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igor_vitrenko [27]
2 years ago
7

Which fractiona are equivalent to 2/12+4/12

Mathematics
1 answer:
Sphinxa [80]2 years ago
7 0

Answer:

6/12 or 3/6 or 1/2

Step-by-step explanation:

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Approximate pi as 3.4*
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Answer:

C

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4 0
2 years ago
Consider the limaçon with equation r = 3 4cos(θ). how does the quotient of a and b relate to the existence of an inner loop?
Delicious77 [7]

If the equation is r = 3 +4cos(θ) then because b/a>1 the curve is a limacon with an inner loop.

Given limacon with equation r=3+4cos(θ) and we have to answer how the quotient of a and b relate to the existence of an inner loop.

Equation is like a relationship between two or more variables expressed in equal to form and it is solved to find the value of variables.

formula of polar graph is similar to r= a+ b cos (θ).

Case 1. If a<b or b/a>1

then the curve is a limacon with inner loop.

Case 2. If a>b or b/a<1

Then the limacon does not have an inner loop.

Here given that r=3+4cos (θ)

It is observed that , a<b or b/a>1

Therefore the curve is limacon with an inner loop.

Hence because b/a>1 the curve is a limacon with an inner loop.

Learn more about limacon at brainly.com/question/14322218

#SPJ4

5 0
2 years ago
Need answer please ty
klemol [59]

Answer:

More detail needed

Step-by-step explanation:

The question is not clear, therefor there is no way to solve it.

8 0
2 years ago
Tali is trying to find the thickness of a page of his telephone
Dmitry_Shevchenko [17]
The answer really would be.. how life begins and end in the long run.. think about that! But 223 inches
4 0
2 years ago
Erin deposited $400 in a saving account earning 3% yearly compound interest. write an equation to represent how much money (y) E
bekas [8.4K]

~~~~~~ \textit{Compound Interest Earned Amount \underline{in 15 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$400\\ r=rate\to 3\%\to \frac{3}{100}\dotfill &0.03\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yeary, thus once} \end{array}\dotfill &1\\ t=years\dotfill &15 \end{cases}

A=400\left(1+\frac{0.03}{1}\right)^{1\cdot 15}\implies A=400(1.03)^{15}\implies A\approx 623.19 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y=400(1.03)^x~\hfill

4 0
2 years ago
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