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3 years ago

Are line segments infinitely long?

1 answer:

4 0

No,

A line is considered a "line segment" when a line connects to 2 points.

If a line was infinite length, it would never end up ending at one point.

Hope this was clear for you

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Is a function that is not continuous

stepan [7]

It appears that your answer contains either a link or inappropriate words. Please correct and submit again!

Answer:

See below. The answer is incomplete, I couldn't post it.

Step-by-step explanation:

Considering a function $f$ , it is said to be discontinuous when it has a hole or breaks, it means places where $f(x)$ cannot be evaluated. For example, when the denominator equals to 0, it is not defined.

5 0

3 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

Francesca wants to convert 1 meter to feet. Use what you know about customary measures to explain how she could do this.

Volgvan

Turn the feet into meters 1 meter = 100cm and if there is 12cms in a foot than divide 100cm by 12cm and you get 8.3ft

5 0

2 years ago

Please help me with this question!!!\<br> please please

strojnjashka [21]

The mean book rating will decrease

Add all the numbers together:
5 + 5 + 6 + 7 + 7 + 7 + 7 + 8 + 8 + 8 + 9 + 9 + 10 + 10 + 10 + 10 + 10= 136

divide the total with the total amount of numbers

136/17 = 8

Now add all the numbers again WITH the one

5 + 5 + 6 + 7 + 7 + 7 + 7 + 8 + 8 + 8 + 9 + 9 + 10 + 10 + 10 + 10 + 10 + 1= 137

Divide the total amount with the amount of numbers there are

137/18 = 7.61

7.61 < 8

The mean book rating will decrease by 0.39

hope this helps

5 0

3 years ago

What is a budget? Describe the four-step process of figuring out your monthly budget.

Len [333]

I believe the answer is C. It is the only one that makes sense

3 0

3 years ago