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3 years ago

Match each whole number with a rational, exponential expression. 1. 216 2. 512 3. 8 4. 729 5. 27 6. 343

Mathematics

1 answer:

Crank3 years ago

6 0

Answer:

216 - 6^3

512 - 2^9

8 - 2^3

729 - 3^6

27 - 3^3

343 - 7^3

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What is -9x-(2x/-3) When x equals 7

yuradex [85]

$-9x- \frac{2x}{-3}$

$-9*(7) - \frac{2*7}{-3}$

$9*7 = 63$

$\frac{2*7}{-3} = \frac{14}{-3}$

Apply the fraction rule : $\frac{a}{-b} = - \frac{a}{b}$

$= - \frac{14}{3}$

$-63-( -\frac{14}{3} )$

Apply rule - ( - a) = a

$-63+ \frac{14}{3}$

Convert element to fraction :

$63 = \frac{63}{1}$

$-\frac{63}{1} + \frac{14}{3}$

Find the LCD 1*3 = 3

$- \frac{63*3}{3}+ \frac{14}{3}$

$\frac{-3+63+14}{3} = \frac{-175}{3}$

<span>Apply the fraction rule : $\frac{a}{-b} = - \frac{a}{b}$ </span>

$= - \frac{175}{3}$

hope this helps!

4 0

3 years ago

Marsha is buying a new cell phone. She has a 45% off coupon from the store. She pays $275 for the cell phone. What is the price

jekas [21]

Answer:

The price of the cell phone without the coupon= $500

Step-by-step explanation:

Step 1: Express discounted amount

The discounted amount can be expressed as a function of the original cost of the phone as follows;

D=r×A

where;

D=discounted amount

r=coupon rate

A=original price of the cell phone before the coupon

In our case;

r=45%=45/100=0.45

A=a

replacing;

Discounted amount=(0.45×a)=0.45 a

Step 2: Amount she pays up

Amount she pays=Original cost of cell phone-discounted amount

where;

Amount she pays= $275

original cost of cell phone=a

discounted amount=0.45 a

replacing;

$275=a-0.45 a

0.55 a=275

a=275/0.55

a=500

The price of the cell phone without the coupon= $500

7 0

3 years ago

Plz help. Thank youuu

Natalija [7]

Answer:

Step-by-step explanation:

cause

3 0

2 years ago

¯¯¯ , BD¯¯¯¯¯ , and CD¯¯¯¯¯ are angle bisectors of the sides of △ABC . CF=8 m and CD=17 m. What is DE ?

34kurt

I just took this quiz. Its 15

3 0

3 years ago

A^3+a+a^2+1 2a^2+2ab+ab^2+b^3

Vikentia [17]

$\frac{a^3+a+a^2+1}{a^3+a^2+ab^2+b^2} \cdot \frac{2a^2+2ab+ab^2+b^3}{a^3+a+a^2b+b} =\frac{a(a^2+1)+1(a^2+1)}{a^2(a+1)+b^2(a+1)} \cdot \frac{2a(a+b)+b^2(a+b)}{a(a^2+1)+b(a^2+1)} =\\\\=\frac{(a^2+1)(a+1)}{(a+1)(a^2+b^2)} \cdot \frac{(a+b)(2a+b^2)}{(a^2+1)(a+b)} = \frac{2a+b^2}{a^2+b^2}$

6 0

3 years ago