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(2500[.25(2500/139.931)+ 0.0004(32.8)+0.0001 (4.3722)(1.5))/(1+ 0.0004(32.8)+0.0001 (4.3722)(1.5))

Missing "]"

Use the pythagorus theorem

a²+b²=c²

29/20 that "Guest" gave you is 1.45 in decimal

Since 6 7/8 equals ((6*8)+7)/8=55/8, 6 7/8 x 12= 55/8 x 12. you can divide the denominator on 55/8 by 4 to get 55/2, and divide 4 on the numerator 12 to get 3. (Notice that I divided the denominator on one number, and the numerator on the other, by the same number), to get 55/8 x 12=55/2 x 3=55/2 x 3/1. The product of the numerators are 55 x 3= 165, and the product of the denominators are 2 x 1=2. Therefore, 6 7/8 x 12=165/2, which is 82.5 in decimal.

No.  1.33333..... etc is one and a third.   0.33333... is 1/3

To see this let x = 0.3333........

Then 10x = 3.333333

so 10x - x = 3

       9x = 3

         x = 3/9

         x = 1/3

.

Vertical motion.  Distance given by:  s_vert = 11.5*sin(35°)*t - (1/2)*9.81*t²

 This is zero at times t = 0 and t = 2*11.5*sin(35°)/9.81 seconds.  The latter time is the time at which the rock comes back to the ground.

Horizontal motion.   Distance given by:  s_horiz = 11.5*cos(35°)*t

So, total horizontal distance travelled = 11.5*cos(35°)*2*11.5*sin(35°)/9.81 metres

I'll leave you to crunch the numbers.

.

I more basic terms it simply means the 'sum' of the equation

For example.

1+2=a

2+4=b

Σab = 9 

Yes

And 1 is not a prime number.

Hello guest!

At t = 0 a ball is thrown straight upward from ground level at speed v0. At the same instant, a distance D above, a ball is thrown straight downward, also at speed v0.Where do the b***s collide (in terms of only D, v0, and g)?

Tonight came a new realization:
The b***s meet twice!

H = D - v₀t - (g/2)t²

(g/2)t² + v₀t + H - D = 0

  a          b         c

\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(t = {-v \pm \sqrt{v^2-2g(H-D)} \over g}\)

H =  v_ot - (g/2)t²

(g/2)t² - vt + H = 0

   a        b     c

\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(t = {v \pm \sqrt{v^2-2gH} \over g}\)

\({\color{blue} {-v \pm \sqrt{v^2-2g(H-D)} \over g}}\)= \({\color{blue}\frac { v \pm \sqrt{v^2-2gH}}{ g}}\) 

2v\({\color{blue}\pm \sqrt{v^2-2gH}}\) = \({\color{blue}\pm\sqrt{v^2-2gH+2gD}}\)

H must be calculated from this.
Maybe someone can help me.
For given sizes, the thing would be simple.

Greetings asinus :- )  !

Greeting asinus :- )   !

I use Khan Academy but I have not seen this video yet, thanks, I take a look.

y=1+1 

y^2 = (1+1)^2 

y^2 = (1^2 + 2(1)(1) + 1^2) 

y^2 = 1+ 2+ 1 

y^2 = (1+1) + 2 

y^2 = y +2 

y^2 - y = 2 

y^2 - y + 1/4 = 2 + 1/4 

(y - 1/2)^2 = 9/4 

y - 1/2 = sqrt(9/4), y>0 

y - 1/2 = 3/2 

y = 1/2 + 3/2 

y = (1+3)/2 

y = 4/2 

y = 2 

1+1 = 2

OR

e^iπ = -1 
2^57,885,161 - 1 is the largest known Mersenne prime number 
2^43,112,609 - 1 is the second largest known Mersenne prime number 
the greatest common divisor of any two integers which are coprime is 1 so 
gcd(2^57,885,161 - 1, 2^43,112,609 - 1) =1 

tan(π/4) = 1 
So ((e^iπ/ gcd((2^57,885,161 - 1, 2^43,112,609 - 1))^2) 
= (-1/1)^2 +1 
= 1 + 1 

 Either way your answer is... (drumroll please) Two.

x(13+6x)=28

Which of the following functions has the solution (3, -2)?

