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Marina CMI [18]

2 years ago

8

6. If F(x) = -6(2x2 + 5x) - 9, what is the value of F(-4)?

1 answer:

Soloha48 [4]2 years ago

5 0

Answer:

-81

Step-by-step explanation:

-6(2x² + 5x) - 9

substitute -4 for 'x'

-6[2(-4)² + 5(-4)] - 9

-6[2(16) + (-20) - 9]

-6(32 - 20) - 9

-6(12) - 9

-72 - 9 = -81

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Joaquin is constructing the perpendicular bisector of ab. What should be his first step?

natali 33 [55]

Answer:

D. Open the compass so that the distance from the two points of the compass is wider than half the length of $\over{AB}$ .

Step-by-step explanation:

To construct a perpendicular for $\over{AB}$ , we must first take a compass & take the distance of its arms wider than half the length of $\over{AB}$ .

This is done in order to get two intersecting arcs in the top & bottom of $\over{AB}$ so that a perpendicular bisector can be drawn through it.

After two intersecting lines are drawn below & above $\over{AB}$ , draw a line joining these 2 points through their points of intersection. The point where it intersects $\over{AB}$ is the middle-most point of $\over{AB}$ & now a perpendicular bisector of $\over{AB}$ is constructed.

$\rule{150pt}{2pt}$

3 0

1 year ago

Convert 13pi/6 to a degree measure <br><br> A=390<br> B=2450.44<br> C=30<br> D=780

Burka [1]

Answer:

390 degrees

Step-by-step explanation:

The conversion factor is 180/pi

13 pi /6 * 180/pi

13/6 *180

390

5 0

3 years ago

What is 0.7 in simplest form

dolphi86 [110]

7/10 you can not simplify tht fraction any more. 7 does not go into
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7 0

3 years ago

Look at the figure. A 10 foot tall flagpole cast a shadow that is 24 feet long what is the approximate length of the shadow cast

masya89 [10]

Approx 60 feet see pic for work

7 0

3 years ago

Calculate s f(x, y, z) ds for the given surface and function. g(r, θ) = (r cos θ, r sin θ, θ), 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π; f(x, y, z)

Triss [41]

$g(r,\theta)=(r\cos\theta,r\sin\theta,\theta)\implies\begin{cases}g_r=(\cos\theta,\sin\theta,0)\\g_\theta=(-r\sin\theta,r\cos\theta,1)\end{cases}$

The surface element is

$\mathrm dS=\|g_r\times g_\theta\|\,\mathrm dr\,\mathrm d\theta=\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

and the integral is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\int_0^{2\pi}\int_0^4((r\cos\theta)^2+(r\sin\theta)^2)\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle2\pi\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac\pi4(132\sqrt{17}-\sinh^{-1}4)$

###

To compute the last integral, you can integrate by parts:

$u=r\implies\mathrm du=\mathrm dr$

$\mathrm dv=r\sqrt{1+r^2}\,\mathrm dr\implies v=\dfrac13(1+r^2)^{3/2}$

$\displaystyle\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac r3(1+r^2)^{3/2}\bigg|_0^4-\frac13\int_0^4(1+r^2)^{3/2}\,\mathrm dr$

For this integral, consider a substitution of

$r=\sinh s\implies\mathrm dr=\cosh s\,\mathrm ds$

$\displaystyle\int_0^4(1+r^2)^{3/2}\,\mathrm dr=\int_0^{\sinh^{-1}4}(1+\sinh^2s)^{3/2}\cosh s\,\mathrm ds$

$\displaystyle=\int_0^{\sinh^{-1}4}\cosh^4s\,\mathrm ds$

$=\displaystyle\frac18\int_0^{\sinh^{-1}4}(3+4\cosh2s+\cosh4s)\,\mathrm ds$

and the result above follows.

4 0

3 years ago