All categories

4 years ago

(6^-8)^4

Mathematics

1 answer:

KIM [24]4 years ago

5 0

Your Answer is A.

(-8) x 4

= -8 x 4

= -32

= 1/6 ^32

HOPE THIS HELPS!!

PLEASE MARK ME AS BRAINLIEST!!

You might be interested in

Pls help me

Vladimir [108]

<h3>Learning Task 1</h3>

Correct responses;

$\displaystyle 1. \hspace{0.3 cm} 7^{-1} = \frac{1}{7}$

2. (14·a·b·c)⁰ = 1

$\displaystyle 3. \hspace{0.3 cm} 10^{-9} = \frac{1}{10^9}$

4. 5·(x·y)⁰ = 5

5. 0¹⁵ = 0

$\displaystyle 6. \hspace{0.3 cm} \frac{24 \cdot x^8 \cdot y^4}{6 \cdot x^5 \cdot y^4} = 4 \cdot x^3$

$\displaystyle 7. \hspace{0.3 cm} \left(\frac{5 \cdot x^0}{y} \right)^{-1} = \frac{y}{5}$

$8. \hspace{0.3 cm}\displaystyle \frac{120 \cdot a^5 \cdot b^6 \cdot c^{-5} }{12 \cdot a^{-2} \cdot b^0 \cdot c^{2}} = \frac{10 \cdot a^7 \cdot b^6}{c^7}$

$\displaystyle 9. \hspace{0.3 cm} \frac{(4 \cdot x)^0 \cdot y^{-5} \cdot z^{-2} }{(234 \cdot x \cdot y \cdot z)^0} = \frac{1}{y^{5} \cdot z^2}$

$\displaystyle 10. \hspace{0.3 cm} \left(\frac{(5 \cdot x \cdot y)^0}{10} \right)^{-2} = 100$

$11. \displaystyle \hspace{0.3 cm} \left(\frac{(a \cdot b)}{9} \right)^{-1} = \frac{9}{a \cdot b}$

$\displaystyle 12. \hspace{0.3 cm} \frac{9}{x^{-2}} = 9 \cdot x^2$

$13. \displaystyle \hspace{0.3 cm} \frac{12 \cdot b^{-3}}{a^{-4}} = \frac{12 \cdot a^4}{b^3}$

$14. \displaystyle \hspace{0.3 cm} \frac{1}{a^{-5 \cdot n}} =a^{5 \cdot n}$

$15. \displaystyle \hspace{0.3 cm} \left(\frac{1}{2} \right)^{-5} = 32$

<h3>Methods by which the above expressions are simplified</h3>

The given expressions expressed into non zero and non negative exponents are;

$\displaystyle 1. \hspace{0.3 cm} \mathbf{7^{-1}} = \underline{\frac{1}{7}}$

2. (14·a·b·c)⁰ = 14⁰ × a⁰ × b⁰ × c⁰ = 1 × 1 × 1 × 1 =<u> 1</u>

$\displaystyle 3. \hspace{0.3 cm} \mathbf{ 10^{-9}} = \underline{ \frac{1}{10^9}}$

4. 5·(x·y)⁰ = 5 × x⁰ × y⁰ = 5 × 1 × 1 = <u>5</u>

5. 0¹⁵ = <u>0</u>

$\displaystyle 6. \hspace{0.3 cm} \mathbf{\frac{24 \cdot x^8 \cdot y^4}{6 \cdot x^5 \cdot y^4}} = \frac{ 4 \times 6 \times x^5 \times x^3 \times y^4}{6 \times x^5 \times y^4} = \underline{4 \cdot x^3}$

$\displaystyle 7. \hspace{0.3 cm} \mathbf{\left(\frac{5 \cdot x^0}{y} \right)^{-1} }= \frac{1}{\frac{5 \times 1}{y} } = \underline{\frac{y}{5}}$

$8. \hspace{0.3 cm}\displaystyle \mathbf{\frac{120 \cdot a^5 \cdot b^6 \cdot c^{-5} }{12 \cdot a^{-2} \cdot b^0 \cdot c^{2}}} = 10 \cdot a^{5 - (-2)} \cdot b^{6 - 0} \cdot c^{-5 -2} = \underline{\frac{10 \cdot a^7 \cdot b^6}{c^7}}$

$\displaystyle 9. \hspace{0.3 cm} \mathbf{\frac{(4 \cdot x)^0 \cdot y^{-5} \cdot z^{-2} }{(234 \cdot x \cdot y \cdot z)^0}} = y^{-5} \cdot z^{-2} = \underline{\frac{1}{y^{5} \cdot z^2}}$

