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aleksandr82 [10.1K]

3 years ago

14

NEED HELP ASAP!!! An airplane flies across the country. Its distance from Portland, as a function of time, is modeled by y = 500

x. Which statement best describes the function?

1 answer:

babunello [35]3 years ago

5 0

Answer:

BRI ISH?!

Explanation:

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Simplify: (sin θ − cos θ)^2 − (sin θ + cos θ)^2 A) −4sin(θ)cos(θ) B) 2 C) sin^2 θ D) cos^2 θ

N76 [4]

Answer:

-4sinθcosθ

Step-by-step explanation:

Note:

1. (a + b)^2 = a^2 + 2ab + b^2

2. (a - b)^2 = a^2 - 2ab + b^2

3. sin^2θ + cos^2θ = 1

(sinθ -cosθ)^2 - (sinθ + cosθ)^2

= sin^2θ - 2sinθcosθ + cos^2θ - (sin^2θ + 2sinθcosθ + cos^2θ)

= sin^2θ + cos^2θ - 2sinθcosθ - (sin^2θ + cos^2θ + 2sinθcosθ)

= 1 - 2sinθcosθ - (1 + 2sinθcosθ)

= 1- 2sinθcosθ -1 - 2sinθcosθ

= - 2sinθcosθ - 2sinθcosθ

= -4sinθcosθ

3 0

3 years ago

a bookcase is to be constructed as shown in the figure below. the height of the bookcase is 2 feet longer than the length of a s

sp2606 [1]

The answer is D. 3 feet and 5 feet

7 0

3 years ago

Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is

$\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6$

4 0

3 years ago

A rectangular garden has an area of 2/3 square miles and a length of 8/9 mile. What is the width of the garden? *HINT: Area form

Rufina [12.5K]

Answer:

the answer is 3/4

Step-by-step explanation:

please make me Brainliest

7 0

2 years ago

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There is a tree in Athens,Georgia, that OWNS itself and an 8 foot radius of land.

oee [108]

Answer:

True

Step-by-step explanation:

8 0

3 years ago

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