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3 years ago

Shown here is the graph of which inequality? Please can’t figure it out I need SOME HELP

Mathematics

2 answers:

Aleks [24]3 years ago

8 0

<h2>D. 3x+2y≥6 </h2><h2>Hope this helps, sorry if not tho</h2><h2></h2><h2></h2>

Annette [7]3 years ago

5 0

Hey there!

Three points to keep in mind!

The given options are in standard form: ax+by=c
The line is a solid line, hence, the possible outcomes would be ≤ or ≥
The shaded part is above the line, hence, the symbol would be ≥

First, let's find the slope:

Slope : (y2-y1)/(x2-x1)

Slope : (0-3)/(2-0)

Slope : -3/2

Looking at line, we know that the y-intercept is 3 b=3

Slope intercept form:

Let's now convert this into a standard form:

<em>Multiplying all the sides with 2: </em>

<em>Remember that the coefficient of 'x' shall not be negative or a fraction!</em>

<em>3x+2y≥6 </em><em>is the final answer!</em>

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PLS HELP IF YOU CAN!THIS IS DUE IN 30 MINUTES :(

ExtremeBDS [4]

Answer:

Length: 20 or 40 feet

Width: 40 or 20 feet

Step-by-step explanation:

Area = 800 = length*width = x*(60-x)= 60x - x^2

- x^2 + 60x - 800 = 0

x = 20

Or x = 40

So length can be either 20 feet or 40 feet and width can be 40 (60-20) or 20 (60-40)

6 0

3 years ago

Help me Please....................

MakcuM [25]

9514 1404 393

Answer:

1.3363

Step-by-step explanation:

The basic idea here is to find an expression for the direction vector between a point on L1 and a point on L2. Then, solve for the points on L1 and L2 that make that vector perpendicular to both lines L1 and L2. (The dot product of direction vectors is zero.) The distance between the points found is the shortest distance between the lines.

Let P be a point on L1. Then the parametric equation for P is ...

P = (6t, 0, -t) . . . . . . origin + t × direction vector

Let Q be a point on L2. The direction vector for L2 is given by the difference between the given points. It is (4-1, 1-(-1), 6-1) = (3, 2, 5). Then the parametric equation for Q is ...

Q = (3s+1, 2s-1, 5s+1) . . . . (1, -1, 1) + s × direction vector

The direction vector for PQ is ...

Q -P = (3s+1-6t, 2s-1, 5s+1+t)

The dot product of this and the two lines' direction vectors will be zero:

(3s+1-6t, 2s-1, 5s+1+t)·(6, 0, -1) = 0 = 13s -37t +5 . . . perpendicular to L1

(3s+1-6t, 2s-1, 5s+1+t)·(3, 2, 5) = 0 = 38s -13t +6 . . . perpendicular to L2

The solution to these equations is ...

s = -157/1237

t = 112/1237

Then (Q-P) becomes (94, -1551, 564)/1237, and its length is ...

|PQ| = √(94² +1551² +564²)/1237 ≈ 1.3363

The distance between the two lines is about 1.3363 units.