the question in English is
<span>In the previous figure, segment BC is parallel to segment AD. The trapezoidal area ABCD, in square centimeters, is
</span>
we know that
area of trapezoid=(1/2)*[(BC+AD)*AB]
BC=8 cm
AD=12 cm
AB=2 cm
so
area of trapezoid=(1/2)*[(8+12)*2]-------> 20 cm²
the answer is
20 cm²
the answer in Spanish
Sabemos que
area del trapecio=(1/2)*[(BC+AD)*AB]
BC=8 cm
AD=12 cm
AB=2 cm
area del trapecio=(1/2)*[(8+12)*2]-------> 20 cm²
La respuesta es
20 cm²
The answer is no because after each 3 there is one more zero added
to be repeating the number of zeros would have to stay the same
Answer:
csc(q)
Step-by-step explanation:
You could plug in values for q for the problem and the choices and see which choice gives the same outputs. Of course, that would mean you need to know that cot() is cos()/sin() or 1/tan() and csc()=1/sin()
So anyways you can also use identities to rewrite the given expression
sin(q)+cos(q)cot(q) [given ]
sin(q)+cos(q)cos(q)/sin(q) by quotient identity
sin(q)+cos^2(q)/sin(q) [simplify]
sin(q)sin(q)/sin(q)+cos^2(q)/sin(q) multiply first term by 1
sin^2(q)/sin(q)+cos^2(q)/sin(q) [simplify ]
(sin^2(q)+cos^2(q))/sin(q) [combined fractions ]
1/sin(q) by Pythagorean identity
csc(q) by reciprocal identity
Answer:
4.27ft
.91in
2.99L
7.24km
Step-by-step explanation:
Answer:
The answer is 1:2
explaination in comment section