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4 years ago

What is the focus of the parabola? y=−14x2−2x−2 Enter your answer in the boxes.

Mathematics

2 answers:

Sloan [31]4 years ago

7 0

Answer:

Step-by-step explanation:

Given is a parabola

$y=-14x^2-2x-2$

Let us rewrite this in vertex form

$y=-14(x+\frac{1}{14})^2-\frac{27}{14}$

Hence vertex = $(\frac{-1}{14} ,\frac{-27}{14} )$

Axis of symmetry is

$x=-\frac{1}{14}$

a= $\frac{1}{(14)4} =\frac{1}{56}$

Hence focus would lie on the line of symmetry at a distance of a from vertex

So focus = $(-\frac{1}{14} ,-\frac{27}{14} -\frac{1}{56} )\\=(-\frac{1}{14} , -\frac{109}{56} )$

julia-pushkina [17]4 years ago

6 0

Answer:

If this is for k12 I took the test and the right answer was (-4, 1)

Step-by-step explanation:

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Describe how to find the domain of a square root function just by looking at the function.

astra-53 [7]

Answer:

Step 1: Set the expression inside the square root greater than or equal to zero. Step 2: Solve the equation found in step 1. In this case, we divided by a negative number, so had to reverse the direction of the inequality symbol. Step 3: Write the answer using interval notation.

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3 years ago

Can someone please help me with my math homework!!

larisa [96]

1.)11x^2-3x
2.)8x+1-x^2
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3 years ago

Which of the following products is negative? Select all that apply. *

AlekseyPX

Answer:

(-4)(9)
(6)(-6)

Step-by-step explanation:

$(-4)(-9)\\\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a,\:-\left(-a\right)=a\\=4\times\:9\\= 36$

$(-4)(9)\\\\\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\\=-4\times\:9\\\\\mathrm{Multiply\:the\:numbers:}\:4\times\:9=36\\=-36$

$(6)(6)\\\\\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\\=6\times\:6\\\\\mathrm{Multiply\:the\:numbers:}\:6\times\:6\\=36$

$(-6)(-6)\\\\\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a,\:-\left(-a\right)=a\\=6\times 6\\\\\mathrm{Multiply\:the\:numbers:}\:6\times\:6\\=36$

$(6)(-6)\\\\\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\\=-6\times 6\\\\\mathrm{Multiply\:the\:numbers:}\:6\times \:6=36\\=-36$

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3 years ago

Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in t

Tresset [83]

Answer: The volume of largest rectangular box is 4.5 units.

Step-by-step explanation:

Since we have given that

Volume = $xyz$

with subject to $x+2y+3z=9$

So, let $z=\dfrac{9-x-2y}{3}$

So, Volume becomes,

$V=xyz\\\\V=xy(\dfrac{9-x-2y}{3})\\\\V=\dfrac{9xy-x^2y-2xy^2}{3}$

Partially derivative wrt x and y we get that

$9-2x-2y=0\implies 2x+2y=9\\\\and\\\\9-x-4y=0\implies x+4y=9$

By solving these two equations, we get that

$x=3,y=\dfrac{3}{2}$

So, $z=\dfrac{9-x-2y}{3}=\dfrac{9-3-3}{3}=\dfrac{3}{3}=1$

So, Volume of largest rectangular box would be

$xyz=3\times \dfrac{3}{2}\times 1=\dfrac{9}{2}=4.5$

Hence, the volume of largest rectangular box is 4.5 units.

4 0

3 years ago

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Anna [14]

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3 years ago