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2 years ago

The graphs of the linear functions f and g are shown above. If h(x)=f(x)+g(x), then h′(x)=

Mathematics

1 answer:

SpyIntel [72]2 years ago

7 0

Based on the graph,

f(x) = 1/2x+3

g(x) = -x+4

so, h(x) = -1/2 x +7

Since h'(x) is the derivative or slope of h(x) and h is a line,

h'(x) = -1/2, Answer C.

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5.(03.06)

maw [93]

Answer:

$y = 3 \div 4x + 10$

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2 years ago

Leon drew a life cycle of Insect M as shown below.

tankabanditka [31]

Step-by-step explanation:

Larval mosquitoes breath through trachea in their siphons. This is a structure analogous to the snorkel on a diesel electric submarine. It allows for gas exchange with the atmosphere while the remainder of the insect is submerged. The opening of the siphon is hydrophobic so it won’t get wet and blocked by water. This works well to keep the siphon and the trachea open under normal conditions. Oil when poured on water forms a thin film. When there are mosquitos in that water, when their siphons contact the oil layer, the oil wets and blocks their siphons and suffocates the mosquitoes. This works against most, but not all mossies, as evolution is an amazing process. Some mosquitoes (Mansonia, Coquilletidia) have a siphon designed to penetrate the air vessels in aquatic plants and they don’t need to come to the surface to breath like other mossies. So oil won’t work on these genera.

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2 years ago

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

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1 year ago

Let f(x) = 8x3 + 16x2 − 15 and g(x) = 2x + 1. Find f of x over g of x.

andrey2020 [161]

Answer:

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1 year ago

Pleeeaaasseeee help (questions in photo) will mark brainliest

KIM [24]

Answer:

a. 0.8727rad

b. 5.585rad

that should be correct, if not, i apologize T_T

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1 year ago