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Papessa [141]
2 years ago
10

Find the area of the circle when the circumference is 26 in

Mathematics
1 answer:
andrew11 [14]2 years ago
3 0

Answer:

"None of the answers are correct."

Step-by-step explanation:

The formula for Area given the circumference is A = \frac{C^{2} }{4\pi }

\frac{26^{2} }{4\pi } = 53.79437

So none of the above are correct

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Combine like terms to create an equivalent expression.
Brut [27]

Question..

Combine like terms to create an equivalent expression.

½ −⅙q +⅚q - ⅓

Answer:

½ −⅙q +⅚q - ⅓ is equivalent to ⅔q + ⅙

Step-by-step explanation:

Given

½ −⅙q +⅚q - ⅓

Required

Equivalence

½ −⅙q +⅚q - ⅓

We start by collecting like terms.

⅚q - ⅙q + ½ - ⅓

Factorize

(⅚ - ⅙)q + ½ - ⅓

((5 - 1)/6)q + ½ - ⅓

(4/6)q + ½ - ⅓

Reduce 4/6 to lowest term

⅔q + ½ - ⅓

Evaluate fraction

⅔q + (3 - 2)/6

⅔q + ⅙

Hence, ½ −⅙q +⅚q - ⅓ is equivalent to ⅔q + ⅙

7 0
3 years ago
Read 2 more answers
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nika2105 [10]

D. f(x)=4 (5/2)^x

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3 years ago
Help i really need help plss
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Answer:

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Answer:

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2 years ago
Dan and Jake have new stamp collecting books. Dan puts 5 stamps in his book every day, and Jake puts 6 stamps in his book every
Ghella [55]
Jake puts 6 stamps in his book every day.

30/6= 5 

Jake put stamps in the book for 5 days.

So, Dan puts 5 stamps in the book every day.

5*5= 25

Dan has 25 stamps when Jake has 30.

I hope this helps!
~kaikers


4 0
3 years ago
How to solve linear and nonlinear equations graphically?
Ede4ka [16]
Plot the equation. If you wish to solve a polynomial, let y= polynomial and plot the graph. Best set up a table of values first.
Where the graph crosses the x axis there is a solution for x. There are also solutions for other horizontal lines (y values) by looking at intersections of the graph with these lines. This technique works for linear and non linear equations. You can also use graphs to solve 2-variable systems of equations by examining where the graphs intersect one another. The disadvantage is that you may not be able to have sufficient detail for high degrees of accuracy because of the scale of the graph and drawing inaccuracies.
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