What is the structure of a major scale?
The structure of a major scale can be represented by: whole,whole, half, whole, whole, whole, half. whole=2 half notes away. half= 1 half note away. To understand what this means we need to look at a chromatic scale. This scale shows the procession of half notes from the previous note. So basically, it is all halves. A C chromatic scale will go start with a C and end at the next C (an octave). The scale is: C, C#, D, D#, E, F, F#, G, G#, A, A#, B, C. So if we use the structure: whole,whole, half, whole, whole, whole, half; we can outline a C Major scale. Note that half is= to one note change on a chromatic scale. Ex. C--> C# is a half. and C-->C#-->D is 2 half notes. Which means C--> D is a whole. A C Major is: C,D,E,F,G,A,B,C
What evidence is there for the endosymbiotic theory?
The evidence is in the DNA of organelles like mitochondria and chloroplast. The organelles have their own circular DNA and contain small ribosomes like ones found in prokaryotes. Also, the organelles replicate through binary fission which is how prokaryotes divide. In order to answer this, you need to understand what the endosymbiotic theory is. It is theorized that two prokaryotes developed a symbiotic relationship and formed a eukaryotic cell. The prokaryotes that were engulfed by protoeukaryotes became the organelles of that cell. With that knowledge, we need to prove that the organelles are prokaryotic. In order to do this, we need to understand the features of a prokaryote. Prokaryotes are characterized by their circular DNA, small ribosomes, a lack of organelles, and a nucleoid( no membrane-bound nucleus). They also divide through a process called binary fission. If we can determine that organelles like mitochondria are prokaryotic, then we can support the endosymbiotic theory. It is known that mitochondria contain circular DNA and small ribosomes. The fact that it has circular DNA is the mist prominent piece of evidence. Also, the fact that it divides through binary fission is also a notable piece of evidence.
At what points are the X-intercepts for the function: f(x)= x^2-3x-18 located at?
The answer to this problem is (-3,0) and (6,0). You can arrive at these points in one of two ways: by factoring the equation or plugging it into the quadratic formula. 1: To factor quadratic equations you will need to recognize the basic structure of it. It can be written as (ax^2+bx+c). Now to factor, you will need to find two numbers that when multiplied equal (a × c) and when added equal (b). This will break up the equation into two binomials. From there, you set the binomials equal to zero and solve for (x). For this specific problem, you can identify the coefficient a, b, and c as a=1, b= -3, and c= -18. Now (a × c) can be substituted as (1 × -18)= (-18). Now to find two numbers that add to (b=-3) and multiply to -18, you can focus on what numbers commonly multiply to -18. To do this, we factor out two numbers that multiply to (-18). The numbers (6 and3),(9 and 2), and (18 and 1) are factors that lead to -18 when one of them is negative and multiplied with the other. For ex: (-6 × 3)= (-18) and (6 × -3)= (-18) . From here we can choose a set of numbers that add to (-3). The pair (-6 x 3) is the best choice because those two numbers can add to (-3). (-6+3)= (-3) Since we have found our numbers, we can break it up into two binomials. They will be broken up as (x+3) ×(x+(-6))=0 Now we solve individual binomials to equal 0. (x+3)= 0----> x= (-3) to solve for x. you single out x. TO do this you subtract the (3) to the other side thus getting (x=-3) You do the same for the other binomial. (x+(-6)=0 Here we can add (-6) because it is negative. This results in x=6. We solve for the zero because the binomials are multiplied. So, to get 0 as our answer we need one of them to equal 0 because (0 × anything)=0. The points are (-3,0) and (6,0). 2. Plug the coefficients into ( -b ± √b2 − 4ac)/( 2a) Remember: a=1, b=-3 and c=-18 = (-1)(-3)= 3 so 3+ √((-3^2)-(4× 1× -18))/ (2× 1) =( 3+ √(9-(-72))/2= (3 + √81)/2= (3+9)/2=(12/2)=6 since we know √b2 − 4ac= 9. we can substitiute 9 in for the subtraction portion. (3-(9))/2= -6/2=(-3). So your answers are also (-3,0) and (6,0).