Sin(5x 3 + 20) = cos(2x 3 + 14)

Give two rational numbers whose product equals (-1)

motikmotik

Product is times so try that as it’s the product of a number

6 0

3 years ago

Mr.William orders 2 feet of trim to decorate pillows. If he needs 1inch of trim for each pillow, how many pillows can he decorat

navik [9.2K]

2 feet = 24 inches

24 inches divided by 1 inch = 24

Answer is 24 pillows.

3 0

2 years ago

AT&T:$0.08 per minute,$5 monthly fee VOIP:(cable phone):free calls,$35 monthly fee Cell:$40 for first 400 minutes,$0.25 per

Luden [163]

Make 2 graphs and with the second graph graph the cell fee and with the first Do the minutes

4 0

2 years ago

WILL GIVE BRAINLIEST AND 20 POINTS! Which function is the inverse of f(x) = -5x -4?

inn [45]

Answer:

-1/5x -4/5 = y

Step-by-step explanation:

f(x) = -5x -4

y = -5x-4

Switch x and y

x = -5y -4

Solve for y

Add 4

x+4 = -5y

Divide by -5

-1/5x - 4/5 = -5y/-5

-1/5x -4/5 = y

6 0

2 years ago

Read 2 more answers

(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

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

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

2 years ago