Answer:
RADIUS
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PROBLEM
Mary’s bicycle wheel has a circumference of 226.08 cm². What is its radius?
SOLUTION
We can solve this problem using the circumference formula in which π stands for ( 3.14 ), C stands for circumference itself and r stands for radius.
\bold{Formula \: || \: C = 2πr}Formula∣∣C=2πr
\tt{226.08 = 2(3.14) r}226.08=2(3.14)r
'Now to find the radius,Substitute 226.08 for c which is circumference in the formula.
\begin{gathered} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{C = 2πr} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{226.08 = 2(3.14)\red{r}} \\ \\ \: \: \: \: \: \: \: \: \large \tt{ \frac{226.08}{6.28} = \cancel\frac{6.28 \red{r}}{6.28} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt\green{C = 36}}\end{gathered}
C=2πr
226.08=2(3.14)r
6.28
226.08
=
6.28
6.28r
C=36
To check:
\begin{gathered} \small\begin{array}{|c|}\hline \bold{circumference }\\ \\ \tt{C = 2πr} \\ \tt{C = 2(3.14) (36\:cm) } \\ \tt{C = 2(113.04\:cm) } \\ \underline{\tt \green{C = 226.08\:cm }} \\ \hline \end{array} \end{gathered}
circumference
C=2πr
C=2(3.14)(36cm)
C=2(113.04cm)
C=226.08cm
FINAL ANSWER
If Mary's Bicycle has a circumference of 226.08 cm then the radius is 36.
\boxed{ \tt \red{r = 36}}
r=36
==================================
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Answer:
The value of y would be 45.5
Step-by-step explanation:
To solve this problem, start with the base form of direct variation.
y = kx
Now we can use our original values to model the equation and find k.
35 = k(2.5)
14 = k
Now we can model the equation as:
y = 14x
Now to find y, when x = 3.25, simply put 3.25 into the equation.
y = 14(3.25)
y = 45.5
Answer:
1) 
2) ![\sqrt[3]{y^5}=y^{\frac{5}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7By%5E5%7D%3Dy%5E%7B%5Cfrac%7B5%7D%7B3%7D)
3) ![\sqrt[5]{a^{12}}=a^{\frac{12}{5} }](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Ba%5E%7B12%7D%7D%3Da%5E%7B%5Cfrac%7B12%7D%7B5%7D%20%7D)
4) ![\sqrt[4]{z^{9}}=z^\frac{9}{4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bz%5E%7B9%7D%7D%3Dz%5E%5Cfrac%7B9%7D%7B4%7D)
Step-by-step explanation:
1) 
We know that 
So,
2) ![\sqrt[3]{y^5}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7By%5E5%7D)
We know that ![\sqrt[3]{x}=x^{\frac{1}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%3Dx%5E%7B%5Cfrac%7B1%7D%7B3%7D)
So,
3) ![\sqrt[5]{a^{12}}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Ba%5E%7B12%7D%7D)
We know that ![\sqrt[5]{x}=x^{\frac{1}{5}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%7D%3Dx%5E%7B%5Cfrac%7B1%7D%7B5%7D)
So,
4) ![\sqrt[4]{z^{9}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bz%5E%7B9%7D%7D)
We know that ![\sqrt[4]{x}=x^{\frac{1}{4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%7D%3Dx%5E%7B%5Cfrac%7B1%7D%7B4%7D)
So,
Answer:
TO MAKE A NUTTY CEREAL ........
