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3 years ago

Which of the following is the correct solution to the linear inequality shown below y 1/3x-2

1 answer:

4 0

We are asked to determine the solution to the linear inequality y <1/3x -2. In this case, we can answer this by equating the equation to zero, transposition and cross multiplication.

y = 0 = 1/3x-2
2 = 1/3x
6 = x

If this is the case, x>6

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B.HELPPPP MEEEE PLSSSS AND THANK YOUU

Gemiola [76]

Answer:

A. (3, 8) and $2\sqrt{2}$

B. (2, 4) and $2\sqrt{10}$

Step-by-step explanation:

Midpoint formula: $(\frac{x1+x2}{2} ,\frac{y1+y2}{2})$

Distance formula: $\sqrt{(x_{2}-x_{1}) ^2+(y_{2}-y_{1}) ^2 }$

A. plug in the points in the formulas

(4, 7) and (2, 9)

Midpoint:

$(\frac{4 + 2}{2} , \frac{7+9}{2} )=6/2, 16/2 = (3,8)$

Length:

$\sqrt{(2-4)^2+(9-7)^2}=\sqrt{4+4} =\sqrt{8} =2\sqrt{2}$

(5, 5) and (-1, 3)

Midpoint:

$(\frac{5-1}{2} , \frac{5+3}{2} )=4/2,8/2=(2,4)$

Length:

$\sqrt{(-1-5)^2+(3-5)^2} =\sqrt{-6^2+-2^2} =\sqrt{36+4} =\sqrt{40} = 2\sqrt{10}$

Can you consider marking my answers as brainliest lol it would mean a lot. Hope this helped.

4 0

1 year ago

ABC IS a right angled triangle at B and D is a point on BC. If AD=18cm , BD=9cm and CD = 4 cm, find AC

leva [86]

$AB = \sqrt{ AD^{2} - BD^{2} } = \sqrt{243} = 9 \sqrt{3} 
$
$AC = \sqrt{ AB^{2} + BC^{2} } = \sqrt{412} = 2 \sqrt{103}$

4 0

2 years ago

The sum of mZAOB and mZBOC IS 128º. Find the measure of ZBOC.

amm1812

128 - 33 = 95
Answer C. 95

4 0

2 years ago

Read 2 more answers

Answer this question please shown in the image below.

Oksana_A [137]

Rectangle , think about what cutting the cylinder in half would make the inside image look like !!

8 0

3 years ago

A bin is constructed from sheet metal with a square base and 4 equal rectangular sides. if the bin is constructed from 48 square

kondaur [170]

This is a problem of maxima and minima using derivative.

In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:

V = length x width x height

That is:

$V = xxy = x^{2}y$

We also know that the <span>bin is constructed from 48 square feet of sheet metal, s</span>o:

Surface area of the square base = $x^{2}$

Surface area of the rectangular sides = $4xy$

Therefore, the total area of the cube is:

$A = 48 ft^{2} = x^{2} + 4xy$

Isolating the variable y in terms of x:

$y = \frac{48- x^{2} }{4x}$

Substituting this value in V:

$V = x^{2}( \frac{48- x^{2} }{x}) = 48x- x^{3}$

Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:

$\frac{dv}{dx} = 48-3x^{2} =0$

Solving for x:

$x = \sqrt{\frac{48}{3}} = \sqrt{16} = 4$

Solving for y:

$y = \frac{48- 4^{2} }{(4)(4)} = 2$

Then, <span>the dimensions of the largest volume of such a bin is:
</span>
Length = 4 ft
Width = 4 ft
Height = 2 ft

And its volume is:

$V = (4^{2} )(2) = 32 ft^{3}$

8 0

2 years ago