Answer:
A) is correct
Step-by-step explanation:
Some basic formulas involving triangles
\ a^2 = b^2 + c^2 - 2bc \textrm{ cos } \alphaa 2 =b 2+2 + c 2
−2bc cos α
\ b^2 = a^2 + c^2 - 2ac \textrm{ cos } \betab 2=
m_b^2 = \frac{1}{4}( 2a^2 + 2c^2 - b^2 )m b2 = 41(2a 2 + 2c 2-b 2)
b
Bisector formulas
\ \frac{a}{b} = \frac{m}{n} ba =nm
\ l^2 = ab - mnl 2=ab-mm
A = \frac{1}{2}a\cdot b = \frac{1}{2}c\cdot hA=
\ A = \sqrt{p(p - a)(p - b)(p - c)}A=
p(p−a)(p−b)(p−c)
\iits whatever A = prA=pr with r we denote the radius of the triangle inscribed circle
\ A = \frac{abc}{4R}A=
4R
abc
- R is the radius of the prescribed circle
\ A = \sqrt{p(p - a)(p - b)(p - c)}A=
p(p−a)(p−b)(p−c)
Answer:
The table C correctly shows the ratio 8:1 for each grade
Step-by-step explanation:
Let
x ----> the number of students
y ----> the number of adults
we know that

<u><em>Verify each table</em></u>
Table A
grade 6

Multiply in cross
----> is not true
Table B
grade 6

Multiply in cross
----> is not true
Table C
<u><em>grade 6</em></u>
\frac{96}{12}=\frac{8}{1}
Multiply in cross

----> is true
<u><em>grade 7</em></u>

Multiply in cross

----> is true
<u><em>grade 8</em></u>
\frac{136}{17}=\frac{8}{1}
Multiply in cross

----> is true
therefore
The table C correctly shows the ratio 8:1 for each grade
Table D
<u><em>grade 6</em></u>

Multiply in cross
----> is not true