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2 years ago

A quadrilateral has vertices E(–4, 2), F(4, 7), G(8, 1), and H(0, –4). Which statements are true? Check all that apply.

Mathematics

1 answer:

lys-0071 [83]2 years ago

8 0

Answer:

D) Quadrilateral EFGH is a parallelogram because both pairs of opposite sides are parallel.

B) The slopes of EF and GH are both 5/8

Step-by-step explanation:

The following inference can be made about each answer choices:

For option A

A) The slope of EH is 3/2 and not -8/5

For EH, slope = (2--4)/(0--4) = 6/4 =3/2

For option B

B) The slopes of EF and GH are both 5/8

slope of a graph can be determined by change in y axis/ change in x axis

For EF, slope = (7-2)/(4--4) = 5/8

for GH, slope = (1--4)/(8-0) = 5/8

For option C

C) FG is perpendicular to GH, its can't be determined since it is concluded that the quadrilateral is a parallelogram, therefore this option is wrong.

For option D

D) Quadrilateral EFGH is a parallelogram because both pairs of opposite sides are parallel.

Slope of a graph can be determined by rise divided by run change in y axis/ change in x axis

For EF, slope = (7-2)/(4--4) = 5/8

for GH, slope = (1--4)/(8-0) = 5/8

For EH, slope = (2--4)/(0--4) = 6/4 =3/2

for FG, slope = (7-1)/(8-4) = 6/4 =3/2

EF and GH give the same slope, likewise EH and FG and they are opposite to each other, this make them parallel. Therefore, EFGH is a parallelogram.

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brainly.com/question/18006680

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1 Point

I am Lyosha [343]

Option C

The ratio for the volumes of two similar cylinders is 8 : 27

<h3><u>Solution:</u></h3>

Let there are two cylinder of heights "h" and "H"

Also radius to be "r" and "R"

$\text { Volume of a cylinder }=\pi r^{2} h$

Where π = 3.14 , r is the radius and h is the height

Now the ratio of their heights and radii is 2:3 .i.e

$\frac{\mathrm{r}}{R}=\frac{\mathrm{h}}{H}=\frac{2}{3}$

<em><u>Ratio for the volumes of two cylinders</u></em>

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{\pi r^{2} h}{\pi R^{2} H}$

Cancelling the common terms, we get

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{\mathrm{r}}{R}\right)^{2} \times\left(\frac{\mathrm{h}}{\mathrm{H}}\right)$

Substituting we get,

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{2}{3}\right)^{2} \times\left(\frac{2}{3}\right)$

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{2 \times 2 \times 2}{3 \times 3 \times 3}$

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{8}{27}$

Hence, the ratio of volume of two cylinders is 8 : 27

7 0

3 years ago

You are given g(x)=4x^2 + 2x and

Strike441 [17]

Answer:

324

Step-by-step explanation:

Given:

$g(x)=4x^2+2x\\ \\f(x)=\int\limits^x_0 {g(t)} \, dt$

Find:

$f(6)$

First, find f(x):

$f(x)\\ \\=\int\limits^x_0 {g(t)} \, dt\\ \\=\int\limits^x_0 {(4t^2+2t)} \, dt\\ \\=\left(4\cdot \dfrac{t^3}{3}+2\cdot \dfrac{t^2}{2}\right)\big|\limits^x_0\\ \\=\left(\dfrac{4t^3}{3}+t^2\right)\big|\limits^x_0\\ \\= \left(\dfrac{4x^3}{3}+x^2\right)-\left(\dfrac{4\cdot 0^3}{3}+0^2\right)\\ \\=\dfrac{4x^3}{3}+x^2$