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so your saying the answer is 12.5 ? because 1*a = 12.5

1. C

2. A

Spring force = k l    k = 50 N/m 

    stretch it .25 m     .25 m x 50 N/ m = 12.5 Newtons

Now remember   f= ma ?     You have found f = 12.5   and you are given mass = 1 kg    ....you should be able to find 'a'

hmm, im not quite sure then but i as well thought it was 0.70 hz

what about this one?

A 1.0 kg weight suspended from a spring is pulled to 0.25 m below its equilibrium point.

If the spring has a spring constant (k) of 50.0 N/m , at what rate will the mass accelerate when it is released?

200 m/s2

12.5 m/s2

6.5 m/s2

2.7 m/s2

2.7 ?

thank you so much! this helped me a lot :)

We're glad to have you, CoolStuffYT   !!!

6^x  =  5           Raise both sides of this equation to the power of 3

( 6^x )³   =   5³           Simplify the left side using the rule  ( x^a )^b  =   x^ab   and simplify the right side.

6^3x   =  125           Multiply both sides of the equation by  6²

6^3x  ·  6²  =  125 · 6²           Simplify the left side using the rule  x^a · x^b  =  x^a+b   and simplify the right side.

6^3x+2  =  4500

EP's  answer is correct....

See  this, Jenny   :  https://www.answers.com/Q/What_is_the_frequency_of_a_pendulum

Period = 2 pi sqrt(l/g) = 2 pi sqrt (.5/9.81) = 1.419 sec

Frequency = 1/period = 1/1.419 = .70 hz

First one   

Two of the numbers are > 4

We have 6 total numbers

What's the probability  ???

Second one

We have 3 evens

We hav 6 total numbers

What's the probability  ???

my teacher said it isnt 0.70 , i know that forsure

false!

If you list all the possibilite outcomes

H H H

H H T

H T H

H T T

T H H

T H T

T T H
T T T

There are 3 cases where exactly one head appears, but there is only 1 case where no heads appear.

So it is more likely to get exactly one head than to get no heads.

Correct !!!

C. 0

There are no orange  cars.....so.....what's the probability  ???

a b c d e f g h i j k  l  m  n   o  p  q  r  s t u v w  x y  z

bvrf    

Because of the "a"....shift the  b back  one letter   = a

Because of the d, shift the v back 4 letters  =   r

Because of the e, shift the r back 5 letters  = m

Because of the g, shift the  f back 7 letters = y

" a  r   m  y "

log _1/2  16 -  log ₉  27      

Note that  in exponential form

(1/2)^(-4)  =  1 / (1/2)^4  =  1  /(1/16)   = 16

So....the first expression evaluates to -4

And

9^x  = 27

(3^2)^x  = 3^3

(3)^(2x)  = 3^3      the bases are the same.....solve for the exponents

2x= 3    divide both sides  by  2

x  = 3/2

So.....the second expression evaluates to 3/2

So

-4  - (3/2)   =

-8/2  - 3/2  =

-11/2  = -5.5

Second one

P( no 3 appears on either face)  =   5/6  * 5/6   =  25 / 36

First one

We have 36 outcomes

P(2)  = 1/36

P(3)  = 2/36

P(4)  = 3/36

P(5) = 4/36   =  1/9

P ( passing first test)  *  P (passing second test)  =  P(passing both tests)

.42   *  P(passing second test )  =  .25          divide both sides by .42

P(passing second test)  =  .25 /.42  ≈  .595  ≈  60%

The first is false

P( no heads appear)  =  1 / 8

P( exactly one head appears)  = 3/8

Second one

P(three heads appear)  =  1/8

( x^2 + y^2) / (xy)  =  8      (1)

(x^6 + y^6) / ( x^3y^3)   =   ( x^2 +y^2) ( x^4 - x^2y^2 + y^4) / (x^3y^3)   (2)

Rearrange (1)  as    (x^2 + y^2) = 8(xy)   sub into  (2)

8 ( xy)  ( x^4 - x^2y^2  + y^4) / ( x^3y^3)  =   8 ( x^4 - x^2y^2 + y^4) / (x^2y^2)      (3)

And  

(x^2 + y^2)  = 8xy       square both sides

x^4 + 2x^2y^2 + y^4  =  64x^2y^2

2x^2y^2  =   64x^2y^2 - x^4 - y^4        add  -3x^2y^2 to both sides

-x^2y^2  =  61x^2y^2  - x^4 - y^4     sub this into  (3)   and we have that  

8 ( x^4 + (61x^2y^2 - x^4 - y^4) + y^4)  / ( x^2y^2)  

8  ( 61x^2 y^2)  / (x^2y^2)  =

8 * 61  =

488