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Artemon [7]
3 years ago
11

What value of x is in the solution set of 3(x – 4) ≥ 5x + 2? –10 –5 5 10

Mathematics
2 answers:
emmasim [6.3K]3 years ago
7 0

<u>Answer:</u>

The value of x is in the solution set of 3(x – 4) ≥ 5x + 2 is -10

<u>Solution:</u>

Need to determine which value of x from given option is solution set of 3(x – 4) ≥ 5x + 2

Lets first solve 3(x – 4) ≥ 5x + 2

3(x – 4) ≥ 5x + 2

=> 3x – 12 ≥ 5x + 2

=> 3x – 5x ≥ 12 + 2

=> -2x ≥ 14

=> -x ≥ 7

=> x ≤ -7

All the values of x which are less than or equal to -7 is solution set of 3(x – 4) ≥ 5x + 2. From given option there is only one value that is -10 which is less than -7  

Hence from given option -10 is solution set of 3(x – 4) ≥ 5x + 2.

Misha Larkins [42]3 years ago
6 0

Answer:

A (-10)

Step-by-step explanation:

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2 years ago
The diameter of a circle is 23 inches and the circle has two chords of length 14 inches what is the distance from each chord to
Svetradugi [14.3K]

Answer:

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Diameter = 23 inches (given)

Radius = 11.5 inches

2 Chords of length = 14 inches ( You didn't specify if the 14 inches is for both chords or for a single cord. I'll assume it's for two cords 14 and 14inches apart.

To clearly solve this, we'll make some mild assumptions.

Let the perpendicular distance of the chords from the center of the circle to represented as " x and y"

Therefore:

x^2 + 7^2 = 11.5 ^ 2

x^2 + 49 = 132.25

x^2 = 132.25 - 49

x^2 = 83.25

x = √ 83.25

x = 9.12 inches

Since the cords have thesame length (Assumed from the way the question was structured, the distance would still be thesame)

y^2 + 7^2 = 11.5 ^ 2

y^2 + 49 = 132.25

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6 0
4 years ago
Creating an Exponential Model
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Answer:

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r = 0.15

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$544.61  (to the nearest cent)

P(1+r)^t

$524.70  (to the nearest cent)

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P = principal amount = $300

r = annual interest rate in decimal form = 15% = 15/100 = 0.15

n = number of times interest is compounded per unit t = 12

<u>How much she'll owe in 4 years</u>

P = 300

r = 0.15

n = 12

t = 4

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P(1+r)^t

<u>How much she'll owe in 4 years at yearly compounding interest</u>

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