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3 years ago

What is the common difference for this arithmetic sequence?

Mathematics

2 answers:

OverLord2011 [107]3 years ago

8 0

Answer:

B. -6

Step-by-step explanation:

The common difference here is the difference between each term. Since it is an arithmetic sequence the difference between each term will be the same value.

So, from 45 (the 1st term) to 39 (the 2nd term), the difference is -6

From 39(the 2nd term) to 33, the difference is also -6.

Therefore, we know that the common difference fro this arithmetic sequence is -6.

Hope this helps and ask me any question if it does not :)

Effectus [21]3 years ago

8 0

Answer:

-6.

Step-by-step explanation:

39 - 45 = -6

33 - 39 = -6

27 - 33 = -6

21 - 27 = -6.

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Write the fraction 7/8 as an equivalent fraction with the given denominator 40

Mekhanik [1.2K]

Answer:

35/40

Step-by-step explanation:

7/8, with the given denominator 40

So how do we get 8, to turn into 40?

Simple multiplication.

8x5 = 40, and whatever we multiply times the denominator, we multiply for the numerator.

7x5 = 35

The answer should be 35/40

8 0

2 years ago

If f(X)=2x-1/2x+5’<br> Findf (-3)

ankoles [38]

Answer:

0.5.

Step-by-step explanation:

Substitute x = -3 into the guiven function:

= 2(-3) - 1/2(-3) + 5

= -6 + 1.5 + 5

= -4.5 + 5

= 0.5

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1 year ago

The total number of people at a football game was 5600. Field-side tickets were 40 dollars and end-zone tickets were 20 dollars.

rewona [7]

Answer:

1100 field-side tickets and 4500 end-zone tickets.

Step-by-step explanation:

Let x represent number of field side tickets and y represent number of end-zone tickets.

We have been given that the total number of people at a football game was 5600. We can represent this information in an equation as:

$x+y=5600...(1)$

$y=5600-x...(1)$

We are also told that Field-side tickets were 40 dollars and end-zone tickets were 20 dollars.

Cost of x field side tickets would be $40x$ and cost of y end-zone tickets would be $20y$ .

The total amount of money received for the tickets was $134000. We can represent this information in an equation as:

$40x+20y=134000...(2)$

Upon substituting equation (1) in equation (2), we will get:

$40x+20(5600-x)=134000$

$40x+112000-20x=134000$

$20x+112000=134000$

$20x+112000-112000=134000-112000$

$20x=22000$

$\frac{20x}{20}=\frac{22000}{20}$

$x=1100$

Therefore, 1100 field side tickets were sold.

Upon substituting $x=1100$ in equation (1), we will get:

$y=5600-1100$

$y=4500$

Therefore, 4500 end-zone tickets were sold.

3 0

3 years ago

Question regarding logarithms.

Eddi Din [679]

$5^{x-2}-7^{x-3}=7^{x-5}+11\cdot5^{x-4}\\\\5^{x-2}-7^{x-2-1}=7^{x-2-3}+11\cdot5^{x-2-2}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\5^{x-2}-\dfrac{7^{x-2}}{7^1}=\dfrac{7^{x-2}}{7^3}+11\cdot\dfrac{5^{x-2}}{5^2}\\\\5^{x-2}-\dfrac{1}{7}\cdot7^{x-2}=\dfrac{1}{343}\cdot7^{x-2}+\dfrac{11}{25}\cdot5^{x-2}\\\\-\dfrac{1}{7}\cdot7^{x-2}-\dfrac{1}{343}\cdot7^{x-2}=\dfrac{11}{25}\cdot5^{x-2}-5^{x-2}\\\\\left(-\dfrac{1}{7}-\dfrac{1}{343}\right)\cdot7^{x-2}=\left(\dfrac{11}{25}-1\right)\cdot5^{x-2}$

$\left(-\dfrac{49}{343}-\dfrac{1}{343}\right)\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\\\\-\dfrac{50}{343}\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\qquad\text{multiply both sides by}\ \left(-\dfrac{25}{14}\right)\\\\\dfrac{50\cdot25}{343\cdot14}\cdot7^{x-2}=5^{x-2}\qquad\text{divide both sides by}\ 7^{x-2}\\\\\dfrac{25\cdot25}{343\cdot7}=\dfrac{5^{x-2}}{7^{x-2}}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}$

$\dfrac{5^2\cdot5^2}{7^3\cdot7}=\left(\dfrac{5}{7}\right)^{x-2}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\\dfrac{5^4}{7^4}=\left(\dfrac{5}{7}\right)^{x-2}\\\\\left(\dfrac{5}{7}\right)^4=\left(\dfrac{5}{7}\right)^{x-2}\iff x-2=4\qquad\text{add 2 to both sides}\\\\\boxed{x=6}$

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inysia [295]

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Hey there!

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