Your answer should be 157.5 Inches
Answer:
190 Degrees Celsius
Step-by-step explanation:
so first you multiply the
Rate per minute X Minutes
15 X 8 = 120
120 this is the temperature increase so you add this increase to the initial temperature which was 70 degrees Celsius
temperature Increase + Initial temperature
120 + 70 = 190 degrees Celsius
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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#CarryOnLearning
Answer:
1200-928 = 272 for the rest of the month Clarissa has to spend.
Step-by-step explanation:
So we don't want her to go over her budget so we wouldn't choose a greater than option for her so eliminate the first option and the last option. Now we just have the middle two. We want to keep it under 1200 but theres nothing wrong with spending that exact amount each month so I would pick x less than or equal to 272, since we can still equal the total of less than the total and not go over. Option 2 or B.
