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2 years ago

How long does it take an SR-71 Blackbird to make a surveillance run of 10,026 miles if it travels at an average of 2,228 mph

1 answer:

4 0

This is simple division:
$t = \frac{10026}{2228}$

So, plug it into your calculator and you get:
$t = 4.5$

Your answer's that it takes 4.5 hrs for the surveillance run.

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ASAPPP!!!! HELP I WILL GIVE 100POINTS AND BRAINLIST Question 9 (Essay Worth 10 points)

Luda [366]

Answer:

Part A)

The x-intercepts are (-1/4, 0) and (4, 0).

Part B)

The vertex is a maximum because the leading coefficient is negative.

The vertex is (1.875, 72.25).

Part C)

We can plot the zeros and the vertex, and connect them with a curve.

Step-by-step explanation:

The function given is:

$f(x)=-16x^2+60x+16$

Part A)

To find the x-intercepts of the function, set the function equal to 0 and solve for x:

$0=-16x^2+60x+16$

We can divide both sides by negative four:

$0=4x^2-15x-4$

Factor:

$0=(4x+1)(x-4)$

Zero Product Property:

$x-4=0\text{ or } 4x+1=0$

Solve for each case:

$\displaystyle x=-\frac{1}{4}\text{ or }x=4$

Hence, our zeros are:

$\displaystyle \left(-\frac{1}{4}, 0\right)\text{ and } \left(4, 0\right)$

Part B)

Note that the leading coefficient of our function is negative.

So, our function will be concave down.

Hence, our vertex will be the maximum.

The vertex is given by:

$\displaystyle \left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$

In this case, a = -16, b = 60, and c = 16.

Find the x-coordinate of the vertex:

$\displaystyle x=-\frac{(60)}{2(-16)}=\frac{15}{8}=1.875$

Substitute this back into the function to find the y-coordinate:

$f(1.875)=-16(1.875)^2+60(1.875)+16=72.25$

Hence, our vertex is:

$(1.875, 72.25)$

Part C)

Since we already determined the zeros and the vertex, we can plot the two zeros and the vertex and draw a curve between the three points.

The graph is shown below. Again, to do this by hand, simply plot the three points and connect them with a parabola. If necessary, we can also find the y-intercept.

7 0

3 years ago

ANSWER CHOICES LISTED!!!

Kipish [7]

Answer:f(x) = x becomes g(x) = (1/5)x through vertical compression by a factor of 5.

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$