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3 years ago

Simplify: 3r+n²-r+5-2n+2

2 answers:

7 0

3r+n2−r+5−2n+2=3r+n2+−r+5+−2n+2=3r+n2+−r+5+−2n+2=(n2)+(−2n)+(3r+−r)+(5+2)=n2+−2n+2r+7
=n2−2n+2r+7

nikitadnepr [17]3 years ago

4 0

3n+n^2-r+5-2n+2
5+2=7
3n+2n=5n
n^2 -r +5n +7
((IDK what order the variables are suppose to go in put i think I got it in the right order.))

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Determine the perimeter of a rectangle that is 10 inches long and 9 inches wide. inches

Alex777 [14]

Answer:

Step-by-step explanation:

2(l + b)

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3 years ago

Read 2 more answers

The linear function f(x) = 0.9x + 79 represents the average test score in your math class, where x is the number of the test tak

Levart [38]

Given:

The linear function for average test score in your math class is

$f(x)=0.9x+79$

where x is the number of the test taken.

The linear function g(x) in the given table represents the average test score in your science class, where x is the number of the test taken.

To find:

Part A: The test average for your math class after completing test 2.

Part B: The test average for your science class after completing test 2.

Part C: Which class had a higher average after completing test 4?

Solution:

Part A:

We have,

$f(x)=0.9x+79$

Put x=2 in the above function, to find the test average for your math class after completing test 2.

$f(2)=0.9(2)+79$

$f(2)=1.8+79$

$f(2)=80.8$

Therefore, the test average for your math class after completing test 2 is 80.8.

Part B:

From the given table it is clear that g(x) =79 at x=2.

Therefore, the test average for your science class after completing test 2 is 79.

Part C:

Put x=4 in f(x), to find the test average for your math class after completing test 4.

$f(4)=0.9(4)+79$

$f(4)=3.6+79$

$f(4)=82.6$

From the table it is clear that the value of g(x) is increased by 1 as x is increased by 1. So, the value of function g(x) at x=4 must be 1 more than 80.

$g(4)=81$

It is clear that,

$82.6>81$

Therefore, math class had a higher average after completing test 4.

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yulyashka [42]

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3 years ago

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Mademuasel [1]

Answer:

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Value of investment after 10 year = 600(1+8%)^10 = $ 1,295

Value of investment after 25 year = 600(1+8%)^25 = $ 4,109

Time after which investment amount double (n)

1200 = 600 (1.08)^n

Log 2 = n log 1.08

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