$\displaystyle 10. \hspace{0.3 cm} \mathbf{\left(\frac{(5 \cdot x \cdot y)^0}{10} \right)^{-2}} = \left(\frac{1}{10} \right)^{-2} = \frac{1}{\left(\frac{1}{10} \right)^2 } = \frac{1}{\frac{1}{100} } = \underline{100}$

$11. \displaystyle \hspace{0.3 cm} \mathbf{\left(\frac{(a \cdot b)}{9} \right)^{-1}} = \frac{1}{\left(\frac{(a \cdot b)}{9} \right) } = \underline{\frac{9}{a \cdot b}}$

$\displaystyle 12. \hspace{0.3 cm} \mathbf{\frac{9}{x^{-2}}} = \frac{9}{\frac{1}{x^2} } = \underline{9 \cdot x^2}$

$13. \displaystyle \hspace{0.3 cm} \mathbf{\frac{12 \cdot b^{-3}}{a^{-4}}} = 12 \cdot \frac{b^{-3}}{a^{-4}} =12 \cdot \frac{1}{\frac{b^3}{a^4} } = 12 \cdot \frac{a^4}{b^3} = \underline{\frac{12 \cdot a^4}{b^3}}$

$14. \displaystyle \hspace{0.3 cm} \mathbf{\frac{1}{a^{-5 \cdot n}}} = \frac{1}{\frac{1}{a^{5 \cdot n}} } = \underline{a^{5 \cdot n}}$

$15. \displaystyle \hspace{0.3 cm} \mathbf{\left(\frac{1}{2} \right)^{-5}} = \frac{1}{\left(\frac{1}{2} \right)^5} = 2^5 = \underline{32}$

Learn more about the laws of indices here:

brainly.com/question/8959311

6 0

2 years ago

Shelley has a collection of dimes, nickels and quarters that is worth $5.25. There are a total of 48 coins. If there are 7 more

Mumz [18]

Answer:

25 dimes

Step-by-step explanation:

Given

Coins: Nickels (N), Quarters (Q) , and Dimes (D)

$0.05 = 1 nickel, $0.10 = 1 dime and $0.25 = 1 quarter

So, the amount implies:

$0.05N + 0.10D + 0.25Q = 5.25$

The number of coin implies:

$N + D + Q = 48$

and

$N = 7 + Q$

Required

Determine the number of Dimes

$0.05N + 0.10D + 0.25Q = 5.25$

$N + D + Q = 48$

$N = 7 + Q$

Substitute 7 + Q for N in the first and second equation

$0.05N + 0.10D + 0.25Q = 5.25$

$0.05(7 + Q) + 0.10D + 0.25Q = 5.25$

$0.35 + 0.05Q+ 0.10D + 0.25Q = 5.25$

Collect Like Terms

$0.10D + 0.25Q+0.05Q = 5.25 - 0.35$

$0.10D + 0.30Q = 4.90$ ----- (1)

$N + D + Q = 48$

$7 + Q + D + Q = 48$

Collect Like Terms

$D + Q + Q= 48 - 7$

$D + 2Q= 41$

Make Q the subject

$2Q = 41 - D$

$Q = \frac{41}{2} - \frac{D}{2}$

$Q = 20.5 - 0.50D$

Substitute 20.5 - 0.50D for Q in (1)

$0.10D + 0.30Q = 4.90$

$0.10D + 0.30(20.5 - 0.50D) = 4.90$

$0.10D + 0.30*20.5 - 0.30*0.50D = 4.90$

$0.10D + 6.15 - 0.15D = 4.90$

$0.10D - 0.15D = 4.90 - 6.15$

$-0.05D = -1.25$

Solve for D

$D = \frac{-1.25}{-0.05}$

$D = 25$

<em>Hence, there are 25 Dimes</em>

4 0

4 years ago

I’m confused on this one

RoseWind [281]

In this question, AE is the length of half of one diagonal in the rectangle. DB is the length of the other diagonal. 2(4x-12) would be the total length of the other diagonal, AC. You can set 2(4x-12) equal to 7x-18 and solve for x.

2(4x-12)=7x-18

8x-24=7x-18

x-24=-18

x=6

6 0

3 years ago

Read 2 more answers

Help me on this plzzzz :((((((

Reil [10]

Answer:

A. A

Step-by-step explanation: