For which equation is x=5 a solution?

A winter coat was originally $65 and is now marked up by 2% . What is the price after the markup?

ratelena [41]

It would be 66.3 because 2% Coverts to .02 then you multiply that to 65 and get 1.3 and then you add that to 65

5 0

3 years ago

Which linear function represents the line given by the point slope equation y + 7= 2/3(x+6)

Andrews [41]

Hello from MrBillDoesMath!

Answer: f(x) = (2/3)*x -3 -- which is not a provided answer..

Discussion:

We are given that y + 7 = (2/3) * (x + 6). Multiply out the right hand side:

y + 7 = (2/3) * x + (2/3) * 6 = (2/3) *x + 12/3

= (2/3)*x + 4

Subtract 7 from both sides:

y + 7 - 7 = (2/3)*x + 4 - 7

or

y = (2/3)*x -3

Thank you,

MrB

y + 7= 2/3(x+6)

7 0

3 years ago

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

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

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

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

as

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

6 0

3 years ago

What is a luxury 1,200 room hotel's nightly sales revenues with a room price of $850 and 925 rooms rented?

kipiarov [429]

Answer:32

Step-by-step explanation:

5 0

3 years ago

The mean temperature for the first 4 days in January was -7°C.

Mkey [24]

Answer:

3ºC

Step-by-step explanation:

3 0

2 years ago