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3 years ago

What is the gradient of the blue line?? Mathswatch

Mathematics

2 answers:

kirill [66]3 years ago

6 0

<em> $answer \\ - 5 \\ please \: see \: the \: attached \: picture \: for \: full \: solution \\ hope \: it \: helps$ </em>

yaroslaw [1]3 years ago

3 0

Answer:

gradient = -5

Step-by-step explanation:

A line passes (x1, y1) and (x2, y2) has the gradient (slope) which is calculated by:

gradient = (y2-y1)/(x2-x1)

As shown in picture, this blue line passes (-1, 0) and (0, -5)

=> gradient = (-5 - 0)/(0 - -1) = -5/1 = -5

Hope this helps!

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Which two lines appear as if they will never intersect? Select all that apply.

Nastasia [14]

C, and D. They’re parallel so they’re never going to cross.

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3 years ago

Read 2 more answers

How do you find the volume of the solid generated by revolving the region bounded by the graphs

d1i1m1o1n [39]

Answer:

About the x axis

$V = 4\pi[ \frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}$

About the y axis

$V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}$

About the line y=8

$V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}$

About the line x=2

$V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi$

Step-by-step explanation:

For this case we have the following functions:

$y = 2x^2 , y=0, X=2$

About the x axis

Our zone of interest is on the figure attached, we see that the limit son x are from 0 to 2 and on y from 0 to 8.

We can find the area like this:

$A = \pi r^2 = \pi (2x^2)^2 = 4 \pi x^4$

And we can find the volume with this formula:

$V = \int_{a}^b A(x) dx$

$V= 4\pi \int_{0}^2 x^4 dx$

$V = 4\pi [\frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}$

About the y axis

For this case we need to find the function in terms of x like this:

$x^2 = \frac{y}{2}$

$x = \pm \sqrt{\frac{y}{2}}$ but on this case we are just interested on the + part $x=\sqrt{\frac{y}{2}}$ as we can see on the second figure attached.

We can find the area like this:

$A = \pi r^2 = \pi (2-\sqrt{\frac{y}{2}})^2 = \pi (4 -2y +\frac{y^2}{4})$

And we can find the volume with this formula:

$V = \int_{a}^b A(y) dy$

$V= \pi \int_{0}^8 2-2y +\frac{y^2}{4} dy$

$V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}$

About the line y=8

The figure 3 attached show the radius. We can find the area like this:

$A = \pi r^2 = \pi (8-2x^2)^2 = \pi (64 -32x^2 +4x^4)$

And we can find the volume with this formula:

$V = \int_{a}^b A(x) dx$

$V= \pi \int_{0}^2 64-32x^2 +4x^4 dx$

$V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}$

About the line x=2

The figure 4 attached show the radius. We can find the area like this:

$A = \pi r^2 = \pi (\sqrt{\frac{y}{2}})^2 = \pi\frac{y}{2}$

And we can find the volume with this formula:

$V = \int_{a}^b A(y) dy$

$V= \frac{\pi}{2} \int_{0}^8 y dy$

$V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi$

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3 years ago

Consider the equation

earnstyle [38]

a) true

b) false

c) true

Step-by-step explanation:

<h3>let's determine the first statement</h3><h3>to determine x-intercept </h3><h3>substitute y=0</h3>

so,

8x-2y=24

8x-2.0=24

8x=24

x=3

therefore

the first statement is <u>true</u>

let's determine the second statement

<h3>to determine y-intercept </h3><h3>substitute x=0</h3>

so,

8x-2y=24

8.0-2y=24

-2y=24

y=-12

therefore

the second statement is <u>False</u>

to determine the third statements

<h3>we need to turn the given equation into this form</h3><h2>y=mx+b</h2><h3>let's solve:</h3>

8x-2y=24

-2y=-8x+24

y=4x-12

therefore,

the third statement is also <u>true</u>

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3 years ago

A room with a length of 270 cm and width 150 cm is to be covered with square tiles .What is the largest size of tiles to be used

Damm [24]

Find the greatest common factor of both numbers:

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Factors for 270:

1, 2, 3, 5, 6 , 9, 10, 15, 18, 27, 30, 45, 54, 90, 135, 270

The largest number they both have in common is 30, so the largest tile could be 30 cm^2.

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jenyasd209 [6]

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