All Answers 356458 Answers

v_o = 1/2 (9.8)(4.5) = 22.05 m/s           (Just like asinus found !)

Yes, and this is demonstrable evidence that dumbness is contagious.  

This equation will work if the total time of flight is used.  4.5 seconds is half   (½) of the total flight time. 

Or take the direct approach and use the equation Alan presented, below ...

--. .-

The correct equation is \(v = v_0 + at\)

Here \(v = 0m/s; a = -9.81m/s^2;t=4.5s \)

57.3% of (√32+√49)

.573(4sqrt2 + 7) = 7.252

v = v_o - 1/2 at      v = 0    a = 9.8 m/s²       t = 4.5 s

0  = v _o  - 1/2 (9.8)(4.5)

v_o = 1/2 (9.8)(4.5) = 22.05 m/s           (Just like asinus found !)

Using line FB, we can find that angle FBC = EFB = 180 - 147 = 33, from the parallel lines. Next, we can see that 4x = x + 33. Simplifying this equation, we then get x = 11.

Oh ok good. Thanks for double checking Pavlov!

60 feet / 0.5 sec = 120 feet per second         as Weiart  found

Lol 

Bus 1 speed = x

Bus 2 speed = x + 20

Rate * time = distance

(x   +   x+20) * ( 18 ) = 1270 km    solve for 'x'    the slower bus, the other bus is x + 20 km/hr

5 x 4.      20

------ = ---------

7 x 9.      63

And there is no way to simplify this.

I'm not 100% sure about this, but if it's per second, then just multiply 60 by 2, and 0.5 by 2 and you get 120 feet per second.

Asinus, this is NOT the correct equation for this question.  

--. .-

An object (has x coordinate of -3 cm) is starting motion from rest along x axis with uniform acceleration. If its x coordinate 3 sec later is -5 cm, then its acceleration is:

Hello Guest!

\(s=x_0-x_1=\color{BrickRed}-3cm+(-5cm) \)

\(s=-2cm\)

\(t=3\ sec\)

\(s=\frac{1}{2}at^2 \)

\(a=\frac{2s}{t^2}\\ a=\frac{2\cdot (-2cm)}{(3sec)^2}\)

\(a=-0.\overline{44}m/sec^2\)

The object experiences a negative acceleration (backwards) of  \(a=- 0.\overline{44}m / sec ^ 2.\)

A baseball is hit so that it travels straight upward after being struck by the bat. If the ball takes 4.5 s to reach its maximum height, then the ball's initial velocity is 

Hello Guest!

\(h=v_0\cdot t-\frac{1}{2}gt^2\\ v_0\cdot t=\frac{1}{2}gt^2\\ v_0= \frac{1}{2}gt\\ v_0=\frac{1}{2}\cdot9.81\frac{m}{sec^2}\cdot4.5sec\)     That is the wrong approach. sorry.

\(v_0=22.0725m/sec\)

  !

1 is answered here

I get the same answer,  7 and a third

22/3

I just said  6/44 = 1/OA

Hi    

Thanks Hugo, that is a really good answer.      

This cannot be simplified, it can only be approximated (with a calculator)

Find the largest number δ such that if |x − 3| < δ, then |4x − 12| < ε, where ε = 1

Repeat and determine δ with ε = 0.1.

Hint:

Start by finding x in terms of  ε  

|4x − 12| < ε

then substitute that value of x into the first inequality and solve

Do not replace  ε with 1 until the end.

Have a go and if you want more help display, or talk about what you have done.

No EP

No child would be taught to multiply by 3.5 !      The maths is too unfriendly

They'd multiply by 3, just like I said. 

Subtract,

solve for y

then sub the y value back into either original formula and solve for x.

Suppose that we have an equation y = ax^2 + bx + c whose graph is a parabola with vertex (-4,7), vertical axis of symmetry, and contains the point (2,-1).  What is (a,b,c)?

Hello Guest!

\(V (-4,7)\\ P_1(2,-1)\)

Because of the symmetry around \( x = -4\), is

\(P_2(-10,-1)\)

The coordinates of the 3 points are used in equation \(y = ax^2 + bx + c\), and the resulting equations are solved for a, b, c.

I. \(7=16a-4b+c\)

II. \(-1=4a+2b+c\)

III. \(-1=100a-10b+c\)

III. - II. \(0=96a-12b\)

I. - II. \(8=12a-6b\)

III. - II.           \(0=96a-12b\)

- 2 * (I. - II.) \(\underline{16=24a-12b}\)

                \(-16=72a\)

                      \(a=-\frac{2}{9}\)

 III. - II.          \(0=96\cdot (-\frac{2}{9})-12b\\ 0=-21.\overline{3}-12b\)

                      \(b=-1.\overline{7}\)

III.              \(-1=100a-10b+c\)

                  \(-1=-100\cdot \frac{2}{9} +10\cdot 1.7+c\)

                      \(c=3.\overline{4}\)

                        !

The equation \(a^7xy-a^6y-a^5x=a^4(b^4-1)\) is equivalent to the equation \((a^mx-a^n)(a^py-a^2)=a^4b^4\)
for some integers \(m\), \(n\), and \(p\). Find \(mnp\).

\(\begin{array}{|rcll|} \hline \mathbf{a^7xy-a^6y-a^5x} &=& \mathbf{a^4(b^4-1)} \\ a^7xy-a^6y-a^5x &=& a^4b^4-a^4 \\ a^7xy-a^6y-a^5x + a^4 &=& a^4b^4 \\ a^7xy-a^5x -a^6y + a^4 &=& a^4b^4 \quad | \quad \mathbf{a^4b^4 = (a^mx-a^n)(a^py-a^2)} \\ a^7xy-a^5x -a^6y + a^4 &=& (a^mx-a^n)(a^py-a^2) \\ a^7xy-a^5x -a^6y + a^4 &=& a^ma^pxy- a^ma^2x-a^na^py+a^na^2 \\ a^7xy-a^5x -a^6y + a^4 &=& a^{m+p}xy - a^{m+2}x-a^{n+p}y+a^{n+2} \\\\ \text{compare} && \text{compare} \\\\ \mathbf{m+2} &=& \mathbf{5} \\ m &=& 5-2 \\ \mathbf{m} &=& \mathbf{3} \\\\ \mathbf{m+p} &=& \mathbf{7} \\ p &=& 7-m \\ p &=& 7-3 \\ \mathbf{p} &=& \mathbf{4} \\\\ \mathbf{n+2} &=& \mathbf{4} \\ n &=& 4-2 \\ \mathbf{n} &=& \mathbf{2} \\\\ \mathbf{n+p} &=& \mathbf{6} \\ p &=& 6-n \\ p &=& 6-2 \\ \mathbf{p} &=& \mathbf{4} \\\\ mnp &=& 3*2*4 \\ \mathbf{mnp} &=& \mathbf{24} \\ \hline \end{array}\)