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It is multiply, not add, catmg .....

I'd go for 12+15 = 27

ok Thanks for answering.

Set these questions out very logically and methodically.  It will help you think each one through properly.

Start with one or more pics to illustrate what is happening.

Here is my initial condition pic with working.  A little more elaborate than what you need.

v is vertical velocity

u is horizontal velocity

initial height = +3feet

acceleration is gravity and it stays the same all the time at 32feet/sec^2  but it is pulling down so it is negative.

And it only affects vertically movement.

\(\frac{dv}{dx} =-32\)

Wind and air resistance is ignored for you so there is no horizontal forces that you need to worry about so what is horizontal acceleration?

Here's one way of tackling this:

here is a clip that might help

I have a bag with red and green marbles in it.
At the moment, the ratio of red marbles to green marbles is 3:2.
If I add 5 red marbles and 5 green marbles, the ratio will be 4:3.
How many red marbles will be in the bag after I add more?

\(\begin{array}{|lrcll|} \hline (1)& \dfrac{r}{g} &=& \dfrac{3}{2} \\\\ & \mathbf{r} &=& \mathbf{\dfrac{3}{2}g} \\ \hline (2)& \dfrac{r+5}{g+5} &=& \dfrac{4}{3} \\\\ & r+5 &=& \dfrac{4}{3}(g+5) \\\\ & \dfrac{3}{2}g +5 &=& \dfrac{4}{3}(g+5) \\\\ & \dfrac{3}{2}g-\dfrac{4}{3}g &=& \dfrac{20}{3}-5 \\\\ & g\left(\dfrac{3}{2}-\dfrac{4}{3}\right) &=& \dfrac{5}{3} \\\\ & g\left(\dfrac{1}{6}\right) &=& \dfrac{5}{3} \\\\ & g &=& \dfrac{5}{3}* 6 \\\\ & \mathbf{g} &=& \mathbf{10} \\ \hline & \mathbf{r} &=& \mathbf{\dfrac{3}{2}g} \\ & r &=& \dfrac{3}{2}*10 \\ & \mathbf{r} &=& \mathbf{15} \\ & \mathbf{r+5} &=& \mathbf{20} \\ \hline \end{array}\)

Convince yourself that the decimal representation of $n/30$ terminates if and only if $n$ is divisible by 3.   If you believe this, then counting those $n$ between 1 and 474 is a simple matter.

I don't understand the acceleration part. Can you please explain?

Oh! You wanted me to find the initial condition, height, vertical speed, horizontal speed, and accelerations.

I thought you wanted me to use these values to find my answer. I really wasn't trying to ignore your questions. I'm sorry if it came out that way.

Initial speed: v = 70 feet per second

height: h = 3 feet

vertical speed: 35

horizontal speed: 60.62

Once again, I am extremely sorry. It was a big misunderstanding.

Alternatively
\[
8a+4b+2c+d
=a\cdot 2^2 + b\cdot 2^2 + c\cdot 2+d
= (3\cdot 2^2-5\cdot 2 +4)(7-2)
= 6\cdot 5
= 30.
\]
 

\(z^4 = -4. \) Find all complex numbers z.

Hello Guest!

\(z^4+4=0\\ z=-1-i\\ z=-1+i\\ z=1-i\\ z=1+i\)

  !

Oh yea you can write it like that then use the log rules and division. Thanks for the help!

So far, you're doing really well..

After \(y = {log128\over log8}\)  you don't really need a calculator to solve this 

We can write this as

   \(y= {log 2^7\over log 2^3}\)  right?

then \(y= {7log2\over 3log2}\)

     ⇒\(y={7\over 3}\)

     \({log}_{8}x = {7\over 3}\)

Hope you got it :)

\((3x^2-5x+4)(7-x) = -3x^3+26x^2-39x+28\)

a = -3, b = 26, c = -39 and d = 28

8a + 4b + 2c + d = 30

Question has been solved already: https://web2.0calc.com/questions/i-m-so-lost

No. of black cards = No. of red cards = 26

4 black cards can be drawn in = \(^{26}C_4\) ways

1 red card card can be drawn in = \(^{26}C_1\) ways 

No. of ways  = \(^{26}C_4 × {^{26}C_1}\)

It would be easier for us if you'd enclose a diagram with question especially geometry. Anyway, 

2) ∠BDE + ∠CBD  + ∠CED + ∠BCE = 360 

    ∠BDE + 31 + 47 + 85 = 360 

     ∠BDE = 360 - 163 = 197°

Digits = 1, 2, 3, 4, 5, 6, 7, 8 and can be repeated. 

1's digit has = 4 possibilities

Rest all digits have 8 possibilities 

No. of 5-digit numbers = 8 × 8 × 8 × 8 × 4 = 16384

If you have problems doing those things then talk to us.

We usually respond really well when people interact with us.

We just don't like to feel that we are doing people's homework and they are taking our answers and often not even trying to understand them.

Post the diagram along with the question..

This is a really question so I'll give you some hints... 

There're 14 females and 13 males, so total = 27

Commitee of 9 can be chosen in total  = \(^{27} C_9\) ways

And, 3 females and 6 males can be chosen = \(^{14} C_3 ×{ ^{13}C_6}\) ways

Hope you got it :) 

This is blatantly a homework question so I'll just give a few hints...

First find the no. of routes to (5,7) and (4,3) and then subtract the latter from the former got it?

So, to get to (5,7) Marvin must take 5 steps right and 7 up in any order.

Similarly to get to (4,3) he does 4 right and 3 up in any order.

Now, no. of routes to (5,7)\(={12!\over 5!7!}\)

Similarly you can find the no. of routes to (4,3). 

There you go, I've given enough hints for this one..😏

Let  \(f(a) = a^3+2a^2-9a-18\)

⇒ (a + 2) is a factor of f(a) 

So dividing f(a) by (a+2) we get 

\(f(a) = (a+2)(a^2-9)\)

         \(=(a+2)(a+3)(a-3)\)

∴ Factors = -2, 3, -3

But I don't know what you mean by bc-inversion, since I have never studied inversion.