All Answers 356462 Answers

Could you attach the plot that the problem mentions?

The quadratic formula is, \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\). First let's focus on the -7. To make -b/2 = -7, b would have to be b = 14 to satisfy. So already have,

\(z^2 + 14z + b = 0\)

Now that we know b, we need to have \(\frac{\sqrt{196-4c}}{2} = 3i\). Square both sides and further solve, 

\(\frac{196 - 4c}{4} = -9\)

\(196 - 4c = -36\)

\(4c = 232\)

\(c = 58\)

So completing the equation we have, \(z^2 + 14z + 58 = 0\), and we can check using the quadratic formula to see that the roots are \(-7 \pm 3i\). So,\( a + b = 14 + 58 =\) 72

Add 2 pi to each of your answers ...

Is there a solution without inversion? I haven't learned it. 

idk if this is right, here's my best shot.

this is just a system of equations:

\(\begin{pmatrix}x=5+\sqrt{69},\:&y=5-\sqrt{69}\\ x=5-\sqrt{69},\:&y=5+\sqrt{69}\end{pmatrix}\)

(b) There are 108/3 = 36 unique sets.

(c) For the first card, there are 3 ways to choose the number.  For the second, there 2 choices, and then there is only 1 choice.  So there are 3*2*1 = 6 ways that the numbers can be chosen.  Doing this for the other attributes, we get 6*6*24*6 ways.  But the order of cards doesn't matter in a set, so we divide by 3!: 6*6*24*6/3! = 864.  So there are 864 sets for part (c).

(d) The cards can have the same number, color, shape, or shading.  If all the colors are the same, then there are 6*24*6/3! = 144 sets.  If all the numbers are the same, then there are 144 sets.  We get the same number for shape and shading, so there are 4*144 = 576 sets for part (d).

(e) We need to choose two of the attributes.  There are C(4,2) = 6 ways of choosing two attributes.  For each of these two attributes, there are 3 options.  For the other two attributes, there are 3 ways of assigning the choices, so there are 6*3*3*3*3 = 486 sets for part (e)

(f) First we choose which attributes are the same.  There are C(4,3) = 4 ways of choosing which attributes are the same.  There are then 3*3*4 = 36 ways to assign which is which for each of these three attributes, and there are 4 ways to assign the choices for the fourth attribute, so there are 4*36*4 = 576 sets for part (f).

answer pls i need to do this quick

Hey there, Guest!

So...

An n number is a 1 followed by 2^(n-1) combinations of n-1 0's and ones, symmetrically distributed, hence (n-1)2^(n-2) zeros.

In total,

1+2*2+3*4+4*8+5*16+6*32+7*64+8=777.

so that for numbers between 0 and 2^n, the number of zeroes is

(n−2)2^(n−1)+n+1

Hope this helped! :)

( ﾟдﾟ)つ Bye

(3x  +   4 )(x     +  2  )       Finish.....

Hey there, Guest!

The answer is 66 ways.

And here is the explanation:

Hope this helped! :)

( ﾟдﾟ)つ Bye

It's 2. I'll explain when I'm on my computer

correct thanks!

The root given is in the form of the quadratic formula, \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

The part we are interested in is the discriminant, \(b^2-4ac\). Using this,

\(b^2 - 4ac = 169 - 8k = 131\)

\(169 = 131 + 8k\)

\( 38 = 8k\)

\(k = 4.75\)

We can notice that in this case, the mean of the set of integers will be the same as the median, as it is increasing by 1

N cannot be even as, for example with N = 10, 11 through 20 will have a mean of a decimal.

If we make N = 9, 10 through 18 makes the mean 14.

If we make N = 11, 12 through 12 makes the mean 17.

If we make N = 13, 14 through 26 makes the mean 20.

So N = 13

We can break up the planes based on how many vertices of the bottom face they pass through.

$1$ plane passes through exactly $4$ vertices, and $0$ pass through exactly $3$ vertices.

The planes passing through $2$ vertices of the bottom face either pass through an edge or a diagonal of the bottom face. Each edge on the bottom face has $2$ valid planes passing through it and each diagonal has $3$ valid planes, for a total of $4 \cdot 2 + 2 \cdot 3 = 14$.

The valid planes passing through exactly $1$ vertex of the bottom face must intersect the top face on a specific diagonal, so there are $4$ such planes.

There is $1$ plane passing through $0$ vertices, so this makes a total of $1 + 14 + 4 + 1 =$ 20 total planes.

For it to have one solution, the discriminant must be equal to 0. Applying it,

\(b^2-4ac = 144 - 16p\)

\(144-16p = 0\)

\(16p = 144\)

\(p = 9\)

Start off with Wilma's equation. We can use Vieta's to find that c/a = 8*1, so c = 8.

Next Greg's equation. Using Vieta's again we can find that -b/a = -4+1 = 3

So piecing it together, the actual equation is,

\(x^2+3x+8\)

Apply quadratic formula to find the roots\(x=−1.5±\frac{\sqrt{23}i}{2}\)

2/3rds of 9 is 6, meaning that he saved $6. 9-6 = 3 dollars spent on supplies

No of ways of selection = 15C3

=455

Total pupils = x 

2x/9 scored A grade. 

No. Of pupils who scored B = y 

Now y/ 2x/9 = 5/3

      9y/2x=5/3 

      y/x   =  10/27 - eq (1) 

Now assume 

   No of pupils who scored C = A

   No of pupils who scored D = B

Now 

   A/B  = 6/5 - eq(2) 

Now given

  y = A+48 -eq (3) 

Now we can write 

2x/9 + A+ B + Y = x

A+ B + Y = 7x/9

Put value of eq (3) 

B+ y-48+y= 7x/9

2y+B-48= 7x /9

Now put value of eq (2) 

5A/6 +2y -48 = 7x/9

(Put A = y-48) 

5(y-48) /6 +2y -48 = 7x/9

5y/6  - 40 +2y -48 = 7x/9

17y/6 -88 = 7x/9

Put value of eq(1)

17(10x/27) /6  -88 = 7x/9

170x/162  -7x/9  = 88  

  (170x - 126x)/162 = 88

  44x/ 162 = 88

    x = 162 * 2

   x = 324 

a)  total pupils (x) = 324

b) no. Of pupils of scores D grade = 

First of all  y/x = 10/27

                    y = 10*324/27  = 120

Now A = y- 48 = 120-48 = 72 

Now

 A/ B = 6/5

B = 72 *5/6

B = 45

So no of pupils who failed = 45

Total students enrolled in atleast any of subjects= students enrolled in algebra+students enrolled in mechanics+students enrolled in physics-students enrolled in algebra & physics-students enrolled in algebra & mechanics-students enrolled in mechanic & physics + students enrolled in all three 

= 48+39+35-20-19-22+10

=71

Therefore students not enrolled in any of three = 100-71

= 29

it says at least 3 

wrong

When the ball hits the ground, the height, y = 0

-16t^2+60t+54 = 0        use quadratic formula or factoring to solve for 't' .....   throw out the negative value....