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3 years ago

Which features are correctly stated? There is more than one answer.

Mathematics

1 answer:

ahrayia [7]3 years ago

4 0

Hope this helps-

y= (x+2)2-3

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What is the value of the sum?

Drupady [299]

Answer:

-1/9

Step-by-step explanation:

-7/9 + 2/3 also equals to 2/3 - 7/9

Now make the denominator same

LCM of 3 and 9 = 9

2/3 = 6/9

∴ 6/9 - 7/9 = -1/9

3 0

3 years ago

What is the inverse of the function f(x) = 2x + 17

IgorC [24]

To find the inverse of a function switch the place of y (aka f(x) ) with x. Then solve for y.

Original equation:

y = 2x + 17

Switched:

x = 2y + 17

Solve for y by isolating it:

x - 17 = 2y + 17 - 17

x - 17 = 2y

(x - 17)/2 = 2y/2

$\frac{1}{2}x-\frac{17}{2}= y$

Hope this helped!

~Just a girl in love with Shawn Mendes

8 0

4 years ago

I need help solving this problem!

melomori [17]

Answer: A

Step-by-step explanation:

The domain is the range of possible x that doesn't make y impossible to find, in this case, all real numbers work

The range is the range of possible y that doesn't make x impossible to find in the inverse function, in this case, all real numbers work

3 0

2 years ago

Read 2 more answers

Using the pattern, give the coefficients of (x + y)^5 and (x + y)^6

musickatia [10]

Answer:

$(x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

$(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$

Step-by-step explanation:

In order to find the values of $(x+y)^5$ and $(x+y)^6$ , you need to apply the binomial theorem (high-level math you most likely don't need to worry about, it's easier than multiplying all the binomials together).

$(x+y)^5$ = $\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$ = $\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$ = $x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$ .

$(x+y)^6$ = $\sum _{i=0}^6\binom{6}{i}x^{\left(6-i\right)}y^i$ = $\frac{6!}{0!\left(6-0\right)!}x^6y^0+\frac{6!}{1!\left(6-1\right)!}x^5y^1+\frac{6!}{2!\left(6-2\right)!}x^4y^2+\frac{6!}{3!\left(6-3\right)!}x^3y^3+\frac{6!}{4!\left(6-4\right)!}x^2y^4+\frac{6!}{5!\left(6-5\right)!}x^1y^5+\frac{6!}{6!\left(6-6\right)!}x^0y^6$ = $x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$ .

5 0

3 years ago

BONUS, Factor 1 - 64x^3 completely,

kvv77 [185]

Rewrite then factor as a difference of cubes:

1 - 64x³ = 1³ - (4x)³

= (1 - 4x) (1² + 4x + (4x)²)

= (1 - 4x) (1 + 4x + 16x²)

6 0

3 years ago