In the anatomical position, the nose and belly button are on the (12 body surface, the calves and shoulder blades are on
the (2) body surface, and the soles of the feet are the most
3 part of the body. The nipples are (4) to the shoulders
and (5) to the upper arms. The heart is (6) to the spine
(7) to the lungs. The knee is (8) to the toes but
to the thigh. In humans, the ventral surface can also be called
the (10) surface; however, in four-legged animals, the ventral
surface is the (11) surface.
and
If an incision cuts the brain into right and left parts, the section is
a(n)_(12) _section, but if the brain is cut so that superior and
inferior parts result, the section is a(n)_(13) section. You are
told to cut an animal along two planes so that the paired lungs
are observable in both sections. The two sections that meet this
requirement are the (14) and (15) sections.
The thoracic cavity is (10_to the abdominopelvic cavity
and (17) to the spinal cavity.

Answer:     the interval on which the curve is concave up is  (-∞, -1/2)======    we find the interval where this curve is concave up by finding the interval in which the 2nd derivative is positive.differentiate the curve equation with respect to x    let the function f denote be the anti-derivative of the inside expression, i.e. a function such that .then use the fundamental theorem of calculus:     differentiate again to get second derivative. we can use chain rule with 1/x as the outside function and 1 + x + x² as the inside function    find critical numbers of the second derivative. these numbers are where the 2nd deriv. is undefined or zero.where the 2nd derivative is zero: set (d²y/dx²) to equal 0 and solve for x    where the 2nd derivative is undefined: if the denominator is equal to zero, the derivative will be undefeined. attempting to solve the equation    1 + x + x² = 0results in a discriminant of      meaning that the equal 1 + x + x² = 0 has no real solutions and therefore will never equal zero.======our only critical number is x = -1/2.this forms two intervals to test the sign for: (-∞, -1/2) and (-1/2, ∞)test values into the 2nd derivative for (-∞, -1/2)if we try x = -1, we get      so the since the 2nd derivative of the curve is positive on the interval  (-∞, -1/2), the curve is concave up on (-∞, -1/2)test values into the 2nd derivative for (-1/2, ∞)if we try x = 0, we get      so the since the 2nd derivative of the curve is positive on the interval  (-∞, -1/2), the  curve is concave down on (-∞, -1/2).the interval on which the curve is concave up is  (-∞, -1/2)

Non of them obstinate means to stubbornly refuse