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Alenkasestr [34]
3 years ago
14

A school 415 third graders and 338 second graders. How many more third graders are there than second graders?

Mathematics
1 answer:
jeka943 years ago
3 0
415 - 338 = 77

There are 77 more 3rd graders than 2nd graders
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Gina made tables of values to solve a system of equations.
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How do you do this question?
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Answer:

(-∞, -2), (-2, -0.618), and (1.618, 3)

Step-by-step explanation:

The red arrows indicate the regions in which the function is decreasing.

The corresponding intervals are (-∞, -0.618) and (1.618, 3),

Let's examine the intervals given.

(-∞, -2): Yes, decreasing.

(-2, -0.618): Yes, decreasing.

(-2, 0.689): No. Decreasing in (-2, -0.618) but increasing in (-0.618, 0.689).

(-1.17, 0.689): No. Decreasing in (-1.17, -0.618) but increasing

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(0.698, 71.457): No. Increasing from (0.698, 1.618), decreasing in (1.618, 3) and increasing in (3, 71.457).

(1.618, 3): Yes, decreasing.

(2.481, ∞): No. Decreasing in (2.481, 3) but increasing in (3, ∞).

The three choices that indicate a decreasing interval are (-∞, -2),

(-2, -0.618), and (1.618, 3).

3 0
3 years ago
It costs $3.55 to make a sandwich at the local deli shop. To make a profit, the deli sells it at a price that is 160% of the cos
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fabiana decides to save the money she is earning from her after school job for college. she makes an initial contribution of 300
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Answer:

A. Total Money Contributed after n months = \$ 3000 + \$ (500\times n)

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Step-by-step explanation:

Given:

Initial contribution = \$\ 3000

each month contribution =\$\ 500

After 1 month contributed = \$\ 3000 + \$\ 500 \times 1= \$ \ 3500

Solving for Part A

let n be the number of months

∴ Total Contribution after n months = Initial contribution + (each month contribution \times Number of months = \$\ 3000 + \$\ 500 \times n

Solving for Part A

Now n= 24 months

∴ Total Contribution after 24 months = \$\ 3000 + \$\ 500 \times 24 = \$\ 3000 + \$\ 12000= \$\ 15000

3 0
3 years ago
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