1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Kay [80]
3 years ago
6

Find an equation, or a set of equations, to describe the set of points that are equidistant from the points p(−9, 0, 0) and q(3,

0, 0). (enter your answers as a comma-separated list of equations.)
Mathematics
1 answer:
wariber [46]3 years ago
7 0
X = -3.  
The distance from p(-9, 0, 0) is
 d = sqrt((x+9)^2 + y^2 + z^2) 
 The distance from q(3,0,0) is
 d = sqrt((x-3)^2 + y^2 + z^2) 
 Let's set them equal to each other.
 sqrt((x+9)^2 + y^2 + z^2) = sqrt((x-3)^2 + y^2 + z^2)
  Square both sides, then simplify
 (x+9)^2 + y^2 + z^2 = (x-3)^2 + y^2 + z^2
 x^2 + 18x + 81 + y^2 + z^2 = x^2 - 6x + 9 + y^2 + z^2
 18x + 81 = - 6x + 9
 24x + 81 = 9
 24x = -72
 x = -3 
 So the desired equation is x = -3 which defines a plane.
You might be interested in
hello! 20 points for this one answered correctly and explained! question: which pair of numbers with sum 8 has the largest produ
Andre45 [30]
4&4 b/c 4+4=8 & 4x4=16
8 0
3 years ago
Where does the last one go???
Nostrana [21]

Answer:

is goes to ¨¨not a function¨¨

Step-by-step explanation:

8 0
3 years ago
Please help!!<br> Write a matrix representing the system of equations
frozen [14]

Answer:

(4, -1, 3)

Step-by-step explanation:

We have the system of equations:

\left\{        \begin{array}{ll}            x+2y+z =5 \\    2x-y+2z=15\\3x+y-z=8        \end{array}    \right.

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}

Now, we can multiply R₂ by -1/5. This yields:

\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}

We can now reduce R₃ by multiply it by -1/4:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

\begin{bmatrix}1+(0) & 2+(-2)& 1+(0)&5+(-(-2))\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 1&7\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

And finally, we can eliminate the second 1 by adding -(R₃):

\begin{bmatrix}1 +(0)& 0+(0)& 1+(-1)&7+(-3)\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 0&4\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Therefore, our solution set is (4, -1, 3)

And we're done!

3 0
3 years ago
A pan is heated to 377°F, then removed from the heat and allowed to cool in a kitchen where the room temperature is a constant 7
saveliy_v [14]
A graph shows the appropriate choice to be
  A. 277.8 °F

_____
Of course, you get the same result by evaluating
  76 + 301*e^(-0.40)

3 0
3 years ago
Geri wants to put 6 stamps on some pages in get stamp book and 9 stamps on other pages.Explain how she could do this with the st
slamgirl [31]
6+9=15
78- 15=63
i hope this help you 

8 0
4 years ago
Read 2 more answers
Other questions:
  • The line with equation −a+3b=0 coincides with the terminal side of an angle θ in standard position and sinθ&lt;0 .
    12·2 answers
  • Why might the range not be the best estimate of variability?
    5·2 answers
  • Somebody please help!!
    11·1 answer
  • Plz help :)))))))))))
    10·1 answer
  • Find the first 4 terms of the geometric sequence a1=40 and r=0.5
    8·2 answers
  • A line passes through the two given points. Is it vertical, horizontal, or neither? (-8, 3), (-7, 3)
    9·1 answer
  • What term best describes the point where the three medians of a triangle intersects
    12·1 answer
  • Rectangles have two properties that are not included in the definition. The opposite sides of a rectangle are always of equal le
    12·1 answer
  • -11x - 72 = 84 + x<br> And I also need to do the check part
    10·1 answer
  • If 9 feet 2 inches of baseboard cost $15 how much would 75 feet cost?
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!