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\(AB=B-A\\ AC=C-A\\ n = AB \times AC\\ \hat{n} = \dfrac{n}{\|n\|}\\ \text{The scaled version of the normal vector the problem is after is $-7\hat{n}$}\)

I leave plugging and chugging to you.

\(\text{We must have the sum the lengths of any two sides greater than the length of the third side}\\ x+13> 37\\ 13 + 37 > x\\ 24 < x < 50\\ x \in \{25, 26, 27, \dots 49\}\\ \text{There are 25 integer values of $x$}\)

I was taught that the conventional way, when the last integer is an exact 5, is to round even down and round odd up. 

Examples:    100.1245  the four is even so this rounds down to 100.124

                    100.1235  the three is odd so this rounds up     to 100.124 

The premise is that there there are an equal number of odd digits as even digits, so the rounding balances out, over the long run.  I see that my method conflicts with CPhill.  You might check with your teacher and see how s/he wants you to do it. 

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I'm Lamboreghini!!

Help me please I need help quick...

whats your username?

Because of the picture, and the name.

AnimalLover111?!?!?! Is that you?!!? If it is, I know you from AoPS!!!

THX srry for the late responce. THANK YOU!!

the diagram shows a sector OPQ of a circle with centre O. the radius of the circle is 18m and the angle POQ is 2π / 3 radians. 

a) find the length of the arc PQ, giving the answer as a multiple of π

b) the tangents to the cirlce at the point P and Q meet at the point T. and the angles TPO AND TQO are both right angles

i) Angle PTQ = ∝ radians. find ∝ in terms of pi 

ii)Find the area of the shaded region bounded by the arc PQ and the tangents TP and TQ

a)   arc length PQ  =  18 *  (2 pi)/3 =     36 pi / 3   =   12 pi

b) i)   Angle  PTQ   will be  supplemental to angle POQ   =   pi  - (2pi)/3   =  pi/3   = 60°

ii )    The area of the circular sector  formed by  arc PQ  =  (1/2)r^2 (2pi/3)  =  (1/2)(18)^2 (2pi)/3 = 108 pi  m^2     (1)

[PTQO]   will from a  kite   ...OQ  = 18     and  triangle TPO will form a 30-60-90 right  triangle with QO = 18 opposite the 30° angle  (angle QTO)  and QT  opposite the 60° angle  (angle QOT)....so  QT  = 18√3

So  the area of this kite =  product of the legs  of triangle OQT =  QO * QT  = 18 * 18√3  = 324√3    (2)

So...the area bounded by arc PQ  and the  tangents TP and TQ   = (2)  - (1)  =  324√3 - 108 pi ≈  221.9 m^2

11 1/12  -  2 2/3   =

133              8

___    -       ___

 12               3

133             8 * 4

___     -      _____

 12               3 * 4

133             32

___  -        ___

 12              12

101

___

 12

8   5/12

3)  x&  = 7 - x       and       &x  =  x - 7       What is   & (12&)

Note that   12&   corresponds to x& = 7 - x....so here, x  must = 12   and we have   7 - 12  =  -5

So

&(12&)  =  & (-5)

And  & (-5)  must correspond to  &x= x - 7  ...so here, x  must =  -5    and we  have  -5 - 7   =  -12

So   

& (12&) =  - 12

A toddler is stacking wooden blocks. He has 4 blocks that are all different colors (red, blue, green, and yellow). How many different towers of blocks can he make?

4 ways to choose the 1st block  * 3 ways to choose the 2nd * 2 ways to choose the  3rd * 1 way to choose the last  =  4 * 3 * 2 * 1   = 24

A committee of 3 people is to be selected from a group of 10 people. The order in which people are selected doesn’t matter. How many different committees can be formed?

We have 10 people and we  wish to choose any  3 of them....so...10C3  = 120  different  committees

An ice cream shop has 20 flavors. If Louis wants a double scoop cone, how many different flavor combinations can he get?

We  can choose   2 flavors  from 20  and order these in 2 ways...so

20C2  * 2   =  190 * 2   = 380 combos

THX for the help All of the answers were correct. THX SO MUCH

I still am having troblue with question 3.

1)Add and simplify. \(1\frac{1}{3} + 2\frac{3}{4}\)

the answer is E, \(4\frac{1}{12}\)

2)Subtact and simplify.

\(11 \frac{1}{12} − 2 \frac{2}{3}\)

the answer is B, \(8\frac{5}{12}\)

3) Multiply and simplify. \( 1 \frac{3}{12} \cdot 2 \frac{2}{3}\)

The answer is D, \(3\frac{1}{3}\)

HOPE THIS HELPS

\(\text{The remainder theorem says that the remainder}\\ \text{when dividing $f(x)$ by $(x-c)=f(c)$} \)

\(f(-3) = 11\)

\(\text{To find the remainder when dividing by $(x+1)$ we evaluate the polynomial at $x=-1$}\\ (-12)(-1)^{17} + 3(-1)^5 - 9(-1)^2 -1 = \\ 12-3-9-1=\\ -1\\~\\ \text{D is the correct choice. Well done!}\)

\(\dfrac{x^2+5}{2} = 7\\ x^2 + 5 = 14\\ x^2 = 9\\ x = \pm 3\)

\(\sqrt{\dfrac{20-5x}{x}}=5\\ \dfrac{20-5x}{x} = 25\\ 20-5x=25x\\ 20=30x\\ x=\dfrac 2 3\)

\(\text{Just as an alternative way to calculate this}\\ z = 1+i = \sqrt{2} e^{i \pi/4}\\ z^4 = (\sqrt{2})^4 e^{4(i\pi/4)} = 4e^{i\pi} = -4\)

E. 3,333.849

28.53 * 1.20   =  $34.24

  10 * 0.33  = $3.30

1) D

2) C