Let EB and CF be two heights. Then quadrilateral EBCF is rectangle, so BC=EF=5 m. It is given that bases are BC=5 m and AD=11 m. Triangles ABE and DCF are congruent, thus,
Consider right triangle ABE, by the Pythagorean theorem
Answer: 
Answer:
i dont even know this language
Step-by-step explanation:
I think it’s b
Not for sure but it makes the most sense
Hey there!!
- The area of a circle = πr²
The area is given as 196π
Hence,
196π = πr²
... r²= 196
... r = √196
... r = 14
<em>Hence, the radius is 14 cm. </em>
<em>Hope it helps!</em>
24q^2 / 8q^-3
lets break this down...
24/8 = 3
q^2 / q^-3....when dividing exponents with the same base, keep the base and subtract the exponents. q^2 / q^-3 = q^(2 -(-3) = q^(2 + 3) = q^5
put them together and u get : 3q^5
