irrational number, any real number that cannot be expressed as the quotient of two integers.
Any number that can be written as a fraction with integers is called a rational number . For example, 17 and −34 are rational numbers. (Note that there is more than one way to write the same rational number as a ratio of integers.
1st Is rational
2nd one is irrational
3rd is irrational
4th is rational
Answer:
8
Step-by-step explanation:
Step-1 : Multiply the coefficient of the first term by the constant 1 • -16 = -16
Step-2 : Find two factors of -16 whose sum equals the coefficient of the middle term, which is 6 .
-16 + 1 = -15
-8 + 2 = -6
-4 + 4 = 0
-2 + 8 = 6 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -2 and 8
p2 - 2p + 8p - 16
Step-4 : Add up the first 2 terms, pulling out like factors :
p • (p-2)
Add up the last 2 terms, pulling out common factors :
8 • (p-2)
Step-5 : Add up the four terms of step 4 :
(p+8) • (p-2)
Which is the desired factorization
Answer:
Neither.
Step-by-step explanation:
The gradients to the line are not co-related.
For problems like this, change the wording into simple algebraic
formulas. We know that the area of a rectangle is its width times its
length, or A=LW. The problem tells you that the area is 48, so LW=48.
Also, if the length = twice the width minus 4, then L=2W-4.
From here, substitute your new L value for the L in the first equation:
48=LW --> 48=(2W-4)W
Now solve this equation by factoring it out into a polynomial:
48=2W^2-4W --> 24=W^2-2W --> W^2-2W-24=0 --> (W+4)(W-6)=0.
Solving this equation gives us W values of 6 and -4, but since the width
of a rectangle cannot be negative, the width must be 6.
Since L=2W-4, then L=2(6)-4 --> L=12-4 = 8. Therefore, the dimensions are 6X8, or 6 feet by 8 feet.
To check our work: The length equals four feet less than twice the width: 8=2(6)-4 --> 8=12-4 --> 8=8. This checks out.
Also, the area is 48 ft^2, and (6)(8) = 48, so this also checks out.
Hopefully this helps. If you need any more help or if I went over something too quickly, just let me know.
