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x^3 + 3x^2 - 25x - 75

Factor by grouping

x^2 ( x +3) - 25 (x + 3)

(x^2 - 25) ( x + 3)

(x + 5) (x - 5) ( x + 3)    are the binomial factors

For the second one we have

x^4 + 5x^3 -x^2  - 5x      factor by grouping

x^3 ( x + 5) - x (x + 5)

(x^3 - x) ( x + 5)

x ( x^2  - 1) (x + 5)

(x - 0 ) ( x + 1) (x - 1) (x + 5)

These are true

f(x)  =    when  x  = -5

f(x)  ivided by ( x + 5)  as a remainder of 0      [ because ( x + 5) is a factor ]

cos  [ 2 arcsin ( x) ]

Let  arcsin (x)  be θ

So...using a trig identity....  we have

cos ( 2θ)  =    1  - 2sin^2(θ)  =  1 - 2 [ sin (θ) ]^2

The sin of θ  =  x / 1  =  x

So...

cos (2θ)  =  1 - 2sin^2 ( θ)  =  1 -  2 [ sin (θ)]^2  =   1  - 2 [x]^2  =  1  - 2x^2

(4r^3)^(1/3)  *  (5r^3)^(1/3)  = 

[ 4 * 5] ^(1/3) *  [ r^3 * r^3 ]^(1/3)  = 

[ 20 r^6 ] ^(1/3)  =

[20]^(1/3)  *  (r^6)^(1/3)  =

20^(1/3) *  r^2  =

r^2 ∛ 20

I like to convert to exponential form  and work with that

We can convert back to radical form at the last

Note that  ∛  =  exponent (1/3)

( 15x)^(1/3) *  (9x^3)^(1/3)  =

( 145 x^4)^(1/3)

Note  ......(145)^(1/3)  =  (5 * 3^3 )^(1/3)  =  5^(1/3) * (3^3)^(1/3)  =

5^(1/3)  *  3        (1)

Also.....note how to simplify  ( x^4)^(1/3)  =  x^(4/3)

Make  4/3  a mixed number  =  1 + 1/3

So  we have  x^(1 + 1/3)  = x^(1) * (x)^(1/3)   =  x * (x)^(1/3)   (2)

So...putting (1)  and (2) together, we have

3 * x *  (5)^(1/3) * (x)^)1/3)  =

3x ∛ [ 5 x ] 

sec (arcos (x) )

The cosine of this angle =  x / 1

So..the sec  is the reciprocal,  1 / x

cot  ( arccos (3x) )

The cosine of this angle  =   [3x] / 1    =  3x

We are looking for the  cot =    cos / sine

So...the sine  =

sqrt  [ 1 - [3x]^2 ]  =  sqrt [ 1 - 9x^2 ]

And the  cotangent  is

[ 3x]  / sqrt [ 1 - 9x^2 ] 

cos ( arcsin (-3/5) )

We are looking for the cosine .... (x / r)..... of an angle whose sin  is  -3/5  =  y / r

So  we need to find x

sqrt [ 5^2  - (-3)^2  ] = sqrt [ 25 - 9 ] =  4  = x

This angle is in the 4th quadrant....the cosine is positive here

So...this evaluates to  x / r  =   4/5

\(\binom73\cdot\binom{10}3 \Rightarrow 35\cdot720 \Rightarrow25200\)

tan  ( arccos (2/3) )

We are looking for the tangent  of an angle whose cosine is  2/3

So..we know  x / r   and we are looking for y/ x...so we have

sqrt  ( 3^2 - 2^2)  =  sqrt 5   =  y

So....y / x  =     sqrt(5) / 2

\(\binom92\cdot\binom{12}3 \Rightarrow 7920\)

Nice, tertre  !!!

Here is the second one.....

Let F  be the fixed charge  and M  be the cost per minute of use

So...we have these equations

500M + F  = 35

1500M + F  = 55

Subtract the first equation from the second and we get that

1000M = 20     divide both sides by 1000

M =  20 / 1000   =   2 / 100   =  2 cents per minute

So...using either equation to determine F, we have

500 (,02) + F  =  35

10 + F  = 35      subtract   10 from each side

F  =   25  (dollars)

You are correct, ANSP  !!!!

At least one card  must contain a "0"....so....the product will be  0

So...there are no ways in which this is possible...

x^4  + 9x^3  + 15x^2  + 9x + 14

We can use a factoring "trick" here....write this as

x^4 +  9x^3 + 14x^2  + x^2  + 9x  + 14     factor by grouping

x^2 ( x^2 + 9x + 14)  +  1 ( x^2 + 9x + 14)

(x^2 + 1) (x^2 + 9x + 14)

(x^2 + 1) (x+ 7) ( x + 2)

(x + 1i) ( x - 1i) (x+ 7) ( x + 2)

Question 3) Six more than the square root of 25.

sqrt (25)  +  6

Question 4) What number does not have a reciprocal?

0

To see why

0 / 1  =  0

But

1 / 0   is undefined

(3/4)  is 60%  of what number

3/4  =   .75

"is"  means  "="

60%  =  .60

"of" means multiply

"number"  =   n

So we have

.75 =  .60 * n       divide both sides by .60

.75 / .60  =  n  =  1.25

Thanks, Knockout.....!!!

"Reduce" means  "subtract from"....so....

43  - l -8 + 6  l

Thanks, tertre   !!!!!

Correct, hoiuu  !!!

C (182, 180)  =

 182!

_________________  =

(182 - 180)! (180 ) !

182 * 181

__________  =

     2!

91 * 181  =

16,741

thx so much

Our keywords: product, increased, quotient 

Now, translating this, we have \(5*6+\frac{30}{6}\) .

Yes, it is! The median is constant!

Hello! Yet again, this question also has been asked before.

https://web2.0calc.com/questions/a-use-the-fundamental-theorem-of-algebra-to-determine

Thank you!

Hi, this question has been asked before. 

https://web2.0calc.com/questions/factor-the-polynomial-function-over-the-complex-number

Thanks!

For any integer \(n\), \(\dbinom{n}{n}\), or \(nCn\), is equal to 1.