Select all that apply.

(1) y=-2x+4

(2) f(x)=-x+2

(3) f(x)=-23x

(4) f(x)=3x−2

(5) f(x)=4x+11

(6) f(x)=6x−20

\(\begin{array}{|rcll|} \hline (3, -2) \quad \text{or } \\ x &=& 3 \quad \text{and }\\ y =f(x) &=& -2 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline (1) & y &=& -2x+4 \qquad & | \qquad x = 3 \\ & y &=& -2\times 3 +4 \\ & y &=& -6 +4 \\ & y &=& -2 \qquad \qquad \mathbf{\text{Yes } y = f(x) = -2} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (2) & f(x)&=&-x+2 \qquad & | \qquad x = 3 \\ & f(x) &=& -3+2\\ & f(x) &=& -1 \qquad \qquad \text{No } f(x) = -1 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (3) & f(x)&=&-23x \qquad & | \qquad x = 3 \\ & f(x) &=& -23\times 3\\ & f(x) &=& -69 \qquad \qquad \text{No } f(x) = -69 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (4) & f(x)&=&3x-2 \qquad & | \qquad x = 3 \\ & f(x) &=& 3\times 3 -2 \\ & f(x) &=& 9 -2 \\ & f(x) &=& 7 \qquad \qquad \text{No } f(x) = 7 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (5) & f(x)&=&4x+11 \qquad & | \qquad x = 3 \\ & f(x) &=& 4\times 3 +11 \\ & f(x) &=& 12 +11 \\ & f(x) &=& 23 \qquad \qquad \text{No } f(x) = 23 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (6) & f(x)&=&6x-20 \qquad & | \qquad x = 3 \\ & f(x) &=& 6\times 3 -20 \\ & f(x) &=& 18 -20 \\ & f(x) &=& -2 \qquad \qquad \mathbf{\text{Yes } y = f(x) = -2} \\ \hline \end{array}\)

Hello guest!

At t = 0 a ball is thrown straight upward from ground level at speed v0. At the same instant, a distance D above, a ball is thrown straight downward, also at speed v0.Where do the b***s collide (in terms of only D, v0, and g)?

H = D - v₀t - (g/2)t²

H =  v_ot - (g/2)t²

D - v₀t - (g/2)t² = v_ot - (g/2)t² 

2v₀t = D

t = D / 2v_o

The b***s meet after the time t = D / 2v₀ . *

H = v_ot - (g/2)t²  = D/2 - (g/2) * D² / 4v₀²

H = D/2 - gD²  / 8v₀²

The b***s meet at height H = D/2 - gD²  / 8v₀² .*

Greeting asinus :- )   !

* b***s means Bälle.

Σ=summation notation

what is the fraction?

34.53/175.90

\({\color{blue}\frac{34.53}{175.90}=0.19630471859}\)  !

what is (-3) squared

\({\color{black} Minus \ times \ minus \ equal \ plus.}\)

\({\color{blue}(-3)^2=(-3)\times(-3) = 9}\)   !

1\2 × -1\3

\({\color{blue}\frac{1}{2}\times(-\frac{1}{3})=-\frac{1}{6}}\)     !

8 divided by 2/3

\({\color{blue}8:\frac{2}{3}= 8\times\frac{3}{2}=\frac{24}{2}=12}\)   !

What is 5,000,000+6,789,654=

\({\color{blue}5,000,000+6,789,654=11,789,654}\)  !

11,789,654

The weight of the rock dosen't change anything, as the initial velocity is already known and acceleration due to gravity is the same for all objects on Earth no matter what the mass is. The question ignores air resistance.

Just simplify 8 into 8/1, then swap the digits of 2/3, then multiply the result, then you've got your answer!