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x^2 + bx + 2

This parabola turns upward....so...the x coordinate of the vertex is given by

-b/ 2  = x

So....we'd ike to solve this

(-b/2)^2 + b(-b/2)  + 2  >  - 2    simplify

b^2/4 - b^2/2 + 4  >  0     

-b^2/4 + 4 > 0

-b^2 + 16 > 0  multiply through by  -1  and change the inequality sign

b^2 - 16 < 0

(b - 4) ( b + 4) < 0

The values  that make this true  are  ( -4, 4)

So...the values that put  -2 out of range  are  (-4, 4)

The distance in the plane, that is: Formula: \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) . We have the points, (2a, a-4) and (4,-1).

\(\sqrt{(4-2a)^2+(-1-a-4)^2}=2\sqrt{10}\), so we get a=1, a=1/5. So, there are two values! Thus, 1*1/5=1/5!

Thank you! Great!

I see only two: 1,000 + 7,000 =8,000.

Here's a pic  (not to scale) :

We have  16  possible outcomes  [2 for each coin = 2^4  = 16 ]

The penny and dime  can both show tails...and there will be 2 * 2  = 4 outcomes for the other two coins

or

The penny and dme  can both show heads...and there will be 2 * 2  = 4 outcomes for the other two coins

So....P ( penny and dime have the same results ) =   is  [ 4  + 4  ]  / 16  =  8 / 16   =  1 / 2

g (2x)  =  ³√ [  (2x + 3) / 4 ]    (1)  

2 (g(x) ) =  2 * ³ √  [ (x + 3 ) / 4 ]    (2)

Let  us  cube  (1) , (2)  and set them equal...so we have...

(2x + 3) / 4  =  8 (x + 3) / 4

(2x + 3)  = 8(x + 3)

2x +  3  = 8x + 24

-21  = 6x

x  = -21/6  =  -7/2

The nickel and quarter are irrelevant in this problem.

So the probability of both the penny and the dime landing on heads is 1/4, and the probability of them landing on tails is also 1/4. The probability of them coming up the same is thus 1/2.

\(g(2x)=\sqrt[3]{\frac{2x+3}{4}}\)

\(2g(x)=2*\sqrt[3]{\frac{x+3}{4}}\)

Cubing both sides, we get \(\frac{2x+3}{4}=2x+6\)

Multiplying by 4, we get \(2x+3=8x+24\)

So \(x=-3.5\)\(\)

This answer is wrong, the answer is 50%

This one was a little difficult...!!!

Note : angle AEB  = angle DEC

and angle CDB  = angle DBA

So...by AA congruency....ΔBEA ~ ΔDEC  →  DC /AB  = DE / BE

Area of  ΔDBA  = area of  ΔADE  + area of ΔABE  = 20 + 50  = 70

So  area of Δ DBA  = (1/2)sin DBA(AB * DB)  = 70

So...sin DBA = 140 / ( AB * DB)

And sin (DBA) = sin CDB

So area of  ΔCDB  = (1/2) sin (DBA)(DB * DC)  = (1/2) (140 / [(AB * DB] ) (DB * DC) =

70 (DC/AB)  = 70 ( DE /BE)

Now

Area ΔAEB  = (1/2)sin AEB ( AE * BE) = 50   (1)

Area Δ ADE  = (1/2) sin DEA (AE * DE)  = 20  (2)

But  angle AEB  is  supplemental to angle DEA  so thier sines are equal

So  sin AEB  = sinDEA

Rearranging (1), (2)  we have that

sin AEB  = 100 / [ AE * BE]      =     sin DEA = 40 / [ AE * DE ]

 So

100 / BE   = 40 / DE  →   DE / BE  = 40/100  = 2/5

But area of ΔCDB  was found to be  70 (DE / BE) = 70 (2/5)  =  28

So...area of trapezoid ABCD  = area of ΔDBA  + area of ΔCDB  =

                                                               70        +        28             =      98

Please help fast!

