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Scorpion4ik [409]
3 years ago
6

HELP ME PLEASEEE!

Mathematics
1 answer:
Molodets [167]3 years ago
5 0
The answer is -1,3 ok ok
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ALGEBRA 2!!!!!!!!! SHOW YOUR WORK!!!!!!!!!!!!!<br> Do f(g(x)) and g(f(x))
bezimeni [28]

\bf f(x)=\cfrac{2x-3}{x+1}~\hspace{10em}g(x)=\cfrac{x+3}{2-x}&#10;\\\\[-0.35em]&#10;\rule{34em}{0.25pt}\\\\&#10;f(~~g(x)~~)\implies \cfrac{2[g(x)]-3}{[g(x)]+1}\implies \cfrac{2\left( \frac{x+3}{2-x} \right)-3}{\left( \frac{x+3}{2-x} \right)+1}\implies&#10;\cfrac{\frac{2x+6}{2-x}-3}{\frac{x+3}{2-x}+1}&#10;\\\\\\&#10;\cfrac{\frac{2x+6-6+3x}{2-x}}{\frac{x+3+2-x}{2-x}}\implies \cfrac{2x+6-6+3x}{2-x}\cdot \cfrac{2-x}{x+3+2-x}\implies \cfrac{5x}{5}\implies x


\bf \rule{34em}{0.25pt}\\\\&#10;g(~~f(x)~~)\implies \cfrac{[f(x)]+3}{2-[f(x)]}\implies \cfrac{\frac{2x-3}{x+1}+3}{2-\frac{2x-3}{x+1}}\implies \cfrac{\frac{2x-3+3x+3}{x+1}}{\frac{2x+2-(2x-3)}{x+1}}&#10;\\\\\\&#10;\cfrac{2x-3+3x+3}{x+1}\cdot \cfrac{x+1}{2x+2-(2x-3)}\implies \cfrac{2x-3+3x+3}{x+1}\cdot \cfrac{x+1}{2x+2-2x+3}&#10;\\\\\\&#10;\cfrac{5x}{5}\implies x


and in case you recall your inverses, when f(  g(x)  ) = x,  or g(  f(x)  ) = x, simply means, they're inverse of each other.

4 0
4 years ago
Of the 2,598,960 different five-card hands possible from a deck of 52 playing cards, how many would contain all black cards.
kumpel [21]

Answer: 65,780

Step-by-step explanation:

When we select r things from n things , we use combinations and the number of ways to select r things = ^nC_r=\dfrac{n!}{(n-r)!r!}

Given : The total number of playing cards in a deck = 52

The number of different five-card hands possible from a deck = 2,598,960

In a deck , there are 26 black cards and 26 red cards.

The number of ways to select 5 cards from 26 cards = ^{26}C_{5}=\dfrac{26!}{(26-5)!5!}

=\dfrac{26\times25\times24\times23\times22\times21!}{21!(120)}=65780

Hence, the number of different five-card hands possible from a deck of 52 playing cards such that all are black cards = 65,780

3 0
3 years ago
Which expression is a factor of 10x^2+11x+3
Ulleksa [173]

Answer:

The factors are (5x + 3) and (2x + 1)

Step-by-step explanation:

When you need to factor a quadratic, and the coefficient of the x² is not 1, use the slide and divide method.

The general form of a quadratic is ax² + bx + c

Factor:  10x² + 11x + 3

Here a = 10, b = 11, and c = 3

Step 1:  Multiply ac, we SLIDE a over to c.  Notice the 10 is gone for now..

    x² + 11x + 30

Step 2:    Factor this   (this step will always factor)

  x² + 11x + 30 = (x + 5)(x + 6)

So the factors are  (x + 5)(x + 6), but we now need to DIVIDE by a, since we multiplied it into c before.  We divide the constants in the factors...

   (x + 5/10 )(x + 6/10 )  

Now reduce the fractions as much as possible...

(x + 1/2 )(x + 3/5)

*If they don't reduce to a whole number, SLIDE the denominator over as a coefficient of x....

(2x + 1)(5x + 3)        *2 slide over in front of x, 5 slide over in front of x, the fractions are gone!

These are our factors!

5 0
3 years ago
Anyone mind helping? Please!
maria [59]

Answer:

vertex: (0,-1)

axis of symmetry: x=0

5 0
2 years ago
How do you solve geometry problems?
topjm [15]
Depends on what equation or question you want to be answered from the whole section of geometry...
7 0
3 years ago
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