All categories

4 years ago

Which of the following proportionally statements is correct?

Mathematics

1 answer:

yulyashka [42]4 years ago

6 0

Answer:

$\frac{AC}{BC}=\frac{DF}{EF}$

Step-by-step explanation:

The figure given has two similar triangles: ΔABC and ΔDEF. Although the triangles are similar, their orientation is different and ΔDEF is flipped. Since the triangles are similar, their side lengths are proportional to each other. Given the orientation of the triangles, we can still see that the diagonal (hypotenuse) for the larger triangle is AC and the smaller is DF. The only answer that matches these up proportionally is the third one. Looking at the second side, BC, we can see that this matches up to the longer leg of EF on the smaller triangle. Final answer being:

$\frac{AC}{BC}=\frac{DF}{EF}$

You might be interested in

HEEEEEEEEEEEEEEEEEEEELP

tatyana61 [14]

Answer:

5:30 PM

Step-by-step explanation:

35 mp ½ hour = 70 mph

245 miles ÷ 70 mph = 3½ hours

2PM + 3½ hours = 5:30 PM

3 0

3 years ago

A reflection will _____________ change the orientation of a figure. * A Always B Sometimes C Never

sleet_krkn [62]

Answer:

a reflection will never change the orientation of a figure

3 0

4 years ago

Please & Thank you!!

lisabon 2012 [21]

Answer:

WOW THIS IS MIDDLE SCHOOL MATH

Step-by-step explanation:

6 0

3 years ago

Adding and Subtracting with Rational Numbers

Pavel [41]

Answer:

-4.6 ft

Step-by-step explanation:

We need to learn to extract only the necessary information to answer question. The surf shop and the restaurant is extra information.

Monday 1 pm 2.2 ft

Monday 7 pm - 2.4 ft

Net change

-2.4 ft - 2.2 ft = -4.6 ft

The ocean went down 4.6 ft from 1 pm to 7 pm on Monday

4 0

3 years ago

Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured across a line (x + 2)/3 = (2y + 3)/4

dimaraw [331]

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

$x_1\ =\ -2$

$y_1\ =\ 3$

$z_1\ =\ -4$

Distance is measure across the line

$\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}$

So, we can write

$\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k$

$=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k$

$=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k$

$=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}$

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

$3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3$

$=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18$

$=>18k-24+12k+9+10k-32\ =\ 18$

$=>\ k\ =\dfrac{13}{8}$

So,

$x\ =\ 3k-4$

$=\ 3\times \dfrac{13}{8}-4$

$=\ \dfrac{7}{4}$

$y\ =\ \dfrac{4k+3}{2}$

$=\ \dfrac{4\times \dfrac{13}{8}+3}{2}$

$=\ \dfrac{19}{4}$

$z\ =\ \dfrac{5k-16}{3}$

$=\ \dfrac{5\times \dfrac{13}{8}-16}{3}$

$=\ \dfrac{-21}{8}$

Now, the distance of point from the plane is given by,

$d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}$

$=\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}$

$=\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}$

$=\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}$

$=\ \sqrt{\dfrac{1177}{64}}$

$=\ 4.28$

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

= 8.56 unit

So, the distance of a point from it's image is 8.56 units.

4 0

3 years ago