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Temka [501]
3 years ago
8

Based on the slopes, what can you say about the relationship between DE and BC?​

Mathematics
1 answer:
solong [7]3 years ago
6 0

Answer:

DE and B⁢C are parallel line segments.

Step-by-step explanation:

Edmentum-

You might be interested in
A lock has two digit numeric code, with the numbers 1 through 7 available for each digit of the code. Note that numbers can be r
lubasha [3.4K]
If there is 2 digits and 7 numbers the combos would be:
1,1 2,1 3,1 4,1 5,1 6,1 7,1
1,2 2,2 3,2 4,2 5,2 6,2 7,2
1,3 2,3 3,3 4,3 5,3 6,3 7,3
1,4 2,4 3,4 4,4 5,4 6,4 7,4
1,5 2,5 3,5 4,5 5,5 6,5 7,5
1,6 2,6 3,6 4,6 5,6 6,6 7,6
1,7 2,7 3,7 4,7 5,7 6,7 7,7

Therefore there are 49 possible combinations :)
7 0
2 years ago
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

6 0
3 years ago
What is the equation of the line, in point-slope form, that passes through the points (4, 8) and (2,-2)?
Annette [7]

Answer:

y = 5x -12

Step-by-step explanation:

point-slope form:

y - y1 = m(x-x1)

m= slope

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

we have (4, 8) and (2,-2)

x1 = 4         y1= 8

x2= 2         y2= -2

m=( -2-8) / (2- 4)

m= -10/ -2

m= 5

so we have:

y - 8 = 5(x-4)

y - 8= 5x -20

y= 5x -20 +8

y = 5x -12

3 0
3 years ago
Please help me simplify these expressions
Fantom [35]

Answer:

1.  -18x¹¹

2.  3n⁷

Step-by-step explanation:

For these problems, there are two things you need to worry about: negative signs and exponents.

1. Let's look at the signs first.  There is only one value with a negative sign, meaning that the negative sign will stay.

When multiplying with exponents, you have to add up the exponents.  Don't forget the numerical coefficients.

-3x² · 3x · 2x³ · x⁵ = -18x¹¹

2.  There are two negative signs in this probem, meaning that they will cancel out.  Multiply the rest like we did in the first problem.

3n² · -n² · -n³ = 3n⁷

7 0
2 years ago
Determine the smallest integer value of x in 5x + 3 ≥ 10
Ivanshal [37]

5x + 3 ≥ 10

3 is adding on the left, then it will subtract on the right

5x ≥ 10 -3

5x ≥ 7

5 is multiplying on the left, then it will divide on the right

x ≥ 7/5

x ≥ 1.4

Then, the smallest integer value of x is 2

6 0
1 year ago
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