Use the variable d to represent the unknown quantity. the product of 4 and the depth of the pool

Pleaseee Help. Don't answer if you aren't going to take it serious.

Nadya [2.5K]

Stem-and-leaf plots are used for showing the frequency of which certain classes of values occur. use a stem-and-leaf plot and let the numbers themselves to show pretty much the same information. In short, they are meant to show frequencies of which numbers or values occur. Not for colors or objects.

Yes, a frequency table is a perfect way to solve the problem! These are meant to help you find the number of something that occurs, and is a simple and clean way to show your word.

6 0

3 years ago

A parabola has a line of symmetry x = -5. The minimum value of the quadratic function that it represents is -7. Find a possible

Ilya [14]

hello :
the parabola's equation is : f(x) = a(x-h)²+k
the verex is (h,k)
line of symmetry x = h
The minimum or maximum value is : k
a possible equation of this parabola is : f(x) = a(x+5)²-7

6 0

3 years ago

Factor completley x2-4x-12

Artist 52 [7]

Step-by-step explanation:

The answer is 4x

6 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

You are taking a multiple-choice test that has six questions. Each of the questions has three answer choices, with one correct a

Nostrana [21]

Answer:

729 ways

Step-by-step explanation:

Given

Number of questions = 6

Choices = 3

Required

Number of ways of answering the questions

For each of the 6 questions;

There are 3 possible ways of selecting an answer;

The possible ways for the 6 questions is:

$Possible\ Ways = 3 * 3 * 3 * 3 * 3 * 3$

$Possible\ Ways = 729$

Hence, there are 729 ways of answering the question

4 0

3 years ago