Is 6/23 minus square root 2 divided by 2 rational or irrational ?

Mathematics

2 answers:

Novosadov [1.4K]3 years ago

7 0

Irritational . Because they both can’t be a ratio I believe correct me if i’m wrong!

Marysya12 [62]3 years ago

6 0

Yes it’s irrational cuz the answer is like 1.739...and it continues
Here’s a tip, if it ends or repeats itself in a pattern than it’s rational, but if it’s infinite then it’s irrational

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NEED HELP ASAP!! I'm doing trigonometry and I've been stuck on this problem I don't really understand it, if you could help that

Tems11 [23]

Answer:

6.1

Step-by-step explanation:

Draw a picture of an equilateral triangle. Cut the triangle in half, so that you get two 30-60-90 triangles. The area of these smaller triangles is 8 square inches.

The short leg of these triangles (the base) is half the side length: ½ s.

According to properties of 30-60-90 triangles, the long leg (the height) is √3 times the short leg: ½ s√3.

Area of a triangle is half the base times the height:

A = ½bh

8 = ½ (½ s) (½ s√3)

8 = ⅛ s²√3

64 = s²√3

s² = 64/√3

s = √(64/√3)

s ≈ 6.1

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3 years ago

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Daniel [21]

see the attached figure with the letters

1) find m(x) in the interval A,B
A (0,100) B(50,40) -------------- > p=(y2-y1(/(x2-x1)=(40-100)/(50-0)=-6/5
m=px+b---------- > 100=(-6/5)*0 +b------------- > b=100
mAB=(-6/5)x+100

2) find m(x) in the interval B,C
B(50,40) C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
m=px+b---------- > 40=(6/5)*50 +b------------- > b=-20
mBC=(6/5)x-20

3) find n(x) in the interval A,B
A (0,0) B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
n=px+b---------- > 0=(6/5)*0 +b------------- > b=0
nAB=(6/5)x

4) find n(x) in the interval B,C
B(50,60) C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
nBC=(3/5)x+30

5) find h(x) = n(m(x)) in the interval A,B
mAB=(-6/5)x+100
nAB=(6/5)x
then
n(m(x))=(6/5)*[(-6/5)x+100]=(-36/25)x+120
h(x)=(-36/25)x+120
find h'(x)
h'(x)=-36/25=-1.44

6) find h(x) = n(m(x)) in the interval B,C
mBC=(6/5)x-20
nBC=(3/5)x+30
then
n(m(x))=(3/5)*[(6/5)x-20]+30 =(18/25)x-12+30=(18/25)x+18
h(x)=(18/25)x+18
find h'(x)
h'(x)=18/25=0.72

for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72

 h'(x) = 1.44 ------------ > not exist