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Problem 1:  Fill in the boxes below to solve the statement.

Find f(2) for the function f(x) = |3x |

f(x) = |3x|

f(2) = |3 *    2     |   = |   6      |   =  6

JG: That is very nice. Since you have a good understanding of deriving mathematical formulas, it would serve you very well to learn a programming language such as Python, C, C++.......etc. You could, for instance, program the above formula using very short one-line code to convert any number into binary, almost instantly, such as this line: n=123456789;m=2; a=n%m;printa,", ",;b=int(n/m);n=b;if(n>m, goto2, discard=0;printb,;

It will instantly convert 123456789 into binary like this:

1 , 0 , 1 , 0 , 1 , 0 , 0 , 0 , 1 , 0 , 1 , 1 , 0 , 0 , 1 , 1 , 1 , 1 , 0 , 1 , 1 , 0 , 1 , 0 , 1 , 1 , 1 [read from right to left].

Nope ok i understand now

Ah, now i see...

Thanks EP

btw for the middle number AB in red im assuming you Multiply (A+B) to get the AB?

if not... then i might be a bit confused

Sum of cubes ....true   (the cubes are 4x  and 5)

Her ya go

Thank you, Omi

In triangle ABC, AB= c = 3, AC= b = 6, BC= a = 8, and D lies on BC such that AD bisects \(\angle BAC \). Find  \(\cos\angle BAC=cos( \alpha) \)
 

\(cosine:\\ a^2=b^2+c^2-2bc\cdot cos(\alpha)\\ cos(\alpha)= \frac{b^2+c^2-a^2}{2bc}=\frac{36+9-64}{2\cdot 6\cdot 3}\)

\(cos(\alpha)= \cos\angle BAC=-0,52\overline 7\)

  !

can you multiply two matrices together?

The top question wants you to multiply A times B, drag that to the box, then multiply that by C, and drag that product to the box.

The bottom question wants you to multiply B times C first, drag that into the box, then multiply A times that and put it in the box.

Well presumably the velocity vs. time graph is actually a group of graphs, one for each directional component of the velocity vector.

Simply look up the value for each component on the graph at the time you want the speed,

then take the square root of the sum of the square of each directional component.

That will be the instantaneous speed.

how do i solve it with that then ? i dont know what to plug in

I am pretty sure it means gravity... since gravity is like 9.81 m/s2 (・∀・ )

well in my kinetamic equations it say g=9.8 but im not sure

Nice job, Guest!

Good job!

Apparently you are supposed to have a graph with all the data you need.

g - a/sin beta

I think a is 9.8 m/s^2 where it is the acceleration of gravity.

I don't know, this is kind of confusing for me too. Do you know what the g is supposed to mean? Gravity?

Does this problem require trigonometry? This is very hard problem!            This problem should use similar triangles i believe

MN = x with triangle AMN similar to triangle ABX

Triangle ANC is proportional to AXC?

So set up proportion

\(\frac{x}{14}=\frac{5}{5+y}\)

With y being length of NX

How do we go from here????

If M is the mid point, then can't we just do \(\frac{x}{14}=\frac{1}{2}\)

x = 7           So MN = 7??????

Nice job guest! Cool how you can easily program something that can solve problems like this.

a=1; b=1;c= gcd((a^3 + b^3)/(a + b), a * b); if(c==9, goto4, goto5);printc, a, b; a++;if(a<100, goto2, 0);a=1;b++;if(b<100, goto2, discard=0;

a = 9    and   b = 3

\(\text{any function $f(x)$ can be decomposed into its even and odd parts}\\ f(x) = e(x)+o(x)=\dfrac{6}{x+2}+x^2+2^x\\ e(x) = \dfrac 1 2 \left(f(x)+f(-x)\right)\\ o(x) = \dfrac 1 2 \left(f(x) - f(-x)\right)\\~\\ o(1) = \dfrac 1 2(f(1) -f(-1))\\~\\ \text{I leave you to plug and chug}\)

\(a-b=6\)

\(\text{gcd}\left(\frac{a^3+b^3}{a+b}, ab\right) = 9\)

\(\frac{(a+b)(a+b)^2}{(a+b)}\)

\((a+b)^2\)

So now it is \(\text{gcd}\left((a+b)^2, ab\right) = 9\)

So now we can plug in a - b = 6

a = 6 + b

\(\text{gcd}((2b+6)^2,b^2+6b)\)

Ok notice how b^2 + 6b must be divisible by 9, so we just keep guess and checking until we hit our first integer that is divisible by nine for b.

1 = 7

2 = 16

3 = 27

So b=3 works for b^2 + 6b,

now we test for (2b+6)^2

We get 144, which is divisible by 9.

So the answer should be 3.

My solution is very bogus, it may be wrong, or it may be right! But most likely wrong. Can someone check?

v_f = v_i + a t     v_f = fnal velocity (final speed)      v_i = initial velocity  (15 m/s)    a = 2 m/s²

When it reaches the peak......it stops moving upwards and begins to fall back down.....so at the peak upward velocity = 0  (it may still have horizontal velocity....whch you will likely learn about later )

v_f = v_{i +}  a t          

v_i = initial velocity  (zero in this case....it starts at rest)      v_f = final velocity   (25 m/s)      t = time (8.6 s)

a = 10 m/s / 25 s

then the displacement will be    v_i t  + 1/2 a t² = distance    

vi = initial velocity (10 m/s)   a = acceleration     t = time (25 s)               I'll let you do the 'plug and chug'

Acceleration is change in velocity / change in time

(28 m/s - 15 m/s ) /5.3 s =====      ?  m/s²