NP Strategy Guide 4th edition In Action Roots Problem Set Chapter 6 Page 83 Question 1 For each of these statements, indicate whether the statement is True or False: (a) If x^2 = 11, then X = √11 On page 77 it says: Rule: Even roots only have a positive value. √4 = 2 NOT +/- 2 The solution state...

(12)^5 = ( 4 . 3 )^5 = (4^5) . 3^5 => Since a^x . b^x = (ab)^x = (2 . 2)^5 . 3 ^5 => Reduce 4 to 2 . 2 = (2^5 . 2^5) . [3^5 ] => Since a^x . b^x = (ab)^x = 2^(5+5) . 3^5 => Since x^a . x^b = x^(a+b) = 2^(10) . 3^5

Hi, For each of these statements, indicate whether the statement is True or False: (a) If x^2 = 11, then X = √11 On page 77 it says: Rule: Even roots only have a positive value. √4 = 2 NOT +/- 2 The above rule applies only to numbers not variables. hence √(x^2) will always of 2 solutions....s...

Hi Please see the below Every non-negative real number x has a unique non-negative square root, called the principal square root , which is denoted with a radical symbol as √, or, using exponent notation, as x^1/2. For example, the principal square root of 9 is 3, denoted √9=3, because 32 = 3 Ã...

Is the sume of integers a and b divisible by 7? (1) a is not divisible by 7 (2) a - b is divisible by 7 I get how (1) and (2) alone by themselves are not sufficient From the answer key: Using the coinmbined values it "tells us that a and b have the SAME REMAINDER when divided by 7 I can't under...

This means that remainders for 'a' and 'b' when divided by 7 are same. This is a conclusion you can draw if 'a' is not divisible by 7 but a-b is. I see the example and understand. I think the following is the mathematical proof for those who want to understand why. (1) Let: a/7 = X + R1 = 7A + R1 =...

ray_serrano Wrote:How does (54 + m) + 0.20(54 + m) becomes 1.2(54 + m)? Where is this covered in the guides?

(54 + m) [ 1 + 0.20 ] => Factor the common (54 + m)
=(54 + m) [1.2]
= 1.2(54 + m)

It might help thinking of it this way: x^2 = 4 Implies x = +/- √ 4 = +/- 2 Notice the +/- is before the √ sign indicating that the value from the √ is always positive. Another example is in the quadratic formula to find the roots Eqn: ax^2 + bx + c = 0 => x = [-b ± √(b2 - 4ac)]/2a Here a...

Q from MGMAT guide 4, page 76, Q14: Three gnomes and three elves sit down in a row of six chairs. If no gnome will sit next to another gnome and no elf will sit next to another elf, in how many different ways can the elves and gnomes sit? First please put the Guide Name and edition as mentioned in ...

Please note that I haven't read what the "SLOT method" is so it could be different.

Question states: "However while we cannot conclude that g is Not divisible by 4, we cannot be certain that g is divisible by 4 either" -- I am confused, my question is dont these two sentences mean the same that basically g is divisible by 4?? Please advise. What it means is the we cannot...

If G^2 < G - which of the following could be G ... The answer is 0 < G < 1 Could someone give me an algebraic solution to this? I am doing something wrong here: G^2 < G => G ^2 - G < 0 => G( G - 1) < 0 => G < 0 , G-1 < 0 => G < 0, G < 1 => G < 1 is obv...

May have found the answer G(G-1) < 0 This means 1) G < 0 and (G-1) > 0 G < 1 and G > 1 -> Invalid or 2) G >0 and (G-1) < 0 G > 0 and G < 1 0 < G and G < 1 0 < G < 1 for quadratic inequalities you need to test all cases E.g. if (x-a)(x-b) < 0 Then we know ...

If AB/7 > 1/14 and A=B which of the following must be greater than 1

The solution states:
(C) 2A^2 since 2A^2 > 1

Fairly obvious.

Can someone please explain why A+B is not greater than 1

Thanks

Number of ways to choose 4 cars from 20 = 20 Choose 4 = 4845 = 3*5*17*19 Number of ways to choose 3 non defective cars if you have chosen 1 defective car 17 Choose 3 = 680 = 17 * 8 * 5 Number of ways to choose 1 defective car = 3 Therefore: P(Choosing exactly 1 defective car) = Number of ways of cho...