√2 is irrational because let's suppose √2 is a rational number. Then, we write
√2=a/b where a and b are whole number, b≠0
From the equality √2=a/b, it follows that 2= a²/b²( Make all of them to the second power), or a²=2×b²
Then, a is 2 times some other whole number. In symbols, a=2k where k is this other number
If we substitute a=2k into the original equation 2=a²/b², this is what we get:
2=(2k)²/b²
2=4k²/b²
2b²=4k²
b²=2k². This means that b² is even, from which follows again that b itself is we . And that is contradiction, and thus our original assumption (that √2 is rational). Therefore, √2 can not be rational
√2=1.41421356237..
√2=1.4( rounded to the nearest tenth). Hope it help!
9514 1404 393
Answer:
a) turning point: (-2, -1)
b) roots: (-3, 0), (-1, 0)
Step-by-step explanation:
The turning point is correctly identified as the vertex in the problem picture.
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The roots are the x-intercept points: (-3, 0) and (-1, 0).
Answer:
Step-by-step explanation:
Triangle ABC is similar to triangle EDC. Therefore,
AB/ED = BC/DC = AC/EC
1) Angle A is congruent to angle E.
2) BC/DC = AC/EC
From the information given,
BC = 6
EC = CE = 20 1/4 = 20.25
AC = 9
Therefore,
6/DC = 9/20.25
Crossmultiplying, it becomes
9 × DC = 6 × 20.25 =
9DC = 121.5
DC = CD = 121.5/9 = 13.5
AB/ED = BC/DC
DE = ED = 22 1/2 = 22.5
Therefore,
AB/22.5 = 6/13.5
Crossmultiplying, it becomes
13.5 × AB = 22.5 × 6
13.5AB = 135
AB = 135/13.5
AB = 10
Answer:
Read it and do your best
Step-by-step explanation:
If the third term of the aritmetic sequence is 126 and sixty fourth term is 3725 then the first term is 8.
Given the third term of the aritmetic sequence is 126 and sixty fourth term is 3725.
We are required to find the first term of the arithmetic sequence.
Arithmetic sequence is a series in which all the terms have equal difference.
Nth term of an AP=a+(n-1)d
=a+(3-1)d
126=a+2d--------1
=a+(64-1)d
3725=a+63d------2
Subtract second equation from first equation.
a+2d-a-63d=126-3725
-61d=-3599
d=59
Put the value of d in 1 to get the value of a.
a+2d=126
a+2*59=126
a+118=126
a=126-118
a=8
=a+(1-1)d
=8+0*59
=8
Hence if the third term of the arithmetic sequence is 126 and sixty fourth term is 3725 then the first term is 8.
Learn more about arithmetic progression at brainly.com/question/6561461
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