All categories

4 years ago

Please help with question #12. Thank you for your assistance

Mathematics

1 answer:

Sav [38]4 years ago

7 0

A = 6

Determine the defined range

Move expression to the left-hand side and change its sign

Write all numerators above the least common denominator

Distribute through the parentheses

Multiply the parentheses

Collect like terms

When adding or subtracting , the quantity does not change

You might be interested in

Debra plans to invest $2,250 for 10 years. She can invest in a savings account that pays 4% simple intrest or a savings account

Diano4ka-milaya [45]

Answer:

$\$180.55$

Step-by-step explanation:

step 1

Simple interest

we know that

The simple interest formula is equal to

$A=P(1+rt)$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest

t is Number of Time Periods

in this problem we have

$t=10\ years\\ P=\$2,250\\r=4\%=4/100=0.04$

substitute in the formula above

$A=2,250(1+0.04*10)$

$A=2,250(1.4)$

$A=\$3,150$

step 2

Interest compounded annually

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=10\ years\\ P=\$2,250\\r=4\%=4/100=0.04\\n=1$

substitute in the formula above

$A=2,250(1+\frac{0.04}{1})^{1*10}$

$A=2,250(1.04)^{10}$

$A=\$3,330.55$

step 3

Find the differences between the two final amounts

$A=\$3,330.55-\$3,150=\$180.55$

5 0

3 years ago

Multiply 3xyz ^ 2 / 6y ^ 4 by 2y / x z ^ 4

Marat540 [252]

$\dfrac{3xyz^2}{6y^4}=\dfrac{3}{6}\cdot\dfrac{xyz^2}{y^4}=\dfrac{1}{2}\cdot\dfrac{xz}{y^3}=\dfrac{xz}{2y^3}\\\\\\\dfrac{3xyz^2}{6y^4}\cdot\dfrac{2y}{xz^4}=\dfrac{xz^2}{2y^3}\cdot\dfrac{2y}{xz^4}=\dfrac{2}{2}\cdot\dfrac{xyz^2}{xy^3z^4}=1\cdot\dfrac{1}{y^2z^2}=\dfrac{1}{y^2z^2}$