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11111nata11111 [884]
2 years ago
6

Find a formula for the nth term in this arithmetic sequence first term -2 common difference an=

Mathematics
1 answer:
malfutka [58]2 years ago
3 0
The formula for the problem is: An = 8(n-1) - 2

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What -7 x4=36? amogus amogus amogus amogus
77julia77 [94]

Based on the given problem of -7x⁴ = 36, the solution to the equation is x = -1.5.

<h3>What is the solution to this problem?</h3>

The equation to find x is given as:

-7x⁴ = 36

Solving it gives:

-7x⁴ = 36

x⁴ = 36/-7

x⁴ = -36/7

x = ⁴√(-36/7)

x = -1.5

In conclusion, x is -1.5.

Find out more questions on solving for x at brainly.com/question/2910769

#SPJ1

3 0
1 year ago
The same price of an item is $48 after a 40% discount what was the original price
Sonja [21]
Let, the original price = x
It would be: x*60% = 48
x = 48 / 0.60
x = 80

In short, Your Answer would be $80

Hope this helps!
8 0
3 years ago
PLEASE HELP!!!!! DUE TODAY!!!
Akimi4 [234]

Answer:

5

Step-by-step explanation:

Rental = $40

Catering = $12 per person

Budget = $100

Amount left after paying for the rental

= 100 - 40

= $60

Maximum number of guests

= 60 ÷ 12

= 5

5 0
3 years ago
Match the expressions with their equivalent simplified expressions.
Tasya [4]

Answer:

\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}


Step-by-step explanation:

\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}} =\sqrt[4]{\frac{(2^4)(x^{6-2})(y^{4-8})}{(3^4)}} =\sqrt[4]{\frac{2^4x^4y^{-4}}{3^4}} =\frac{2xy^{-1}}{3}=\frac{2x}{3y}

\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} =\sqrt[4]{\frac{(3^4)(x^{2-6})(y^{10-6})}{(2^4)}} =\sqrt[4]{\frac{3^4x^{-4}y^{4}}{2^4}} =\frac{3x^{-1}y^1}{3}=\frac{3y}{2x}

\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}} =\sqrt[3]{\frac{(4^3)(x^{8-2})(y^{7-10})}{(5^3)}} =\sqrt[3]{\frac{4^3x^6y^{-3}}{5^3}} =\frac{4x^2y^{-1}}{5}=\frac{4x^2}{5y}

\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}} =\sqrt[5]{\frac{(3^5)(x^{17-7})(y^{16-21})}{(2^5)}} =\sqrt[5]{\frac{3^5x^{10}y^{-5}}{2^5}} =\frac{3x^2y^{-1}}{2}=\frac{3x^2}{2y}

\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} =\sqrt[5]{\frac{(2^5)(x^{12-7})(y^{15-10})}{(3^5)}} =\sqrt[5]{\frac{2^5x^{5}y^{5}}{3^5}} =\frac{2x^1y^{1}}{3}=\frac{2xy}{3}

\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}} =\sqrt[4]{\frac{(2^4)(x^{10-2})(y^{9-17})}{(4^4)}} =\sqrt[4]{\frac{2^4x^{8}y^{-8}}{4^4}} =\frac{2x^{1}y^{-1}}{4}=\frac{x}{2y}

Thus,

\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}

3 0
3 years ago
PLEASE HELP ASAP!! VERY IMPORTANT!!!!! (NUMBER 15!!)
maria [59]
The answer is c cause the answer is 16,741.55ft^2 and c is the closest to it
5 0
3 years ago
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