Answer:
Yes; the corresponding sides are proportional
Step-by-step explanation:
If two figures are similar, the lengths of their sides are proportional. This means if we set up a proportion with the sides we are given and cross-multiply, we get a true statement at the end of it.
Using the similarity statement ABCD~EFGH, we will compare AB to EF and BC to FG:
5/3 = 25/15
Cross multiply:
5(15) = 3(25)
75 = 75
We got a true statement, so the sides are proportional.
13. Is 18
15. Is 86
17 is 5
19 is 15
21 is 24
23 is 29
Multiply $30 by .25 you get, it's $7.50so you subtract $7.50 from $30 and you get $22.50. Then multiply 2 (sweaters) by $22.50 and you get $45,but wait you still have to multiply that sales tax so you multiply .04 by $45 and you get $1.80, later you subtract $1.80 from $45 and you get $43.20(the cost of the two sweaters)
I did part1 only cuz' no I have no time left Sorry for the inconvenience
Answer:
x = pi/2 + 2 pi n x = pi + 2 pi n where n is an integer
x = 5pi /3 + 2 pi n
Step-by-step explanation:
8 cos^2 x + 4 cos x-4 = 0
Divide by 4
2 cos^2 x + cos x-1 = 0
Let u = cos x
2 u^2 +u -1 =0
Factor
(2u -1) ( u+1) = 0
Using the zero product property
2u-1 =0 u+1 =0
u = 1/2 u = -1
Substitute cosx for u
cos x = 1/2 cos x = -1
Take the inverse cos on each side
cos ^-1(cos x) = cos ^-1(1/2) cos ^-1( cos x) =cos ^-1( -1)
x = pi/2 + 2 pi n x = pi + 2 pi n where n is an integer
x = 5pi /3 + 2 pi n
Answer:
B. A teacher compares the pre-test and post-test scores of students
Step-by-step explanation:
the answer is true, because it is a good example to compare the tests between students, we know that a matching pair design is a random model and is used when the experiment allows grouping subjects in pairs based on a variable and each pair will receive randomly a different handling, the answer A is not true because in the example all students are uniformly averaged and the variable is not correlated with subgroups, option c is incorrect because the variable was not randomized and generates classification bias and the option d is incorrect because the teacher compares a small sample as her class with a score of a total sample, but does not intervene on her students when selecting the corresponding group
