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3 years ago

What is the common difference of an arithmetic sequence defined by the formula an=2n−4?

Mathematics

2 answers:

Ainat [17]3 years ago

5 0

Answer:

d = 2

Step-by-step explanation:

Generate the first few terms

a₁ = (2 × 1) - 4 = 2 - 4 = -2

a₂ = (2 × 2) - 4 = 4 - 4 = 0

a₃ = (2 × 3) - 4 = 6 - 4 = 2

a₄ = (2 × 4) - 4 = 8 - 4 = 4

d = 0 - (- 2) = 2 - 0 = 4 - 2 = 2

allsm [11]3 years ago

5 0

Answer:

Common difference = 2.

Step-by-step explanation:

The general formula for an arithmetic sequence is

an = a1 + d(n-1) where d = the common difference. This can be rearranged to:

an = dn - d + a1

So comparing:

an = 2n - 4 with the above we see that d = 2.

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Last week’s and this week’s low temperatures are shown in the table below.

zavuch27 [327]

Answer:

1st, 2nd, 3rd and 5th statement are correct choices.

Step-by-step explanation:

We have been given a table of low temperatures during last week and this week. We are asked to choose the correct options from our given choices about the center and variability of data.

Low Temperatures of this Week (f): 4, 10, 6, 9 ,6.

Low Temperatures of the last Week (f): 13, 9, 5, 8, 5.

Let us check our given choices one by one.

1. The mean of this week's temperatures is greater than 5 degrees.

$\text{Mean of the low temperatures this week}=\frac{4+10+6+9+6}{5}$

$\text{Mean of the low temperatures this week}=\frac{35}{5}$

$\text{Mean of the low temperatures this week}=7$

Since the mean of this week's temperatures is 7 and 7 is greater than 5, therefore, 1st statement is true.

2. The mean of last week's temperatures is greater than 5 degrees.

$\text{Mean of the low temperatures last week}=\frac{13+9+5+8+5}{5}$

$\text{Mean of the low temperatures last week}=\frac{40}{5}$

$\text{Mean of the low temperatures last week}=8$

Since the mean of last week’s temperatures is 8 and 8 is greater than 5, therefore, 2nd statement is true as well.

3. The range of this week's temperatures is greater than 5 degrees.

We know that range is the difference between greatest value and smallest value of data set.

$\text{Range}=\text{ Greatest value of data set-Lowest value of data set}$

$\text{Range of the low temperature this week}=10-4$

$\text{Range of the low temperature this week}=6$

We can see that range of low temperature this week is 6 and 6 is greater than 5, therefore, 3rd statement is true.

4. The mode of last week's temperatures is greater than 5 degrees.

Mode is the most frequently occurring number found in a set of numbers. We can see that 5 is the mode of last week's temperature and it is not greater than 5, therefore, 4th statement is not true.

5. The range of last week's temperatures is greater than 5 degrees.

$\text{Range of the low temperature last week}=13-5$

$\text{Range of the low temperature last week}=8$

We can see that range of last week's temperatures is 8 and 8 is greater than 5, therefore, 5th statement is a correct choice.

8 0

3 years ago

Read 2 more answers

Zoe bought a bike on sale at 15% off the original price.

mestny [16]

$\$420$ original prize
$15\%$ discount
$100\%-15\%=85\%$
$85\%=0.85$
$\$420*0.85=\$357$

6 0

3 years ago

Read 2 more answers

What is the value of 78(84+16)78(84+16) ?

trapecia [35]

Step-by-step explanation:

We can simplify this to be (78+78)(84+16)

We calculate from here

(156)(100)

Bracket means multiplication so we multiply

=15,600

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3 years ago

Please help me! I need help on Part A, B, and C

alexgriva [62]

Part A = 85$
Part B = He pays 25$ a month.

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3 years ago

A pizzeria makes brick-oven pizzas that are shaped like long rectangles with semi-circles at two ends. The pizzas come in three

Digiron [165]

Answer:

Area of pizza whose longer side is of length is 14 inches = 218.5 $inches^{2}$

Area of pizza whose longer side of rectangle is 18 inches=258.5 $inches^{2}$

Area of pizza whose longer side is of length is 29 inches = 368.5 $inches^{2}$

Step-by-step explanation:

Given A pizzeria makes brick-oven pizzas that are shaped like long rectangles with semi-circles at two ends. The pizzas come in three different sizes, which are measured by the longer side of each rectangle: 14 inches, 18 inches, and 29 inches. All of the pizzas are 10 inches wide, so the semi-circles at the ends have diameters of 10 inches. we have to find the area of each pizza.

First, let us find the area of pizza whose longer side of rectangle is 14 inches

Area of 2 semicircles whose diameter is 10 inches

$=2(\frac{1}{2}\pi r^{2})=\pi r^{2}=3.14\times 5 \times 5 = 78.5 inches^{2}$

Length = 14 inches

Breadth = 10 inches

Area of rectangle = $length \times breadth$

= $14\times 10=140 inches^{2}$

∴ Area of pizza whose longer side is of length is 14 inches = Area of rectangle + area of 2 semicircles

=140+78.5=218.5 $inches^{2}$

Now, let us find the area of pizza whose longer side of rectangle is 18 inches

Length = 18 inches

Breadth = 10 inches

Area of rectangle = $length \times breadth$

= $18\times 10=180 inches^{2}$

∴ Area of pizza whose longer side is of length is 18 inches = Area of rectangle + area of 2 semicircles

=180+78.5=258.5 $inches^{2}$

Now, let us find the area of pizza whose longer side of rectangle is 29 inches

Length =29 inches

Breadth = 10 inches

Area of rectangle = $length \times breadth$

= $29\times 10=290 inches^{2}$

∴ Area of pizza whose longer side is of length is 29 inches = Area of rectangle + area of 2 semicircles

=290+78.5=368.5 $inches^{2}$

4 0

3 years ago