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The general quadratic equation for a projectile's free fall is

\(p(t) = -at^2 + v_0t + h,\)

where \(a\) represents acceleration due to gravity, \(v_0\) is the initial (launch) velocity of the projectile, and \(h\) is our starting height.

In terms of feet, the acceleration of a projectile due to gravity (on Earth) comes out to \(32.17 \frac{\text{ft}}{\text{s}^2}\). I'll assume from your statement that there's no initial velocity present. Then, with a starting height of 6 feet, our quadratic equation will take the form

\(p(t) = -32.17t^2 +6.\)

What is the largest angle measure in a triangle with sides of lengths 3, 5, and 4?   

Note that 5² = 3² + 4²     That means it's a right triangle, thus the largest angle is 90^o.  

_.

By the Pythagorean Theorem, \(BD = \sqrt 7\)

Let \(CE = x\) and \(DE = y \)

We have the following system: 

\(x^2+y^2=4\)

\(x^2+(\sqrt7 - y)^2=3\)

Solve for x by solving this system. 

If AB is tangent to the semicircle, angle(CBA) = 90^o.

In triangle(ABC), angle(C) = 30^o.

We have the equation: \(2(n+1)!+6n!=4(n+1)!\)

We can subtract \(2(n+1)!\) from both sides, which yields: \(6n!=2(n+1)!\)

Note that \((n+1)! = n! \times (n+1)\)

Substituting this gives us: \(6n!=2(n+1)n!\)

Now, we just have to divide by \(n!\), then solve for n. 

Can you take it from here?

Note that \(\large{0.125 = {1 \over 8}}\). This can be written as \(\large{1 \over 2^3}\) or \(\large{2^{-3}}\)

We now have \((2^{-3})^3\). We can simplify this by multiplying the exponents. 

Can you solve it now?

Answered 

Let's count by case:

1 x 1 square: 1 square

2 x 2 squares: 4 squares

3 x 3 squares: 4 squares

4 x 4 squares: 4 squares

5 x 5 squares: 1 square

Now, can you tell me how many squares contain the black square?

Adding the two equations cancels out the t. This gives: \(2.5s=2\)

Solving, we find that \(s = 0.8\). 

Now substitute this into one of the 2 equations and solve for t. 

It is 13. Because a prime number has 0 factors, it can't be any bigger. 

What number must be placed in the box?

Hello Guest!

\(9x^2 + 6x + c=0\)

a          b        c

\(x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\ The\ radicant\ must\ be\ Zero\ for\ x\ to\ have\ only\ one\ value.\\ b^2-4ac=0\\ c=\dfrac{b^2}{4a}=\dfrac{6^2}{4\cdot 9}\\ \color{blue}c=1\)

\(x=-\dfrac{b}{2a}=-\dfrac{6}{2\cdot 9}\\ \color{blue} x=-\dfrac{1}{3}\)

\((x+\frac{1}{3})^2=x^2+\frac{2}{3}x+\frac{1}{9}=(3x+1)^2=9x^2+6x+1=0\)

1 must be placed in the box.

  !

Use the latex button and enter your question properly.

Thanks Builderboi.

Thanks Builderboi, you would think they could have found a symbol that didn't already have another meantng....

I thought there must be some obscure symbol, thanks for finding it.

It is 20736 since 16x12x12x9

for 1st 2nd and 3rd rook

I just took the test, this is the answer.

Let C =  (0,0)   A =  (4,0)  and B = (0,3)

We have a 3-4-5 right triangle so BA =  5

We can find the coordinates of the incenter as  (  [ 4(3)  ] / 12  , [  3(4) / 12 ] =  (1 , 1)

Equation of line through   A and I   ....  y = (-1/3)(x - 4)

When x =  0    D = ( 0, 4/3)

Equation of line  through B and I  ......    y = (-2 / 1) x + 3

When y = 0,  E  =  (0, 3/2)

DE =   sqrt [ (4/3)^2  + (3/2)^2]   =   sqrt  [ 16/9 + 9/4 ]  =  sqrt  [  145 ] / 6   

By stars and bars, the answer is C(12,4) = 495.

Only 4

S =  (7 + 8 + 9 ] / 2   =  12

Area = 

sqrt  [ S * (S  - 7) (S - 8) (S - 9) ]  = 

sqrt [ 12 * 5 * 4 * 3  ]   =

sqrt [ 12 * 12 * 5 ]  =

12sqrt  5  

1/3 + 2/3^2  + 1/3^3 + 2/3^4 + 1/3^5 + 2/3^6  + 1/ 3^7 + 2/3^8 .........

We can write this as

(3 + 2)/3^2   + (3 + 2)/3^4  + (3 + 2) / 3^6  + (3 + 2) / 3^8   .....   =

5 [  1/3^2  + 1/3^4  +  1/ 3^6  + 1/3^8  .....  ]    =

5 [ the sum of an infinite series  with  the first term = 1/3^2  and a  common ratio of 1/3^2 ]  = 

5 (1/3^2)  / ( 1 - 1/3^2)  =

(5/9) / ( 8/9)  =  

5 /  8

In triangle(ABC):  x + x + y + y + 30 = 180

Since x = 17:  17 + 17 + y + y + 30 = 180

                                   34 + 2y + 30 = 180

                                           2y + 64 = 180

                                                   2y = 116

                                                     y = 58

Not sure this is possible

To see why

Consider that no one takes both

Then 48 take Spanish and 49 take Chinese.....but we're told that 5 more take Chinese than Spanish

Consider that only  1 takes both

Then  48 + 1  =  49 take Spanish

So  53  take only Chinese and  1 takes both  =  54

But this gives us     53 + 1 + 48    =   102 total students (too  many)

Conslder that  2 take both

Then 50 take Spanish   and  55 would take Chinese   (53 taking only Chinese and 2  taking both)

But this gives us 53 + 2 + 48  = 103

The same thing happens if 3 take both

51 take Spanish   and 56 take Chinese  (53 taking only Chinese and 3 taking both)   =  53 + 3 + 48 = 104

So.....it appears impossible that 48 take only Spanish

Let the total no. of cookies = x

Chocolate cookies = x/5

Remaining = x - x/5 = 4x/5

Nuts cookie = 1/4 * 4x/5 = x/5

Remaining = 4x/5 - x/5 = 3x/5

Butter cookies = 3x/5

Given:

Chocolate cookie = 1

x/5 = 1   x = 5

=> for every 1 chocolate cookie

Nut cookies = 1

Butter cookies = 3

Total cookies = 5

So the answer is 3 shortbread (or butter) cookies

The probablility that you draw \(2\) Black balls from the box is just \(\frac13\cdot\frac58=\frac{5}{24}\)