The first four prime numbers are 2,3,5,7

So for the first game, he chooses either 0 or 1

For the second one, he chooses 0,1, or 2

For the third, he chooses 0,1,2,3 or 4

For the fourth, he chooses 0,1,2,3,4,5, or 6

There are a total of \(2*3*5*7=210\)choices

There is only one case where the sum is zero

So there are 209 cases where the sum is greater than zero

The probability that the sum is greater than zero is thus \(209/210\)

Should be 98, I think...

In need of help here!!

First, find the length of a side. Since it is a rhombus, the side length is \(148/4=37\)

Applying the pythagorean theorem, we get half the diagonal squared plus half 24 squared equals the side length squared.

So \((\frac{1}{2}x)^2+12^2=37^2\) and half the diagonal is equal to \(35\)

Thus the diagonal is equal to \(35*2=70\)

[4! /2!*2!] / 2^4 =0.375 x 100 =37.5% - prob. of the penny and the dime coming up the same.

There are 10 balls of which 5 are black and 5 are white. Now, out of 5 balls drawn from 10 balls if 3 are white, then remaining 2 balls are black. Hence, probability of having exactly 3 white balls is [5C3 x 5C2] / [10C5] = 100/252 = 25/63=50/126
source: https://www.quora.com/There-are-5-black-and-5-white-balls-A-person-picks-5-balls-from-the-box-at-random-What-is-the-probability-that-exactly-3-of-5-balls-are-white.

I get a slightly different result from Daisy

Reasoning :

Let us just choose any two black balls on the first five draws  [so, by default, the other three will be white ]

First possibility :

B B W W W  =  (5/10) (4/9) (5/8) ( 4/7) (3/6)  = 

(5 * 5 * 4 * 4 * 3)  /( 10 * 9 * 8 * 7 * 6)

Second  possibility : 

B W B W  W  = (5/10) (5/9) (4/8) (4/7) (3/6)  =

( 5 * 5 * 4 * 4 * 3) / (10 * 9 * 8 * 7 * 6)

Third possibilty 

B W W B W = ( 5/10) (5/9)(4/8)(4/7)(3/6) =

( 5 * 5 * 4 * 4 * 3) / (10 *9 * 8 * 7 * 6)

etc......

Out of any 5 positions, the black balls can occur  in any 2 of them...so we have  C(5,2)   = 10 possible arrangements  and each of these has the probabillity  (5 * 5 * 4 * 4 * 3) / (10 * 9 * 8 * 7 * 6) associated with it

So...P( 2 black in 5 drraws)  = P (3 white in 5 draws)   =

10 * ( 5 * 5 * 4 * 4 * 3) / ( 10 * 9 * 8 * 7 * 6)  =   25 /63  = 50 / 126

 Sorry but I was hoping for a better explanation and solution because I didn't understand it...

Daisy is correct !!!!...[ the sum of each row  is also 15...if the missing number is 6  ]

There are \(10\) balls in total. \(5\) are white, \(5\) are black. The probability of choosing a black ball is \(\frac{1}{2}\). Then without replacement, choosing another black ball would be \(\frac{4}{9}\). Now, we have chosen all of our black balls. We can go on to finding the probability of choosing \(3\) white balls. Our first ball probability would be \(\frac{5}{8}\), then \(\frac{4}{7}\), and finally \(\frac{3}{6} = \frac{1}{2}\). So now we have \(\frac{1}{2}\)\(\times\)\(\frac{4}{9}\)\(\times\)\(\frac{5}{8}\)\(\times\)\(\frac{4}{7}\)\(\times\)\(\frac{1}{2}\)\( = \)\(\frac{80}{2016} = \frac{5}{126}\). Then we have to multiply by \(10\) for \(10\) total arrangements, which gives us \(\frac{50}{126} = \frac{25}{63}\).

 - Daisy

After 10 minutes...the last square will have side lengths of 1 + 10(2) = 21 cm

The area of the last square = 21^2  = 441 cm^2

The area of the first square  is 1^2  = 1 cm^2

The difference in their areas  is   441   -1    =  440

Thanks! Your right!