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sergij07 [2.7K]
3 years ago
8

Which of the following equations could be used to solve for the tenth term of the following sequence? 15, 13, 11, 9, ...

Mathematics
2 answers:
Darya [45]3 years ago
6 0

Answer:

Equations that could be used to solve for the tenth term of the sequence:

15, 13, 11, 9, ...    is  

b) A(10) = 15 + 9(-2)

Step-by-step explanation:

We are given a sequence:

 15,13,11,9,...

The above sequence is an arithmetic progression with

first term=A(1)=15 and common difference=d=-2

Now, the nth term of an arithmetic progression is determined by the formula

A(n)= A(1)+(n-1)d

⇒  A(10)=  15+9(-2)

Hence, the correct option is:

  b) A(10) = 15 + 9(-2)

omeli [17]3 years ago
4 0

Answer

b) A(10) = 15 + 9(-2)


<u>Explanation</u>

15, 13, 11, 9, ...  This is an arithmetic sequence.

The common difference = 13 - 15 = -2

                                     or = 9 - 11 = -2

The first term of the sequence = 15.

The nth term of an arithmetic sequence is given by:

Tn = a + (n - 1)d

∴ T₁₀ = 15 + (10 - 1)(-2)

        = 15 + (9 × -2)  ⇒ This is the expression you need.

        = 15 + -18

       = -3

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AT first, shop A had 156 kg of rice and shop B had 72 kg of rice. After each shop sold the same quantity of rice, the amount of
aleksandr82 [10.1K]

Answer:

44 kg, and since both sold the same amount so did A.

Let me know if this method doesn't make sense though and I should be able to explain it.

Step-by-step explanation:

Both shops sold the same amount, let's call it x.

At the end, let's say shop B had y left, this means A would have 4y left, since the question says "After each shop sold the same quantity of rice, the amount of rice left in Shop A was 4 times that of Shop B"

So this means A started with 156, sold x and was left with 4y, or in math 156 - x = 4y

B you could write 72 - x = y

Now you have a system of equations.

156 - x = 4y

72 - x = y

There are a number o ways to look at this.  If you graph them you want the (x, y) coordinates where they intersect, or in other words the x and y values that work for both of these equaions.  

The methods all wing up making you solve for one variable and plugging it into the other equation to solve for the other.  The easiest method is probably canceling out.

156 - x = 4y

72 - x = y

Now, for these two the goal is to make one of the variables go away by adding or subtracting one equation from the other.  You could also multiply one by something.  In fact here are all possibilities.  

If you subtract one from the other here is what would happen.

156 - x = 4y

-(72 - x = y)

_________

156 - x = 4y

-72 + x = -y because you distribute the minus sign

_________

84 = 3y because 156-72 = 84, -x + x = 0 so they canceled out and 4y - y = 3y

28 = y because you divide both sides by 3.

It's all just algebra after the addition/ subtraction.  You could also do it the other way subtracting 156 - x = 4y from 72 - x = y.  Either way would cancel the xs.  But now that you know y = 28 you can plug that into either of the equations again to find x.

156 - x = 4y

156 - x = 4(28)

156 - x = 112

156 = 112 + x

44 = x

72 - x = y

72 - x = 28

72 = 28 + x

44 = x

Now, if you wanted to cancel the ys initially you would have to multiply.

156 - x = 4y

72 - x = y

Normally subtracting just got rid of the xs.  You would need to multiply the bottom one by 4 to get rid of ys.

156 - x = 4y

288 - 4x = 4y

Now if you subtracted one from the other the 4ys would cancel.  You could also divide the top one by 4 and get

39 - x/4 = y

72 - x = y

here you have a fraction though and that's gonna make things a little more difficult.

Let me know if this method doesn't make sense though and I should be able to explain it.

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