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
iragen [17]
1 year ago
15

Given that 212basex + 122basex=1111 basex.Find x​

Mathematics
1 answer:
Lisa [10]1 year ago
8 0

Write out the decimal expansion of each number:

212_x \equiv 2x^2 + x + 2

122_x \equiv  x^2 + 2x + 2

1111_x \equiv x^3 + x^2 + x + 1

Note that x must be greater than 2. Solve for x.

212_x + 122_x = 1111_x \iff (2x^2 + x + 2) + (x^2 + 2x + 2) = x^3 + x^2 + x + 1 \\\\ ~~~~ \implies 3x^2 + 3x + 4 = x^3 + x^2 + x + 1 \\\\ ~~~~ \implies x^3 - 2x^2 - 2x - 3 = 0 \\\\ ~~~~ \implies (x^3 - 3x^2) + (x^2 - 3x) + x - 3 = 0 \\\\ ~~~~ \implies x^2(x-3) + x(x-3) + (x-3) = 0 \\\\ ~~~~ \implies (x-3) (x^2+x+1) = 0 \\\\ ~~~~ \implies \boxed{x=3}

You might be interested in
A cell phone company charges $40 per month for unlimited calling, and $0.20 per text message sent. If t represents the number of
kotegsom [21]

Answer: 40 + 0.2t

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
My mystery number has 5 tens and 29 ones?
Anika [276]
79 is your number.....
6 0
3 years ago
Read 2 more answers
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
The fifth graders were given sandwiches for lunch during your field trip Nathan ate 5 out of 6 leroy ate 7 out of 8 of his sandw
Scorpion4ik [409]

Answer:

Step-by-step explanation:

The fifth graders were given sandwiches for lunch during their field trip. Nathan ate 5/6 of his sandwich, Leroy ate 7/8 of his sandwich, and Sofia ate 5/8 of her sandwich. Who ate the greatest amount of their sandwich? ... Nathan ate 20/24, Leroy ate 21/24, and Sofia ate 15/24 meaning Leroy ate the most.

5 0
4 years ago
Read 2 more answers
What is the y-intercept of y=3x-2/3
Gre4nikov [31]

Answer: -2/3

Step-by-step explanation:

5 0
3 years ago
Other questions:
  • Evaluate 3 − 24 ÷ 8 + 4^2
    15·2 answers
  • Describe the transformation of ƒ(x) = log5 x which is given by g(x) = 3 log5 (x−7). Question 16 options: A) g(x) is shrunk verti
    6·1 answer
  • 4x+7y=2<br> help . Pls.
    11·1 answer
  • What is the area of this trapezoid?<br><br><br> Plz help
    12·2 answers
  • What is the ratio of yellow butterflies to total butterflies? Choose the correct option
    11·1 answer
  • What are the exact values of the x-intercepts of the function f(x)=x2-8x-6
    6·1 answer
  • {HELP ASAP }WILL GIVE 50+ POINTS AND BRAINLIEST
    14·1 answer
  • Determine if the expression is greater than 1.
    11·1 answer
  • What is an equation of the line that passes through the point (-2,-3)(−2,−3) and is parallel to the line 5x+2y=14
    8·1 answer
  • Plz help me im vary confused
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!