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3 years ago

Enter an inequality that represents the graph in the box.

Mathematics

2 answers:

Slav-nsk [51]3 years ago

7 0

Since y=-1 when x=0 and we can write the equation (we'll turn it into an inequality later) as y=mx-1 from y=mx+b by plugging (0,1) in. Next, y equals 2 when x=1, so we plug those in to get 2=m*1-1. Adding 1 to both sides, we get 3=m, making our equation y=3x-1 (since y and x stay variables). Lastly, we turn it into an inequality, It seems to be inclusive to the line, so it's either
3x-1≤y or 3x-1≥y. Finding a random point in the inequality (4, 1), we plug it in to get 12-1=11, which is clearly larger than 1, so we get 3x-1≥y.

ziro4ka [17]3 years ago

6 0

The correct answer is

y <= 3x -1

You might be interested in

From the top of a tower, an object which is 97 m away from the base of the tower, is at a 37° angle of depression.

mario62 [17]

<h3>Answer: 73</h3>

==================================================

Work Shown:

Check out the diagram below. Note the pair of alternate interior angles that are congruent (each 37 degrees). Then focus on triangle ABC. With the reference angle being at A, this means we use the tangent function because BC = x is the opposite side and AB = 97 is the adjacent side.

tan(angle) = opposite/adjacent

tan(A) = BC/AB

tan(37) = x/97

97*tan(37) = x

x = 97*tan(37)

x = 73.094742859971

For the last step, you'll need a calculator that can handle trig functions. Make sure the calculator is in degree mode. The result here is approximate. This rounds to 73 when rounding to the nearest whole number.

7 0

3 years ago

Is (-3, -9) a solution of the equation y = 3x ?

makkiz [27]

Answer:

Yes.

Step-by-step explanation:

-3 * 3 = -9

3 0

3 years ago

What is the solution set to the inequality 5(x – 2)(x + 4) > 0?

ankoles [38]

Answer:

Step-by-step explanation:

The Answer is {x| x <-4 or x> 2}

8 0

3 years ago

Read 2 more answers

If 2a/4=b/4 and b≠0, what does 5b equal in terms of a?

dedylja [7]

Step-by-step explanation:

you must have made a typo here.

none of the answer options are correct for the given problem.

let me just show you what you told me, and how I can solve at least this :

2a/4 = b/4

this simply means (multiply both sides by 4) :

2a = b

so, 5b is then (multiplying both sides by 5) :

5×2a = 5b

10a = 5b

but again, "10a" is not among the offered answers.

so, I don't know if you made a mistake with the basic problem or with the answer options.

8 0

2 years ago

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago