All Answers 358089 Answers

NO

\(\frac{2}{5}\times 2 = \frac{4}{5}\\ 1\frac{4}{3}\;\mbox{is not even a proper mixed numeral - the little number has to be on the top of the fraction.}\)

-14  +63 =  +63  -14    = 63-14 =   ???

p^(2/5)-5^(1/5)+6=0

p= 32 and 243

(1/2)x-1=(2/3)y

(20/9)=(1/3)x+(1/3)y

x=(14/3)

y=2

Please help me solve these before tomorrow? My teacher gave me the answers but I can't figure out how to get to that point.

Simplify the following:
32^(2/5)-5^(1/5)+6

32^(2/5)  =  (32^2)^(1/5)  =  (2^10)^(1/5)  =  2^2:
2^2-5^(1/5)+6

2^2 = 4:
4-5^(1/5)+6

4-5^(1/5)+6 = 10-5^(1/5):
Answer: |  10 - 5^(1/5)

Simplify the following:
243^(2/5)-5^(1/5)+6

243^(2/5)  =  (243^2)^(1/5)  =  (3^10)^(1/5)  =  3^2:
3^2-5^(1/5)+6

3^2 = 9:
9-5^(1/5)+6

9-5^(1/5)+6 = 15-5^(1/5):
Answer: |  15 - 5^(1/5)

(1/2)*(14/3)-1=(2/3)*2=4/3=4/3

(1/3)*(14/3)+(1/3)*2=20/9

Simplify the following:
(14/3)/3+2/3

(14/3)/3 = 14/(3×3):
14/(3×3)+2/3

3×3  =  9:
14/9+2/3

Put 14/9+2/3 over the common denominator 9. 14/9+2/3  =  14/9+(3×2)/9:
14/9+(3×2)/9

3×2  =  6:
14/9+6/9

14/9+6/9  =  (14+6)/9:
(14+6)/9
Answer: |  20/9

The century decay factor  = 0.96^100  = about  0.01867   = [ .02 ]  rounded

We can't tell unless we factor out the "i's"  [eyes], first......!!!!

As with melody, I'm not sure where your gettin the exponent 5 from. However, I've checked and double checked the math and looked around for another answer online and you aren't the first to notice this, but yes, you're factually correct according to the internet.

Change of base is this :   log_a b   = log b / log a    .....so....

log₄ 67  = log 67 / log 4  = about   3.033

Your answers to #6 are correct.

 a(b+c) = bc for c     distribute the "a" over the terms in the parentheses on the left

ab + ac  = bc     subtract ac from both sides

ab = bc - ac       factor out "c" on the right

ab = c( b - a)     divide both sides by (b - a)

ab / ( b - a)  =   c

And that's it.....not too bad, huh??

3 < log 4 67 < 4

-4(3x-5)-3(2+7x)+15-10x    =    

-4(3x) + (-4)(-5)  -3(2) - 3(7x) + 15 - 10x   =

-12x + 20  - 6 - 21x  + 15 - 10x      collect like terms

-12x - 21x- 10x +  20 - 6 + 15  =

-43x + 29 

saying that your useing 3.14 (pie) for that. 

I thing 431.75 or 215.87

saying that your useing 3.14 (pie) for that. 

I thing 431.75 or 215.87

yes, it means you can reverse it from a negetive and a postive number

do you have measuments ??

aa+ab

Passing one of the classes is independent of passing the other. However, the probability of passing both classes depends on the probability of passing the two classes seperatly.

is the answer long and pointy

The answer is 972972.

 is the answer 579

Well, you first want to distrubute that -4 over and that -3 over... Then just combine like terms, so here it is semi-step-by-step:

1. -4(3x-5)-3(2+7x)+15-10x

2. -12x +20 - 6 - 21x + 15 - 10x

3. -43x + 29

Your answer would be -43x + 29 after combining like terms! Hope I could help!

-Nick

@Other Guest

Actually, Chris is right. I tried 144 on alcumus and it was wrong- 720 is correct.

y = -5 - x    (1)

x = 5 - x   →  2x = 5    →  -5  = -2x   (2)

Subbing (2) into (1), we have

y = -2x - x

y = -3x   →     y/-3   = x

And using (2)

y/-3 = 5 - y/-3

2y/-3  = 5

[-2/3]y = 5    multiply both sides by -3/2

y = -15/2

And x =  y/-3 =  (-15/2)/ (-3)  = 15/6  =  5/2

So....the solution is   (x, y)  = ( 5/2, -15/2)

a²+ab

If  A+B+C=pi & cosA=cosB*cosC then prove that-

tanA=tanB+tanC

A  = pi - (B + C)     ......therefore.....

sin A = sin [(pi) - ( B + C) ]

sinA  = sin(pi)cos(B +C) - sin(B+ C)cos(pi)      ......    [ sin(pi) = 0,   cos(pi) = -1   ]

sinA  = sin(B + C)

sinA  =  [ sinBcosC + sinCcosB]

Therefore

tanA  = sinA/cosA          [given →   cosA = cosBcosC]

tanA  =  [ sinBcosC + sinCcosB] / [cosBcosC]      

tanA  = [sinBcosB]/ [cosBcosC] + [sinCcosB] / [cosBcosC]

tan A  = [sinB/cosC] + [ sinC/cosC]

tanA = tanB  + tanC