QUESTION 3 Which sequence of transformations will turn figure 1 into figure 2? Figure 1 contains a right isosceles triangle oriented on its base with height on the left. Figure 2 is a larger right isosceles triangle oriented on its hypotenuse. a dilation and a rotation a reflection and a translation a dilation and a translation a reflection and a dilation
hard question i can help you if you like me to
#1:Proportional Medians Theorem
#2:BD/FH = CB/GF
#3: 24/16 27/x
So the first ratio is 3/2 because both are divisible by 8. the ratio stays 3/2. 3*9=27. so 2*9=18
#4: x = 8.5
#5: x = 68
Hope this helps!!
good for the kids fff and the other i.c tes 8x UTC UTC USA is us rd FCC FCC TV un un un in in in in km jvm
DC fcb GB jvm hbkddf ggv FCC FCC TV ggv ggv TV TV ggv th un
HOPE it is HELPFUL for you
Structural components of RNA nucleotide monomer includes the following a nitrogen base(adenine/guanine/cytosine/uracil) , A ribose sugar(with 2’-OH group present) and Three phosphate groups.
The role of the RNA POL is for transcribing DNA to RNA
(1) Structural components of RNA nucleotide monomer are:
a. A nitrogen base(adenine/guanine/cytosine/uracil)
b. A ribose sugar(with 2’-OH group present)
c. Three phosphate groups
Role of RNA POL- RNA pol is the principle enzyme for transcribing DNA to RNA. Bacteria has a single RNAP , while eukaryotes have 3 RNAP(plants have additional 2 polymerases).RNAP catalyses the joining of two ribonucleotides one with another by using nucleophilic attack mecvhanism using the 2 metal ions present in the active site of the RNA pol. The two nucleotides added over the template strand are complementary to it. The RNA pol does this job in 5 to 3 direction
(2) The dependent variable in the experiments is the- Maximum elongation rate the control group in the second experiment is the WILD STRAIN.
Use of control group- Control group consists of individual (here the wild type RNAP) that will show a known result under a particular experimental condition(amanitin treatment). This known response is used as a reference to access the response in a experimental group both qualitatively and quantitatively.
(3). Effect of amanitin-
For wild type- the maximum elongation rate decreased, since the Amanitin binds to the active site of the enzyme. So naturally the maximum elongation rate decreased.
For Mutant- The maximum elongation rate also decreased , but it appears the rate is little more than the wild type. This may be due to the particular amino acid substitution that countered the effect of Amanitin and gave the following result.
The ratio is= 1:6
( 4). The null hypothesis for experiment 1: The maximum elongation rate of the experimental strain will not change due to the amino acid substitution.
Reasoning- The shape of a protein is greatly determined by its amino acid sequence. Even a single amino acid change can cause a big change. So have done the said single amino acid substitution. Since the substitution was in the active site of the enzyme(which is involved in catalyzing the elongation process), so the mutant have definitely caused a change in the shape, that has caused to a reduction in the maximum elongation rate of the enzyme.
the answer is The quadrilateral PQRS and its image P'Q'R'S' is shown in the diagram below.
The transformation that gives the image is the reflection on the line y = x
This is proved by the distance of each coordinate of PQRS to the mirror line equals to the distance of each coordinate of P'Q'R'S' to the mirror line.
hope it helps!
There will be more essay questions
Assume that the number of mcq questions is x and that the number of essay questions is y
We are given that:
1- The total number of questions is 52.
This means that:
x + y = 52
This equation can be rewritten as:
y = 52 - x > equation I
2- Each mcq question is worth 0.5 point and each essay question is worth 2.5 points. Also, the total number of points is 100
This means that:
0.5x + 2.5y = 100 > equation II
Substitute with equation I in equation II and solve for x as follows:
0.5x + 2.5y = 100
0.5x + 2.5(52-x) = 100
0.5x + 130 - 2.5x = 100
-2x = -30
x = 15
Substitute with x in equation I to get y:
y = 52 - x
y = 52 - 15
y = 37
Based on the above:
Number of mcq questions = x = 15 questions
Number of essay questions = y = 37 questions
Therefore, there would be more essay questions which means you should spend more time studying essay questions.
Hope this helps :)
Where's the image of the surfaces?
y=1/4x-2 so it is a
plz mark me as branliest
(2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))
simplify the following:
x/x^2 + 3 x + 2 - (x (x + 1))/(x + 2)
combine powers. x/x^2 = x^(1 - 2):
x^(1 - 2) + 3 x + 2 - (x (x + 1))/(x + 2)
1 - 2 = -1:
x^(-1) + 3 x + 2 - (x (x + 1))/(x + 2)
put each term in 1/x + 3 x + 2 - (x (x + 1))/(x + 2) over the common denominator x (x + 2): 1/x + 3 x + 2 - (x (x + 1))/(x + 2) = (x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)):
(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2))
(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)) = ((x + 2) + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2)):
(2 + x + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))
3 x^2 (x + 2) = 3 x^3 + 6 x^2:
(2 + x + 3 x^3 + 6 x^2 + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))
2 x (x + 2) = 2 x^2 + 4 x:
(2 + x + 6 x^2 + 3 x^3 + 2 x^2 + 4 x - x^2 (x + 1))/(x (x + 2))
-x^2 (x + 1) = -x^2 - x^3:
(-x^2 - x^3 + 3 x^3 + 6 x^2 + 2 x^2 + 4 x + x + 2)/(x (x + 2))
grouping like terms, 3 x^3 - x^3 + 6 x^2 + 2 x^2 - x^2 + 4 x + x + 2 = (3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2:
((3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))
3 x^3 - x^3 = 2 x^3:
(2 x^3 + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))
6 x^2 + 2 x^2 - x^2 = 7 x^2:
(2 x^3 + 7 x^2 + (x + 4 x) + 2)/(x (x + 2))
x + 4 x = 5 x:
answer: (2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))
idk lol xd
ez do it yourself lazy