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3 years ago

If the butcher paper is 21 inches long,what is the area of one circle

Mathematics

1 answer:

sergejj [24]3 years ago

5 0

Answer:

= $12.25$ $\pi$

Step-by-step explanation:

Given:

Length of butcher paper,(diameter of three congruent circle) = 21 inches

Means, diameter of a circle = $\frac{21}{3}$ = 7 inches

Questioned asked:

Area of one circle = ?

Solution:

Radius of a circle ,r = half of of one circle's diameter

r = $\frac{7}{2}$

= $3.5$

Area of circle = $\pi$ $r^{2}$

$3.5\times3.5$ $\pi$

= $12.25$ $\pi$

Therefore, Area of one circle == $12.25$ $\pi$

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Please help!! What Find the Error and explain why it is wrong!

Nata [24]

Answer:

First image attached

The error was done in Step E, because student did not multiply $2\cdot x -8$ by the negative sign in numerator. Step E must be $\frac{2\cdot x -5}{(x+4)\cdot (x-4)}$ .

Second image attached

The error was done in Step C, because the student omitted the $2\cdot a \cdot b$ of the algebraic identity $(a+b)^{2} = a^{2}+2\cdot a\cdot b +b^{2}$ . Step C must be $5\cdot x = x^{2}+4\cdot x + 4$

Step-by-step explanation:

First image attached

The error was done in Step E, because student did not multiply $2\cdot x -8$ by the negative sign in numerator. The real numerator in Step E should be:

$3-(2\cdot x -8)= 3-2\cdot x+8 = 11-2\cdot x$

Hence, Step E must be $\frac{2\cdot x -5}{(x+4)\cdot (x-4)}$ .

Second image attached

The error was done in Step C, because the student omitted the $2\cdot a \cdot b$ of the algebraic identity $(a+b)^{2} = a^{2}+2\cdot a\cdot b +b^{2}$ . Step C must be $5\cdot x = x^{2}+4\cdot x + 4$

And further steps are described below:

Step D

$x^{2}-x+4 = 0$

Which according to the Quadratic Formula, represents a polynomial with complex roots. That is: ( $a = 1$ , $b = -1$ , $c = 4$ )

$D = b^{2}-4\cdot a\cdot c$

$D = (-1)-4\cdot (1)\cdot (4)$

$D = -17$ (Conjugated complex roots)

Step E

$(x-0.5-i\,1.936)\cdot (x-0.5+i\,1.936) = 0$

Step F

$x = 0.5+i\,1.936\,\lor\,x = 0.5-i\,1.936$

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jek_recluse [69]

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