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3 years ago

-3w-2<-17 help me please

1 answer:

4 0

-3w - 2 < -17

add 2 on both sides

-3w - 2 < -17
+2 +2
-------------------
-3w < -15
divide by -3
(the inequality sign changes to the opposite side when you are dividing by negative number)

w > 5

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Find the equation for the parabola that has its vertex at the origin and has directrix at x=1/48

Oksana_A [137]

Answer:

The equation for a parabola with vertex at the origin and a directrix at x = 1/48 is $x= \frac{1}{12}\cdot y^{2}$ .

Step-by-step explanation:

As directrix is a vertical line, the parabola must "horizontal" and increasing in the -x direction. Then, the standard equation for such geometric construction centered at (h, k) is:

$x - h = 4\cdot p \cdot (y-k)^{2}$

Where:

$h$ , $k$ - Horizontal and vertical components of the location of vertex with respect to origin, dimensionless.

$p$ - Least distance of directrix with respect to vertex, dimensionless.

Since vertex is located at the origin and horizontal coordinate of the directrix, least distance of directrix is positive. That is:

$p = x_{D} - x_{V}$

$p = \frac{1}{48}-0$

$p = \frac{1}{48}$

Now, the equation for a parabola with vertex at the origin and a directrix at x = 1/48 is $x= \frac{1}{12}\cdot y^{2}$ .

6 0

3 years ago

You are designing a rectangular poster to contain 7575 in2 of printing with a 33-in margin at the top and bottom and a 11-in m

erma4kov [3.2K]

Answer:

The dimensions of the rectangular poster is 15 in by 5 in.

Step-by-step explanation:

Given that, the area of the rectangular poster is 75 in².

Let the length of the rectangular poster be x and the width of the rectangular poster be y.

The area of the poster = xy in².

$\therefore xy=75$

$\Rightarrow y=\frac{75}{x}$ ....(1)

1 in margin at each sides and 3 in margin at top and bottom.

Then the length of printing space is= (x-2.3) in

=(x-6) in

The width of printing space is = (y-2.1) in

=(y-2) in

The area of the printing space is A =(x-6)(y-2) in²

∴ A =(x-6)(y-2)

Putting the value of y

$\Rightarrow A =(x-6)(\frac{75}{x}-2)$

$\Rightarrow A = 87-\frac{450}{x}-2x$

Differentiating with respect to x

$A '= \frac{450}{x^2}-2$

Again differentiating with respect to x

$A''=-\frac{900}{x^3}$

To find the minimum area of printing space, we set A' = 0

$\therefore \frac{450}{x^2}-2=0$

$\Rightarrow 450 =2x^2$

$\Rightarrow x^2=225$

$\Rightarrow x=\pm 15$

Now putting x=±15 in A''

$A''|_{x=15}=-\frac{900}{15^3}$

$A''|_{x=-15}=-\frac{900}{(-15)^3}=\frac{900}{(15)^3}>0$

Since at x=15 , A"<0 Therefore at x=15 , the area will be minimize.

From (1) we get

$y=\frac{75}{x}$

Putting the value of x

$y=\frac{75}{15}$

=5 in

The dimensions of the rectangular poster is 15 in by 5 in.

4 0

3 years ago

R(x) = -0.21x^3 + x^2 - 8.1x - 3 for x= -1 and x = 2

Minchanka [31]

Answer:

r(-1) = 6.31 and r(2) = -16.88

Step-by-step explanation:

* Lets read the problem and solve it

- Evaluate means find the value, so evaluate r(x) means find the value

of it at the given values of x

∵ r(x) = -0.21x³ + x² - 8.1x - 3

∵ x = -1 and x = 2

- Then find r(-1) by substitute x by -1 and find r(2) by substitute x by 2

# At x = -1

∴ r(-1) = -0.21(-1)³ + (-1)² - 8.1(-1) - 3

∴ r(-1) = -0.21(-1) + (1) - 8.1(-1) - 3

∴ r(-1) = 0.21 + 1 + 8.1 - 3

∴ r(-1) = 6.31

# At x = 2

∴ r(2) = -0.21(2)³ + (2)² - 8.1(2) - 3

∴ r(2) = -0.21(8) + (4) - 8.1(2) - 3

∴ r(2) = -1.68 + 4 - 16.2 - 3

∴ r(2) = -16.88

* r(-1) = 6.31 and r(2) = -16.88

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Answer:

Step-by-step explanation:

1+1=2

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