Answer:
The dimensions of the border are 7 yd by 6.25 yd.
A:b = 8:5 b:c = 3:4
since both sides have b. make b the same. multiply the left by the right side b value of 3 and multiply the right by the left side b value of 5.
3(a:b) and 5(b:c)
3 (8:5) and 5(3:4)
24:15 15:20
now that both sides have the same b value
a:b:c = 24:15:20
50 shirts is equal to 100% of the shirts right? So, to find what percentage 35 shirts is, you can divide 35 by 50 to get the decimal value of the percentage. This gives you 0.7, which is the same as 70%.
35 shirts are 70% of 50 shirts.
let's firstly convert the mixed fractions to improper fractions and then subtract, bearing in mind that the LCD of 4 and 2 is 4.
![\bf \stackrel{mixed}{8\frac{3}{4}}\implies \cfrac{8\cdot 4+3}{8}\implies \stackrel{improper}{\cfrac{35}{4}}~\hfill \stackrel{mixed}{7\frac{1}{2}}\implies \cfrac{7\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{15}{2}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{35}{4}-\cfrac{15}{2}\implies \stackrel{\textit{using the LCD of 4}}{\cfrac{(1)35~~-~~(2)15}{4}}\implies \cfrac{35-30}{4}\implies \cfrac{5}{4}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B8%5Cfrac%7B3%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B8%5Ccdot%204%2B3%7D%7B8%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B35%7D%7B4%7D%7D~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B7%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B7%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B15%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B35%7D%7B4%7D-%5Ccfrac%7B15%7D%7B2%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Busing%20the%20LCD%20of%204%7D%7D%7B%5Ccfrac%7B%281%2935~~-~~%282%2915%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B35-30%7D%7B4%7D%5Cimplies%20%5Ccfrac%7B5%7D%7B4%7D)
Only two real numbers satisfy x² = 23, so A is the set {-√23, √23}. B is the set of all non-negative real numbers. Then you can write the intersection in various ways, like
(i) A ∩ B = {√23} = {x ∈ R | x = √23} = {x ∈ R | x² = 23 and x > 0}
√23 is positive and so is already contained in B, so the union with A adds -√23 to the set B. Then
(ii) A U B = {-√23} U B = {x ∈ R | (x² = 23 and x < 0) or x ≥ 0}
A - B is the complement of B in A; that is, all elements of A not belonging to B. This means we remove √23 from A, so that
(iii) A - B = {-√23} = {x ∈ R | x² = 23 and x < 0}
I'm not entirely sure what you mean by "for µ = R" - possibly µ is used to mean "universal set"? If so, then
(iv.a) Aᶜ = {x ∈ R | x² ≠ 23} and Bᶜ = {x ∈ R | x < 0}.
N is a subset of B, so
(iv.b) N - B = N = {1, 2, 3, ...}
