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DiKsa [7]
2 years ago
12

Write the fraction of 12/15 in its simple form​

Mathematics
2 answers:
marusya05 [52]2 years ago
7 0
Answer:

4/5

Explanation:


Find the Greatest Common Factor (GCF) of 12 and 15,If it exists, and by dividing both the numerator and denominator by it, reduce our fraction. GCF = 3, and our simplified answer

12 ÷ 3 = 4
15 ÷ 3 = 5

So the answer is 4/5
Ugo [173]2 years ago
4 0

Answer:

4/5

Step-by-step explanation:

To get the answer, find the GCF (greatest common factor) of the numerator and the denominator. Once you have the GCF, divide the 12 and 15 by that number. Here, the GCF is 3.

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Y=5x-2 graphed is 1/3 a solution to the equation
ivolga24 [154]

Answer:

1/3 =5x-2

3(1/3)=3(5x-2)

1=15x-6

7=15x

x=7/15   (7/15 , 1/3)

Step-by-step explanation:

8 0
3 years ago
Question c please, it’s about probabilities. i attached photo
Reil [10]

Answer:

200

Step-by-step explanation:

Assuming all of the pie chart sections are equal.

There are 5 equal sections so divide the 1000 spins by 5 and you get 200

6 0
3 years ago
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What is the square footage of a 13 foot by 14 foot circle?
sineoko [7]

{169}ft^{2}


3 0
3 years ago
Whats 15-b=4b-5(5-3)
Gala2k [10]

15-b=4b-5(5-3)

Multiply the bracket by -5

(-5)(5)=-25

(-5)(-3)=15

15-b=4b-25+15

15-b=4b-10

Move 15 to the other side

Sign changes from +15 to -15

15-15-b=4b-10-15

-b=4b-25

Move 4b to the other side.

-b-4b=4b-4b-25

-b-4b=-25

-5b=-25

divide both sides by -5

-5b/-5=-25/-5

Answer: b=5

5 0
3 years ago
Consider the function g(x) = (x-e)^3e^-(x-e). Find all critical points and points of inflection (x, g(x)) of the function g.
Elden [556K]

Answer:

The answer is "cirtical\  points \ (x,g(x))\equiv  (e,0),(e+3,\frac{27}{e^3})"

Step-by-step explanation:

Given:

g(x) = (x-e)^3e^{-(x-e)}

Find critical points:

g(x) = (x-e)^3e^{(e-x)}

differentiate the value with respect of x:

\to g'(x)= (x-e)^3 \frac{d}{dx}e^{e-r} +e^{e-r}  \frac{d}{dx}(x-e)^3=(x-e)^2 e^{(e-x)} [-x+e+3]

critical points g'(x)=0

\to (x-e)^2 e^{(e-x)} [e+3-x]=0\\\\\to e^{(e-x)}\neq 0 \\\\\to (x-e)^2=0\\\\ \to [e+3-x]=0\\\\\to x=e\\\\\to x=e+3\\\\\to x= e,e+3

So,

The critical points of (x,g(x))\equiv  (e,0),(e+3,\frac{27}{e^3})

7 0
3 years ago
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