14)If you multiply John's age by 4 and then

4x+7=23. explain the steps you would use to solve the equation for x.

yaroslaw [1]

Answer:

x=4

Step-by-step explanation:

1. isolate the variable by subtracting 7 on both sides: 4x=23-7

2. simplify: 4x=16

3. divide both sides by 4: x=16/4

4. simplify: x=4

5 0

3 years ago

Read 2 more answers

An item is regularly priced at $90 . leila bought it at a discount of 60% off the regular price. how much did leila pay?

forsale [732]

90(.4) = 36

Leila payed $36

3 0

4 years ago

Robert makes $951 gross income per week and keeps $762 of it after tax withholding. How many allowances has Robert claimed? For

MAXImum [283]

Answer:

Option D,four is correct

Step-by-step explanation:

The tax withholding from the gross income of $951 is the gross income itself minus the income after tax withholding i.e $189 ($951-$762)

The percentage of the withholding =189/951=20% approximately

Going by the multiple choices provided,option with 4,189 dollars seems to the correct option as that is the exact of the tax withholding on Robert's gross income and his earnings fall in between $950 and $960

4 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

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

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

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

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

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

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

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