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2 years ago

Calculate the unit prices for each of the items which is the better buy

Mathematics

1 answer:

Harman [31]2 years ago

8 0

Calculate the unit prices for each of the items which is the better buy

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(3+3x6+(3 with the power of 2)-3

Effectus [21]

You do pemdas and the answer should be 27

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3 years ago

WILL GIVE BRAINLIEST

Alona [7]

Given:

One linear function represented by the table.

Another linear function represented by the graph.

To find:

The greater unit rate and greater y-intercept.

Solution:

Formula for slope (unit rate):

$m=\dfrac{y_2-y_1}{x_2-x_1}$

From the given table it is clear that the linear function passes through (0,5) and (5,15). The function intersect the y-axis at (0,15), so the y-intercept is 15.

$m_1=\dfrac{15-5}{5-0}$

$m_1=\dfrac{10}{5}$

$m_1=2$

So, the unit rate of first function is 2.

From the given graph it is clear that the linear function passes through (0,6) and (-4,0). The function intersect the y-axis at (0,6), so the y-intercept is 6.

$m_2=\dfrac{0-6}{-4-0}$

$m_2=\dfrac{-6}{-4}$

$m_2=\dfrac{3}{2}$

So, the unit rate of first function is $\dfrac{3}{2}$ .

Now,

$2>\dfrac{3}{2}$

$m_1>m_2$

And,

$15>6$

Therefore, the greater unit rate of the two functions is 2. The greater y-intercept of the two functions is 15.

4 0

2 years ago

Read 2 more answers

Which of the following is an arithmetic sequence?

mr_godi [17]

Answer:

222

Step-by-step explanation:

1114-5646

7 0

3 years ago

Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

Bezzdna [24]

Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$