Answer:
1st scenario: 163.3°
2nd scenario: 125.2°
3rd scenario: 142.3°
Step-by-step explanation:
1st scenario: angles 1 and 2 form a linear pair and add up to 180
2nd scenario: angles 1 and 4 are same-side interior and add up to 180
3rd scenario: angles 1 and 8 are corresponding and equal; if angle 7 is 37.7 then angle 8 is 142.3 (they form a linear pair). Since angles 1 and 8 are equal then angle 1 is also 142.3
Answer:
Option C
Step-by-step explanation:
you cancel the whole parentheses.
So you know, x+6 can't equal to 0.
that's why there is a choice that is all real numbers but not -6.
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
(-5/6)+(-5/6) have the same denominator (number at bottom of fraction), so all you have to do is add the numerator(top number). So -5 +-5 =-10 . Now you have -10/6, which simplifies to -5/3.
Hope this helps :)
