Can anyone plz help me with this math problem? E = {√2, π, √2 + 1, π + ½}?
in the above problem List the numbers in the set at are:
a). natural numbers
b). integers
c). Rational numbers
d).irrational numbers
e). real numbers
REFERENCE:Intermediate Algebra by Michael Sullivan chapter-16 number 25.
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Verified answer
a.) 1
b.)
c.) 1/2 and pi
d.) sqrt2
e.) 1, sqrt2, pi, 1/2
a million. For the 1st one, i'm getting the image that the circle has the comparable diameter because of the fact the size of the sq.'s factor. So in case you shrink the circle in 0.5 you get 2 semicircles. Stick one on the two factor and you have what might appear like a racetrack oval. yet wait, a sq. has 4 aspects, so upload 2 greater semicircles. this suggests which you prefer the area of the sq. plus the area of four semicircles or 2 entire circles. in case you are not getting this, then draw the image on a chew of paper and shrink the semicircles off of the sq. and assemble them into 2 circles. If the factor is 6 cm, then the area of the sq. is 6 x 6 = 36 cm². The circles have the diameter of 6 cm and a radius of three cm. the area of a circle is ? r², the place r is the radius. So the area of each and every circle is ? (3)² = 9? = 9 (3.14159) = 28.27 cm². Now upload those mutually and you get 36 cm² + 28.27 cm² + 28.27 cm² = ninety two.fifty 4 or ninety two.5. it particularly is determination d. 2. A circle inscribed in a sq. means that the diameter of the circle is the comparable because of the fact the size of the factor of the sq. (comparable as difficulty a million). back, in basic terms be certain one after the different the climate of the two the sq. and the circle, all you're able to do is subtract one from the different. i won't be in a position to work out the image to inform what area is shaded, yet i'm guessing that it is the area of the sq. it particularly is outdoors the circle. think of in case you drilled a hollow in the sq. it is the size of the circle. The shaded area is what you will possibly have left. So in basic terms subtract off the area of the circle and you gets the area of what's left over. Now, if the area of the sq. is eighty one cm², then the size of the factor of the sq. is the sq. root of eighty one = 9 cm. it particularly is the comparable because of the fact the diameter of the circle. The radius is a million/2 the diameter, or 4.5 cm. the area of the circle is (4.5)² ? = 20.25 ? = 20.25 (3.14159) = sixty 3.6. So the adaptation between the area of the sq. and the area of the circle is eighty one cm² - sixty 3.6 cm² = 17.4 cm² or determination d.
wow! those are some creative numbers you got there