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2 years ago

Please help will name brainiest 6n+11=-13 what in n

Mathematics

1 answer:

tatuchka [14]2 years ago

4 0

Answer:

-4

Step-by-step explanation:

We know the value has to be less than 0 since the answer is negative.

We also know that the value has to be greater than -13.

Now we can try the values -1 through -12

Just try each number and remember to add 11 to each value.

After doing this, you should have noticed that the number -4 had given the correct number. This means the answer is -4

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Ross has to fill up a 300 gallon tank. There are already 75

stepan [7]

Answer: He needs 9 hours to finish the task.

Step-by-step explanation:

Hi, first we have to subtract the gallons of water already in the tank (75) to the total capacity of the tank (300).

300-75 = 225

He needs to fill 225 gallons.

Since the pump Ross uses can fill 25 gallons of water per hour, the proportion is :

25 g / 1h

For 225 gallons:

225 g / x hours

Solving for x

25/1 = 225/x

x= 225/ (25/1)

x = 9 hours

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3 years ago

In November 2010 the average price of 5 gallons of regular unleaded gasoline in the United states was $14.46. what was the price

Lapatulllka [165]

Propecia:

5 gallons ====== $ 14.46.

16 gallons ======= x dollars

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Answer: $ 46.27.

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3 years ago

There is 4 students in the class 3 ran away how many are left?

pochemuha

Answer:

1 student

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Sanjit has a collection of quarters and

Zinaida [17]

12 quarters and 7 dimes

8 0

3 years ago

Help I am very bad at math I need answers by tomorrow if you can help please do

d1i1m1o1n [39]

Answer:

Step-by-step explanation:

(a - b)(a +b) = a² - b²

1 - Sin² A = Cos² A

$LHS = \frac{1}{1- Sin A} + \frac{1}{1 + Sin A}\\\\= \frac{1*(1 + Sin A)}{(1- Sin A)(1 + Sin A)} + \frac{1*(1- Sin A)}{(1 + Sin A)(1- Sin A)}\\\\= \frac{1 + Sin A+ 1 - Sin A}{1^{2}- Sin^{2} A}\\\\= \frac{2}{1 - Sin^{2} A}\\\\= \frac{2}{Cos^{2} A}\\\\= 2 Sec^{2} A$

2) Sec² A - Tan² A = 1

$LHS = \frac{1}{Sec A - Tan A}\\\\=\frac{1*(Sec A + Tan A)}{(Sec A - Tan A)(Sec A + Tan A)}\\\\=\frac{Sec A + Tan A}{Sec^{2} A - Tan^{2} A}\\\\=\frac{Sec A + Tan A }{1}\\\\= Sec A + Tan A = RHS\\\\\\$

3) LHS = Cosec² A + Cot² A

= Cosec² A + Cosec² A - 1

= 2Cosec² A - 1 = RHS

$4) LHS = \frac{Sec A}{Cos A}- \frac{Tan A}{Cot A}\\\\ = Sec A*\frac{1}{Cos A}-Tan A*\frac{1}{Cot A}\\\\ = Sec A * Sec A - Tan A * Tan A\\\\= Sec^{2} A - Tan^{2} A \\\\= 1$

3 0

3 years ago