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riadik2000 [5.3K]
3 years ago
12

Bills paid online on the weekend are seen til the next business day - yes or no

Mathematics
1 answer:
Vinil7 [7]3 years ago
7 0

Answer:

yes, bills paid online over the weekend would not be seen until the next business day, usually Monday.

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"13 more than the product of 4 and a number, n
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Answer:

4n+13

Step-by-step explanation:

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{30 POINTS} For a promotion, a store offers to pay the sales taxes on any item
Luden [163]

--Answer--

1. $85

2. $15

--Explanation--

1. 15% off of 100 just means moving the decimal in 15 over by 2 to the left.

From there, multiply 100 x 0.15.

Then you get 15, so that's how much tax you are paying. Meaning Subtract 15 from 100 to get the price it originally was before tax. 100 - 15 = 85.

2. $15 dollars tax from moving decimal 2 to the left of 15%.

5 0
3 years ago
Question 3, I give thanks and 5 stars so you get more points :P
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PLEASE HELP... What do you do when none of the variables cancel out while solving systems by elimination?
Vitek1552 [10]

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3 0
3 years ago
Prove that: [1 + 1/tan²theta] [1 + 1/cot² thata] = 1/(sin²theta - sin⁴theta]
Stels [109]

Step-by-step explanation:

<h3><u>Given :-</u></h3>

[1+(1/Tan²θ)] + [ 1+(1/Cot²θ)]

<h3><u>Required To Prove :-</u></h3>

[1+(1/Tan²θ)]+[1+(1/Cot²θ)] = 1/(Sin²θ-Sin⁴θ)

<h3><u>Proof :-</u></h3>

On taking LHS

[1+(1/Tan²θ)] + [ 1+(1/Cot²θ)]

We know that

Tan θ = 1/ Cot θ

and

Cot θ = 1/Tan θ

=> (1+Cot²θ)(1+Tan²θ)

=> (Cosec² θ) (Sec²θ)

Since Cosec²θ - Cot²θ = 1 and

Sec²θ - Tan²θ = 1

=> (1/Sin² θ)(1/Cos² θ)

Since , Cosec θ = 1/Sinθ

and Sec θ = 1/Cosθ

=> 1/(Sin²θ Cos²θ)

We know that Sin²θ+Cos²θ = 1

=> 1/[(Sin²θ)(1-Sin²θ)]

=> 1/(Sin²θ-Sin²θ Sin²θ)

=> 1/(Sin²θ - Sin⁴θ)

=> RHS

=> LHS = RHS

<u>Hence, Proved.</u>

<h3><u>Answer:-</u></h3>

[1+(1/Tan²θ)]+[1+(1/Cot²θ)] = 1/(Sin²θ-Sin⁴θ)

<h3><u>Used formulae:-</u></h3>

→ Tan θ = 1/ Cot θ

→ Cot θ = 1/Tan θ

→ Cosec θ = 1/Sinθ

→ Sec θ = 1/Cosθ

<h3><u>Used Identities :-</u></h3>

→ Cosec²θ - Cot²θ = 1

→ Sec²θ - Tan²θ = 1

→ Sin²θ+Cos²θ = 1

Hope this helps!!

7 0
2 years ago
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