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\(\binom{8}{3} + \binom{8}{4} = \binom{9}{n}\\ LHS\\ =\binom{8}{3} + \binom{8}{4} \\ =\frac{8!}{3!\;5!}+\frac{8!}{4!\;4!}\\ =\frac{9! }{3!\;5!*9}+\frac{9!}{4!\;4!*9}\\ =\frac{9! }{3!\;4!*5*9}+\frac{9!}{3!*4\;*4!*9}\\ =\frac{9!}{3!4!*9}\left( \frac{1}{5}+\frac{1}{4} \right)\\ =\frac{9!}{3!4!*9}\left( \frac{9}{20} \right)\\ =\frac{9!}{3!4!}\left( \frac{1}{4*5} \right)\\ =\frac{9!}{4!5!}\\ =\binom{9}{4}\quad or \quad \binom{9}{5}\)

So the largest value of n is 5

These are all concave up parabolas.

they are parabolas becasue the highest power of x (the degree) for each one is 2

They are all concave up becasue the number in front of x^2 (the leading coefficient) is positve.

The higher the leading coefficient the steeper, or narrower, the parabola.

Now you can work if out for yourself.

Thanks Alan.

1/4 is the correct answer

See the answer here; https://web2.0calc.com/questions/help-xd_1

n = 5

8C3 + 8C4 =9C5

56   +  70   = 126 

One-letter words =5P1 = 5 words

Two-letter words =5P2 = 25 words

Three-letter words =5^3 = 125 words

Total =5 + 25 + 125 = 155 words in the language of Mumblians.

\(x^2-4x+3 \text{ factors into }(x-3)\text{ and }(x-1)\text{, so false.}\\ 3x+3\text{ factors into }3(x+1)\text{, so true.}\)

Hey I am doing this problem to and heers my answer,to do this we need to replace the angle of elevation that 67 degrees with  55 degrees.Then we make the equation tan55degrees is equal to 1000/x which is the length from the base of the tower to the surveyor when the angle of elevation is 55 degrees.Then we can multiply both sides by x to get tan55 times x is equal to 1000.Then we divide both sides by tan 55 degrees to get 1000/tan 55 degrees is equal to x.Then we plug that into our calculator to get about 700 ft.Then we compare subtract whatever you get for 67 degrees which would be 425 for me roundede to thenearest foot as 424.47 was the nearest hundred adn 424 to the enarest ones place.so we subtract 424 from 700 ft to get 276.The surveyor would need to move 276 ft back to get an angle of elevation of 55 degrees. JUDGE MY SPELLLING PLEASE LOL also Like this cus I spent so long typing this.

So wouldn't it be 1/4?

Also, I got b<=-2a+2, can you check yours

EDIT: Nevermind, found my error

\((2x^4)(z^4)(2x^2)(y^{-2})(z^4)(x^{-3})(y^3)(x^3)^4=(2x^4)(x^{12})(2x^2)(x^{-3})(y^{-2})(y^3)(z^4)(z^4)=4x^{15}\times y \times z^8\)

You are very welcome!

:P

The first choice and fourth choice. 

We could find the least common multiplier, turn things around and get \(x\left(x+3\right)+2=5\left(x+2\right)\) .

Then, solving, we get \(x=4,\:x=-2\) and \(\boxed{x=4}\)  as    \(x=2\) , creates a zero in the denominator. 

Yes, they are also called complementary angles. 

How can the height be 2?, doesnt\(0 \le a,b \le 1\)mean \(0 \le a \le 1\)and \(0 \le b \le 1\)? So it would have to be inside a 1 by 1 square?

With no frictional forces ----elastic collisions ,the ball will continue to bounce to the same height indef.

RPQ is supplemental to RPS

And the sines of supplemental angles are equal

Thus...... sin RPQ = sin RPS = 7/25

And since RPS is obtuse, the cosine of this angle =  - sqrt[ 25^2 - 7^2 ] / 25 =

- sqrt [ 625 - 49 ] / 25 =

- sqrt [ 576] / 25 =

-24 / 25

[  ( 4 - (-5)]^2  =  [ 4 + 5]^2  = 9^2  =  81     (1)

[ ( -9 - 2 ]^2  = [ -11]^2 = 121    (2)

Add (1) and (2), NSS

Take the square root of this sum and that will be your answer

Excellent, NSS !!!!

Adjacent side to R = 12

Hypotenuse = 13

Do you see the answer ???

Opposite side to R  = 5

Adjacent side to R = 12

Do you see the answer ???

cos(180°-θ) = - cos θ            & sin(180°-θ) = sin θ

So cos of rps = - cos rpq

1-sin^2 = cos^2

1-(7/25)^2 = cos^2

.96 =cos ^2

cos = .96                            - cos = - .96   (-24/25)

is it A