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3 years ago

How do I create a graph that represents the phrase, 5 is greater than all the number in the solution.

1 answer:

7 0

The equation would be x<5

First, graph the line x=5
Then, to determine which side of the line to shade, use a test point. I like (0,0). I plug in 0 for x because that is the x value. 0 does make the equation correct so (0,0) is included inside the shaded area. Shade the side that has (0,0) in it and you're done.

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What is the constant of variation, k,

ValentinkaMS [17]

The K-constant variation is the factor of the increase-decrease relationship between the x-variable and y-variable which can be stated as y=k*x or x=y/k. For example, if y variable equal 5 and the x variable equal 2 then the k-constant variation equal 2 (calculation y=k*x --> 5=2*2.5). This constant shows the relationship between the variables.

6 0

3 years ago

Read 2 more answers

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

Solve: -19 = X - 19 halppppp

Juliette [100K]

Answer:

x = 0

Step-by-step explanation:

Step 1: Write equation

-19 = x - 19

Step 2: Solve for x

Add 19 to both sides: 0 = x
Rewrite: x = 0

Step 3: Check

Plug in x to verify it's a solution.

-19 = 0 - 19

-19 = -19

3 0

2 years ago

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Convert: 42 centigrams 6 milligrams = 1 milligrams

Naddik [55]

Answer:

centigram is 'bigger.'

Well, if you look on the metric scale it is apparent that 1 mg = 0.1 (or 1/10) cg suggesting that a milligram is the smaller value.

6 0

3 years ago

The volume of a rectangular tank is 20 000 cm3 . Given that the length is 40 cm and width is 25 cm, find its height

olga nikolaevna [1]

$answer \\ 20 \: cm \\ solution \\ volume = 20000 \: {cm}^{3} \\ length = 40 \: cm \\ breadth = 25 \: cm \\ height = \\ now \\ v = l \times b \times h \\ \: \: \: \: \: \: or \: 20000 = 40 \times 25 \times h \\ or \: 20000 = 1000 \times h \\ or \: h = \frac{20000}{1000} \\ h = 20 \: cm \\ hope \: it \: helps$

5 0

2 years ago

Read 2 more answers