are real, there is no way it is an imaginary number. False.
Thank you so much guys! Can you help with some more?? I have a deadline in like two hours and have 20 things to do Dx
Sure!
I've got all day!
Sweet, thank you ^^ The minimum value or a function is the smallest y-value of the function True or false
I think true
True. It is the peak of the parabola of the function that is u-shaped.
Alright, And one more true or false: Complex numbers can be graphed on the real xy coordinate plane
False
False. X-axis is for Real Numbers and y-axis for imaginary numbers.
Suppose a parabola has an axis of symmetry at x=-1, a maximum height of 6, and passes through a the point (-2,1). Write the equation of the parabola in vertex form A. y=-0.56(x-1)^2+6 B. y=-5(x+1)^2+6 y+-2.5(x-1)^2+6 C. y=-2(x+6)^2-1
Not sure but i'm thinking b...
Same here
Use the formula\[y=a(x-p)^2+q\] Where q is y-vertex and p a x-vertex. A point is to be replaced at y and x being left with a, which is hat we have to find first. So, making substitutions, we get\[1=a[-2-(-1)]^2+6\rightarrow1=a+6\rightarrow a=1-6=-5=a\] Lastly, we rewrite the same formula, without replacing x and y as the points. That is \[y=-5(x+1)^2+6\]
B shall be the answer.
Thank you so much!! Vertex form of this equation: y=-x^2+12x-4?? and expression in factored form 2x^2+16x+24
First one: The vertex is (-6,-40) Second one: 2(x+6)(x+2)
And the question is...?
Thank you so much everyone!!
No problem @Squeeks1713 .
Join our real-time social learning platform and learn together with your friends!